Need to make a Java program which calculates the number of days betw.pdf
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Need to make a Java program which calculates the number of days between two dates. I am NOT able to use anything too advanced yet such as import.java.util.Calender, import.java.util.Date; or import.java.util.GregorianCalender; Here are a couple of examples of what is should look like when complete. I am having troubles getting my calculations to come out correctly. I’ve also written the code I have so far. EXAMPLE “Enter first date now: Enter Month: 1 Enter Day: 1 Enter Year: 1900 Enter the second date: Enter Month: 1 Enter Day: 1 Enter Year: 2400 The first date is: 1/1/1900 The other date is: 1/1/2400 The total number of days between the dates is 182621” AND “Enter first date now: Enter Month: 2 Enter Day: 23 Enter Year: 2016 Enter the second date: Enter Month: 11 Enter Day: 30 Enter Year: 1955 The first date is: 2/23/2016 The other date is: 11/30/1955 The total number of days between the dates is 22000” Here is what I have so far: Import java.util.Scanner; Public class dates { Public static void main ( String args [] ) { int monthOne; int monthTwo; int dayOne; int dayTwo; intyearOne; intyearTwo; Scanner get = new Scanner (System.in ); System.out.println ( “Enter first date now: “ ); do { System.out.print ( “Enter Month: “); monthOne = get.nextInt(); System.out.print ( “Enter Day: “ ); dayOne = get.nextInt(); System.out.print ( “Enter Year : “ ); yearOne = get.nextInt(); if (monthOne < 1 || monthOne > 12 ) { System.out.printf ( “%d is not a valid month\ ”, monthOne ); } If ( (!(yearOne % 4 == 0 && yearOne & 100 !=0 || yearOne % 400 == 0) && (monthOne == 2 && dayOne >= 29)) || ((yearOne % 4 == 0 && yearOne % 100 != 0 || yearOne % 400 == 0) && (monthOne == 2 && dayOne >= 30))) { System.out.printf ( “%d is not a valid day.\ ”, dayOne ); } If ( (monthOne == 4 && dayOne > 30) || (monthOne ==6 && dayOne > 30) || (monthOne == 9 && dayOne > 30) || (monthOne == 11 && dayOne > 30) || (dayOne < 1 || dayOne > 31)) { System.out.printf ( “%d is not a valid day.\ ”, dayOne ); If ( yearOne < 1900 || yearOne > 2400 ) { System.out.printf ( “%d is not a valid year.\ ”, yearOne ); } } while (( (!(yearOne % 4 == 0 && yearOne % 100 !=0 || yearOne % 400 == 0) && (monthOne == 2 && dayOne >= 29)) || ((yearOne % 4 == 0 && yearOne % 100 != 0 || yearOne % 400 == 0) && (monthOne == 2 && dayOne >= 30 )) || (monthOne == 4 && dayOne > 30) || (monthOne == 6 && dayOne > 30) || (monthOne == 9 && dayOne > 30) || (monthOne == 11 && dayOne > 30) || (monthOne < 1 || monthOne > 12) || (dayOne < 1 || dayOne > 31) || (yearOne < 1900 || yearOne > 2400) )); System.out.printf ( “\ Enter the second date: \ ” ); do { System.out.print ( “Enter Month: “); monthTwo = get.nextInt(); System.out.print ( “Enter Day: “ ); dayTwo = get.nextInt(); System.out.print ( “Enter Year : “ ); yearTwo = get.nextInt(); if (monthTwo < 1 || monthTwo > 12 ) { System.out.printf ( “%d is not a valid month\ ”, monthTwo ); } If ( (!(yearTwo % 4 == 0 && yearTwo & 100 !=0 || yearTwo % 400 == 0) && (monthTwo == 2 && .
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![Need to make a Java program which calculates the number of days between two dates. I am NOT
able to use anything too advanced yet such as import.java.util.Calender, import.java.util.Date; or
import.java.util.GregorianCalender; Here are a couple of examples of what is should look like
when complete. I am having troubles getting my calculations to come out correctly. I’ve also
written the code I have so far.
EXAMPLE
“Enter first date now:
Enter Month: 1
Enter Day: 1
Enter Year: 1900
Enter the second date:
Enter Month: 1
Enter Day: 1
Enter Year: 2400
The first date is: 1/1/1900
The other date is: 1/1/2400
The total number of days between the dates is 182621”
AND
“Enter first date now:
Enter Month: 2
Enter Day: 23
Enter Year: 2016
Enter the second date:
Enter Month: 11
Enter Day: 30
Enter Year: 1955
The first date is: 2/23/2016
The other date is: 11/30/1955
The total number of days between the dates is 22000”
Here is what I have so far:
Import java.util.Scanner;
Public class dates {
Public static void main ( String args [] ) {
int monthOne;
int monthTwo;](https://image.slidesharecdn.com/needtomakeajavaprogramwhichcalculatesthenumberofdaysbetw-230709035629-70c2d983/85/Need-to-make-a-Java-program-which-calculates-the-number-of-days-betw-pdf-1-320.jpg)
Recommended
Hi,I have implemented increment() method. Please find the below up.pdf
Hi,
I have implemented increment() method. Please find the below updated code.
DaysBetween Class:
import java.text.DateFormatSymbols;
import java.util.Calendar;
import java.util.Scanner;
class DateClass {
protected int year;
protected int month;
protected int day;
public static final int MINYEAR = 1583;
// Constructor
public DateClass(int newMonth, int newDay, int newYear)
{
month = newMonth;
day = newDay;
year = newYear;
}
// Observers
public int getYear()
{
return year;
}
public int getMonth()
{
return month;
}
public int getDay()
{
return day;
}
public int lilian()
{
// Returns the Lilian Day Number of this date.
// Precondition: This Date is a valid date after 10/14/1582.
//
// Computes the number of days between 1/1/0 and this date as if no calendar
// reforms took place, then subtracts 578,100 so that October 15, 1582 is day 1.
final int subDays = 578100; // number of calculated days from 1/1/0 to 10/14/1582 November
17, 1858
int numDays;
// Add days in years.
numDays = year * 365;
// Add days in the months.
if (month <= 2)
numDays = numDays + (month - 1) * 31;
else
numDays = numDays + ((month - 1) * 31) - ((4 * (month-1) + 27) / 10);
// Add days in the days.
numDays = numDays + day;
// Take care of leap years.
numDays = numDays + (year / 4) - (year / 100) + (year / 400);
// Handle special case of leap year but not yet leap day.
if (month < 3)
{
if ((year % 4) == 0) numDays = numDays - 1;
if ((year % 100) == 0) numDays = numDays + 1;
if ((year % 400) == 0) numDays = numDays - 1;
}
// Subtract extra days up to 10/14/1582.
numDays = numDays - subDays;
return numDays;
}
Override
public String toString()
// Returns this date as a String.
{
String monthString = new DateFormatSymbols().getMonths()[month-1];
return(monthString + \"/\" + day + \"/\" + year);
}
public class mjd
{
public int mjd()
{
final int subDays = 678941;
int numDays;
numDays = year * 365;
if (month <= 2)
numDays = numDays + (month - 1) * 31;
else
numDays = numDays + ((month -1) * 31) - ((4 * (month-1) + 27)/10);
numDays = numDays + day;
numDays = numDays + (year / 4) - (year / 100) + (year / 400);
if (month < 3)
{
if ((year % 4) == 0) numDays = numDays -1;
if ((year % 100) == 0) numDays = numDays + 1;
if ((year % 400) == 0) numDays -= numDays -1;
}
// Days subtracted up to 10/14/1582
numDays = numDays - subDays;
return numDays;
}
}
public class djd
{
public int djd()
{
final int subDays = 693961; // number of calculated days from 1/1/0 to January 1,1900
int numDays;
// Add days in years.
numDays = year * 365;
// Add days in the months.
if (month <= 2)
numDays = numDays + (month - 1) * 31;
else
numDays = numDays + ((month - 1) * 31) - ((4 * (month-1) + 27) / 10);
// Add days in the days.
numDays = numDays + day;
// Take care of leap years.
numDays = numDays + (year / 4) - (year / 100) + (year / 400);
// Handle special case of leap year but not yet leap day.
if (month < 3)
{
if ((year % 4) == 0) numDays = numDays - 1;
if ((year % 100) == 0) numDays = numDays + 1;
if ((year % .
Write a program to generate the entire calendar for one year. The pr.pdf
Write a program to generate the entire calendar for one year. The program must get two values
from the user: (1) the year and (2) the day of the week for January 1st of that year. The year,
which should be positive, is needed to check for and handle leap years1. The day of the week for
January 1st is needed so that you know where to start the calendar. The user should enter 0 for
Sunday, 1 for Monday, ... or 6 for Saturday. As always, you need to validate the user\'s input. To
actually print the calendar, you must use a single method that prints out the calendar for one
month and then call this function 12 times from main(), once for each month in the year. To
check for a leap year you will need to write another method that takes the year as a parameter
and returns true if it’s a leap year, or false otherwise. Stubs (i.e. method signatures without any
code) for both of these methods have been provided for you.
The calendar should be printed in the form below. In this example, January starts on a Saturday
(day 6). Note that February starts on a Tuesday in this example because January ended on a
Monday.
January
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 2829
30 31
February
1 2 3 4 5
6 7 8 9 10 11 12 ...
The format he gave us is as follows:
public class PA6a {
/**
* Error to output if year is not positive
*/
static final String E_YEAR = \"The year must be positive!\";
/**
* Error to output if the day is not between 0 and 6
*/
static final String E_DAY = \"The day of January 1st must be between 0 and 6!\";
/**
* Determines if an input is a leap year
*
* @param year year in question
* @ereturn true if a leap year
*/
public static boolean isLeapYear(int year) {
return false; // TODO: replace with your code
}
/**
* Outputs a month to the console
*
* @param month title
* @param startDay 0=Sunday ... 6=Saturday
* @param numDays number of days in the month
* @return day of the week of the last day of the month
*/
public static int printMonth(String month, int startDay, int numDays) {
return 0; // TODO: replace with your code
}
/**
* Program execution point:
* input year, day of the week (0-6) of january 1
* output calendar for that year
*
* @param args command-line arguments (ignored)
*/
public static void main(String[] args) {
// TODO: write your code here
}
I am confused especially with the print month method, help is greatly appreciated.
Solution
package org.students;
import java.util.Scanner;
import com.sun.org.apache.bcel.internal.generic.IUSHR;
public class Calander {
//declaring static variables
static final String E_YEAR = \"The year must be positive!\";
static final String E_DAY = \"The day of January 1st must be between 0 and 6!\";
public static void main(String[] args) {
//Declaring variables
int getFirstDay;
int year;
Scanner scanner = new Scanner(System.in);
//This while loop continue to execute until user enters valid year
while (true) {
//Getting the year entered by the user
System.out.print(\"Enter the year:.
Assignment Details There is a .h file on Moodle that provides a defi.pdf
Assignment Details There is a .h file on Moodle that provides a definition for a
WeatherForecaster class. The functionality for that class is similar to the functionality you
implemented in Assignment 5, with a few additional functions. Instead of using an array of
structs and functions to process the array, you will create one WeatherForecaster object that
includes the array of structs as a private variable and public methods to process the data. The
struct for this assignment has an additional member called forecastDay, you will need to store all
of the data this time. struct ForecastDay{ string day; string forecastDay; int highTemp; int
lowTemp; int humidity; int avgWind; string avgWindDir; int maxWind; string maxWindDir;
double precip; }; Methods in the WeatherForecaster class void addDayToData(ForecastDay); •
Takes a ForecastDay as an argument and adds it to the private array stored in the
WeatherForecaster object. • Use the private index variable to control where ForecastDay is
added to the array. void printDaysInData(); • Show the dates in the data set where the day and
the forecast day are the same. void printForecastForDay(string); • Take a date as an argument
and shows the forecast for that date. CSCI 1310 - Assignment 6 Due Saturday, Oct 15, by 12:30
pm void printFourDayForecast(string); • Takes a date as an argument and shows the forecast
issued on that date and for the next three days. For example, for a date of 1- 26-2016, you would
show the forecast for 1-26-2016 issued on 1- 26-2016 as well as the forecast for 1-27, 1-28, and
1-29 issued on 1-26. double calculateTotalPrecipitation(); • Returns the sum of the precipitation
in the data set. void printLastDayItRained(); • Shows the date of the last measureable
precipitation. void printLastDayAboveTemperature(int); • Takes an integer as an argument and
shows the date for the last day above that temperature. If no days are above the temperature,
prints “No days above that temperature.” void printTemperatureForecastDifference(string); •
Takes a date as an argument and shows the temperature forecast for that date for the three days
leading up to the date and the day-of forecast. void printPredictedVsActualRainfall(int); • Shows
the difference between the predicted and actual rainfall total in the entire data set. • The
argument to the function is the number of forecast days away. For example, the forecast for 1-27-
2016 is one day away from 1- 26-2016. string getFirstDayInData(); • Returns the first date in the
data with a day-of forecast, i.e. day = forecastDay string getLastDayInData(); • Returns the last
date in the data with a day-of forecast, i.e. day = forecastDay Challenge functions 1. There is
another header file on Moodle called WeatherForecastChallenge.h that uses a vector to store the
future forecast days. Instead of including all data in the yearData array, you can include only
days where the day = forecast day in the array. The other forecast days are stored in the vecto.
Recommended
Hi,I have implemented increment() method. Please find the below up.pdf
Hi,
I have implemented increment() method. Please find the below updated code.
DaysBetween Class:
import java.text.DateFormatSymbols;
import java.util.Calendar;
import java.util.Scanner;
class DateClass {
protected int year;
protected int month;
protected int day;
public static final int MINYEAR = 1583;
// Constructor
public DateClass(int newMonth, int newDay, int newYear)
{
month = newMonth;
day = newDay;
year = newYear;
}
// Observers
public int getYear()
{
return year;
}
public int getMonth()
{
return month;
}
public int getDay()
{
return day;
}
public int lilian()
{
// Returns the Lilian Day Number of this date.
// Precondition: This Date is a valid date after 10/14/1582.
//
// Computes the number of days between 1/1/0 and this date as if no calendar
// reforms took place, then subtracts 578,100 so that October 15, 1582 is day 1.
final int subDays = 578100; // number of calculated days from 1/1/0 to 10/14/1582 November
17, 1858
int numDays;
// Add days in years.
numDays = year * 365;
// Add days in the months.
if (month <= 2)
numDays = numDays + (month - 1) * 31;
else
numDays = numDays + ((month - 1) * 31) - ((4 * (month-1) + 27) / 10);
// Add days in the days.
numDays = numDays + day;
// Take care of leap years.
numDays = numDays + (year / 4) - (year / 100) + (year / 400);
// Handle special case of leap year but not yet leap day.
if (month < 3)
{
if ((year % 4) == 0) numDays = numDays - 1;
if ((year % 100) == 0) numDays = numDays + 1;
if ((year % 400) == 0) numDays = numDays - 1;
}
// Subtract extra days up to 10/14/1582.
numDays = numDays - subDays;
return numDays;
}
Override
public String toString()
// Returns this date as a String.
{
String monthString = new DateFormatSymbols().getMonths()[month-1];
return(monthString + \"/\" + day + \"/\" + year);
}
public class mjd
{
public int mjd()
{
final int subDays = 678941;
int numDays;
numDays = year * 365;
if (month <= 2)
numDays = numDays + (month - 1) * 31;
else
numDays = numDays + ((month -1) * 31) - ((4 * (month-1) + 27)/10);
numDays = numDays + day;
numDays = numDays + (year / 4) - (year / 100) + (year / 400);
if (month < 3)
{
if ((year % 4) == 0) numDays = numDays -1;
if ((year % 100) == 0) numDays = numDays + 1;
if ((year % 400) == 0) numDays -= numDays -1;
}
// Days subtracted up to 10/14/1582
numDays = numDays - subDays;
return numDays;
}
}
public class djd
{
public int djd()
{
final int subDays = 693961; // number of calculated days from 1/1/0 to January 1,1900
int numDays;
// Add days in years.
numDays = year * 365;
// Add days in the months.
if (month <= 2)
numDays = numDays + (month - 1) * 31;
else
numDays = numDays + ((month - 1) * 31) - ((4 * (month-1) + 27) / 10);
// Add days in the days.
numDays = numDays + day;
// Take care of leap years.
numDays = numDays + (year / 4) - (year / 100) + (year / 400);
// Handle special case of leap year but not yet leap day.
if (month < 3)
{
if ((year % 4) == 0) numDays = numDays - 1;
if ((year % 100) == 0) numDays = numDays + 1;
if ((year % .
Write a program to generate the entire calendar for one year. The pr.pdf
Write a program to generate the entire calendar for one year. The program must get two values
from the user: (1) the year and (2) the day of the week for January 1st of that year. The year,
which should be positive, is needed to check for and handle leap years1. The day of the week for
January 1st is needed so that you know where to start the calendar. The user should enter 0 for
Sunday, 1 for Monday, ... or 6 for Saturday. As always, you need to validate the user\'s input. To
actually print the calendar, you must use a single method that prints out the calendar for one
month and then call this function 12 times from main(), once for each month in the year. To
check for a leap year you will need to write another method that takes the year as a parameter
and returns true if it’s a leap year, or false otherwise. Stubs (i.e. method signatures without any
code) for both of these methods have been provided for you.
The calendar should be printed in the form below. In this example, January starts on a Saturday
(day 6). Note that February starts on a Tuesday in this example because January ended on a
Monday.
January
1
2 3 4 5 6 7 8
9 10 11 12 13 14 15
16 17 18 19 20 21 22
23 24 25 26 27 2829
30 31
February
1 2 3 4 5
6 7 8 9 10 11 12 ...
The format he gave us is as follows:
public class PA6a {
/**
* Error to output if year is not positive
*/
static final String E_YEAR = \"The year must be positive!\";
/**
* Error to output if the day is not between 0 and 6
*/
static final String E_DAY = \"The day of January 1st must be between 0 and 6!\";
/**
* Determines if an input is a leap year
*
* @param year year in question
* @ereturn true if a leap year
*/
public static boolean isLeapYear(int year) {
return false; // TODO: replace with your code
}
/**
* Outputs a month to the console
*
* @param month title
* @param startDay 0=Sunday ... 6=Saturday
* @param numDays number of days in the month
* @return day of the week of the last day of the month
*/
public static int printMonth(String month, int startDay, int numDays) {
return 0; // TODO: replace with your code
}
/**
* Program execution point:
* input year, day of the week (0-6) of january 1
* output calendar for that year
*
* @param args command-line arguments (ignored)
*/
public static void main(String[] args) {
// TODO: write your code here
}
I am confused especially with the print month method, help is greatly appreciated.
Solution
package org.students;
import java.util.Scanner;
import com.sun.org.apache.bcel.internal.generic.IUSHR;
public class Calander {
//declaring static variables
static final String E_YEAR = \"The year must be positive!\";
static final String E_DAY = \"The day of January 1st must be between 0 and 6!\";
public static void main(String[] args) {
//Declaring variables
int getFirstDay;
int year;
Scanner scanner = new Scanner(System.in);
//This while loop continue to execute until user enters valid year
while (true) {
//Getting the year entered by the user
System.out.print(\"Enter the year:.
Assignment Details There is a .h file on Moodle that provides a defi.pdf
Assignment Details There is a .h file on Moodle that provides a definition for a
WeatherForecaster class. The functionality for that class is similar to the functionality you
implemented in Assignment 5, with a few additional functions. Instead of using an array of
structs and functions to process the array, you will create one WeatherForecaster object that
includes the array of structs as a private variable and public methods to process the data. The
struct for this assignment has an additional member called forecastDay, you will need to store all
of the data this time. struct ForecastDay{ string day; string forecastDay; int highTemp; int
lowTemp; int humidity; int avgWind; string avgWindDir; int maxWind; string maxWindDir;
double precip; }; Methods in the WeatherForecaster class void addDayToData(ForecastDay); •
Takes a ForecastDay as an argument and adds it to the private array stored in the
WeatherForecaster object. • Use the private index variable to control where ForecastDay is
added to the array. void printDaysInData(); • Show the dates in the data set where the day and
the forecast day are the same. void printForecastForDay(string); • Take a date as an argument
and shows the forecast for that date. CSCI 1310 - Assignment 6 Due Saturday, Oct 15, by 12:30
pm void printFourDayForecast(string); • Takes a date as an argument and shows the forecast
issued on that date and for the next three days. For example, for a date of 1- 26-2016, you would
show the forecast for 1-26-2016 issued on 1- 26-2016 as well as the forecast for 1-27, 1-28, and
1-29 issued on 1-26. double calculateTotalPrecipitation(); • Returns the sum of the precipitation
in the data set. void printLastDayItRained(); • Shows the date of the last measureable
precipitation. void printLastDayAboveTemperature(int); • Takes an integer as an argument and
shows the date for the last day above that temperature. If no days are above the temperature,
prints “No days above that temperature.” void printTemperatureForecastDifference(string); •
Takes a date as an argument and shows the temperature forecast for that date for the three days
leading up to the date and the day-of forecast. void printPredictedVsActualRainfall(int); • Shows
the difference between the predicted and actual rainfall total in the entire data set. • The
argument to the function is the number of forecast days away. For example, the forecast for 1-27-
2016 is one day away from 1- 26-2016. string getFirstDayInData(); • Returns the first date in the
data with a day-of forecast, i.e. day = forecastDay string getLastDayInData(); • Returns the last
date in the data with a day-of forecast, i.e. day = forecastDay Challenge functions 1. There is
another header file on Moodle called WeatherForecastChallenge.h that uses a vector to store the
future forecast days. Instead of including all data in the yearData array, you can include only
days where the day = forecast day in the array. The other forecast days are stored in the vecto.
Date.h#ifndef DateFormat#define DateFormat#includetime.h.pdf
//Date.h
#ifndef DateFormat
#define DateFormat
#include
#include
class Date {
public:
Date();
Date(int, int);
Date(int, int, int);
Date(char *, int, int);
voidsetMonth(int);
voidsetDay(int);
voidsetYear(int);
voidprintDateSlash(void) const;
voidprintDateMonth(void) const;
voidprintDateDay(void) const;
constchar *monthName(void) const;
intleapYear(void) const;
intdaysOfMonth(void) const;
voidchange(int);
intchange1(void) const;
voidchange2(constchar * const);
constchar *monthList(int) const;
intdays(int) const;
private:
int day;
int month;
int year;
};
class main {
public:
main();
virtual~main();
};
#endif
//Date.cpp
#include\"stdafx.h\"
#include
#include
#include
#include \"Date.h\"
Date::Date()
{
struct tm *ptr;
time_t t = time( 0 );
ptr = localtime( &t );
day = ptr->tm_mday;
month = 1 + ptr->tm_mon;
year = ptr->tm_year + 1900;
}
Date::Date( int ss, int ppp )
{
setYear( ppp );
change(ss);
}
Date::Date( int mm, int day1, int yy )
{
setYear( yy + 1900 );
setMonth( mm );
setDay(day1);
}
Date::Date( char *mPtr, int day1, int ppp )
{
setYear( ppp );
change2( mPtr );
setDay(day1);
}
voidDate::setDay( int d1 )
{ day = d1 >= 1 && d1 <= daysOfMonth() ? d1 : 1; }
voidDate::setMonth( int m ) { month = m >= 1 && m <= 12 ? m : 1; }
voidDate::setYear( int y ) { year = y >= 1900 && y <= 1999 ? y : 1900; }
voidDate::printDateSlash( void ) const
{
cout<< month << \'/\' << day << \'/\' << year << \'\ \'; }
voidDate::printDateMonth( void ) const
{
cout << monthName() << \' \' << day << \", \" << year << \'\ \'; }
voidDate::printDateDay( void ) const
{
cout << convert2() << \' \' << year << \'\ \'; }
constchar *Date::monthName( void ) const { return monthList( month - 1 ); }
intDate::daysOfMonth( void ) const
{ return leapYear() && month == 2 ? 29 : days( month ); }
intDate::leapYear( void ) const
{
if ( year % 400 == 0 || ( year % 4 == 0 && year % 100 != 0 ) )
return 1;
else
return 0;
}
voidDate::convert1( int ss )
{
int dayTotal = 0;
if ( ss < 1 || ss > 366 )
ss = 1;
setMonth( 1 );
for ( int m = 1; m < 13 && ( dayTotal + daysOfMonth() ) < ss; ++m ) {
dayTotal += daysOfMonth();
setMonth( m + 1 );
}
setDay( ss - dayTotal );
setMonth( m );
}
intDate::change1( void ) const
{
int ss = 0;
for ( int m = 1; m < month; ++m )
ss += days( m );
ss += day;
return ss;
}
voidDate::change2( constchar * const mPtr )
{
int flag = 0;
for ( int subscript = 0; subscript < 12; ++subscript )
if (!strcmp( mPtr, monthList( subscript ) ) ) {
setMonth( subscript + 1 );
flag = 1;
break;
}
if ( !flag )
setMonth( 1 );
}
constchar *Date::monthList( int mm ) const
{
char *months[] = { \"January\", \"February\", \"March\", \"April\", \"May\",
\"June\", \"July\", \"August\", \"September\", \"October\",
\"November\", \"December\" };
return months[ mm ];
}
intDate::days( int m ) const
{
constint monthDays[] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
return monthDays[ m - 1 ];
}
//main.cpp
#include \"Date.h\"
#include
using sts::cout
using std::endl
#include
#include
#include \"p7_07..
Python and PostgreSQL: Let's Work Together! | PyConFr 2018 | Dimitri Fontaine
Python is often used to maintain application backends. When the backend should implement user oriented workflows, it may rely on a RDBMS component to take care of the system's integrity.
PostgreSQL is the world's most advanced open source relational database, and is very good at taking care of your system's integrity. PostgreSQL also comes with a ton of data processing power, and in many cases a simple enough SQL statement may replace hundreds of lines of code written in Python.
In this talk, we learn advanced SQL techniques and how to reason about which part of the backend code should be done in the database, and which parf of the backend code is so easier to write as a SQL query.
Date class that represents a date consisting of a year, month, and a.pdf
Date class that represents a date consisting of a year, month, and a day
import java.util.TimeZone;
public class Date {
// constants
private static final int JANUARY = 1;
private static final int FEBRUARY = 2;
private static final int DECEMBER = 12;
private static final int DAYS_PER_WEEK = 7;
private static final int DAYS_PER_YEAR = 365;
private static final int DAYS_PER_LEAP_YEAR = 366;
private static final String[] DAY_NAMES = {\"SUNDAY\", \"MONDAY\", \"TUESDAY\",
\"WEDNESDAY\", \"THURSDAY\", \"FRIDAY\", \"SATRDAY\"};
private static final int[] DAYS_PER_MONTH = { -1,
// 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31,
};
// fields
private int year;
private int month;
private int day;
/**
Constructs a new object representing the given year, month, and day.
@throws IllegalArgumentException if month is not between 1 and 12, or if day is not between 1
and the number of days in that month.
*/
public Date(int year, int month, int day) {
this.year = year;
this.month = month;
this.day = day;
if (month < 1 || month > 12 || day < 1 || day > getDaysInMonth()) {
throw new IllegalArgumentException(\"Invalid day or month: \" + toString());
}
}
/** Constructs a new object representing today\'s date. */
public Date() {
this(2006, july
, 22);
int daysSinceEpoch = (int) ((System.currentTimeMillis() +
TimeZone.getDefault().getRawOffset()) / 1000 / 60 / 60 / 24);
for (int i = 0; i < daysSinceEpoch; i++) {
nextDay();
}
}
/**
Adds the given number of days to this Date.
@throws IllegalArgumentException if days < 0.
*/
public void addDays(int days) {
while (days > getDaysInYear()) {
days -= getDaysInYear();
year++;
}
for (int i = 1; i <= days; i++) {
nextDay();
}
}
/**
Returns whether o refers to a Date object representing the same
year, month, and day as this one.
*/
public boolean equals(Object o) {
Date other = (Date) o;
return day == other.day &&
month == other.month &&
year == other.year;
}
/**
Returns the day represented by this Date object.
@return the day, between 1 and the number of days in this Date\'s month
(which varies from 28 to 31)
*/
public int getDay() {
return day;
}
/** Returns the number of days in the month represented by the current Date. */
public int getDaysInMonth() {
int result = DAYS_PER_MONTH[month];
if (month == FEBRUARY && isLeapYear()) {
result++;
}
return result;
}
public int getDaysInYear() {
if (isLeapYear()) {
return 366;
} else {
return 365;
}
}
/**
Returns a string such as \"Monday\" representing what day of the week
this Date fell on.
@return the day as either \"Sunday\", \"Monday\", \"Tuesday, \"Wednesday\", \"Thursday\",
\"Friday\", or \"Satrday\"
*/
public String getDayOfWeek() {
int index = 1;
Date temp = new Date(1753, JANUARY, 1);
while (temp != this) {
temp.nextDay();
index = (index + 1) % DAYS_PER_WEEK;
}
return DAY_NAMES[index];
}
/**
Returns the month represented by this Date object.
@return the month, between 1 and 12
*/
public int getMonth() {
return month;
}
/**
Returns the yea.
publicclass Date {privatestatic String DATE_SEPARATOR = ;pr.pdf
publicclass Date {
privatestatic String DATE_SEPARATOR = \"/\";
privatestaticintDAYS_PER_WEEK = 7;
//Attributes
privateint day;
privateint month;
privateint year;
/**
* Default Constructor
* Instantiates an object of type Date to 1/1/2000
*/
public Date() {
this.day = 1;
this.month = 1;
this.year = 2000;
}
/**
* Constructs a new date object to represent the given date.
* @param day
* @param month
* @param year
*/
public Date(int day, int month, int year) {
if(isValid(day, month, year)) {
this.day = day;
this.month = month;
this.year = year;
} else
System.out.println(\"Invalid Date.\");
}
/**
* Returns the day value of this date for example, for the date 2006/07/22, returns 22.
* @return
*/
publicint getDay() {
return day;
}
/**
* Returns the month value of this date ,for example, for the date 2006/07/22, returns 7.
* @return
*/
publicint getMonth() {
return month;
}
/**
* Returns the year value of this date, for example , the date 2006/07/22, returns 2006.
* @return
*/
publicint getYear() {
return year;
}
/**
* @param day the day to set
*/
publicvoid setDay(int day) {
this.day = day;
}
/**
* @param month the month to set
*/
publicvoid setMonth(int month) {
this.month = month;
}
/**
* @param year the year to set
*/
publicvoid setYear(int year) {
this.year = year;
}
/**
* Returns true if the year of this date is a leap year.
* A leap year occurs every 4 years , except for multiples of 100 that are not multiples of 400.
* For example, 1956,1844,1600,and 2000 are leap years, but 1983,2002,1700,and 1900 are not.
* @return
*/
publicboolean isLeapYear(){
if(((this.year % 400) == 0) || (((this.year % 4) == 0) && ((this.year % 100) != 0)))
returntrue;
else
returnfalse;
}
/**
* Checks if the date is valid
* @param day
* @param month
* @param year
* @return
*/
publicboolean isValid(int day, int month, int year) {
if((month < 1) || (12 < month))
returnfalse;
else {
if(((month == 1) || (month == 3) || (month == 5) || (month == 7) ||
(month == 8) || (month == 10) || (month == 12)) && ((day < 1) || (31 < day)))
returnfalse;
elseif(((month == 4) || (month == 6) || (month == 9) || (month == 11)) && ((day < 1) || (30 <
day)))
returnfalse;
elseif(month == 2) {
if(isLeapYear() && ((day < 1) || (29 < day)))
returnfalse;
elseif((day < 1) || (28 < day))
returnfalse;
}
}
returntrue;
}
/**
* Returns the maximum number of days in a month
* @return
*/
publicint maxMonthDays() {
if(this.month == 2) {
if(isLeapYear())
return 29;
else
return 28;
} elseif((this.month == 1) || (this.month == 3) || (this.month == 5) || (this.month == 7) ||
(this.month == 8) || (this.month == 10) || (this.month == 12))
return 31;
else
return 30;
}
/**
* Checks if this dat is same as other date
* @param other
* @return
*/
publicboolean isEqual(Date other) {
if((this.day == other.day) && (this.month == other.month) && (this.year == other.year))
returntrue;
else
returnfalse;
}
/**
* Moves this Date object forward in time by the given number of days .
* @param days
*/
publicvoid addDays(int days).
From the proposal to ECMAScript – Step by Step
Do ever wonder how ECMAScript evolves and it's available almost everywhere? We
gonna do a walkthrough of how TC39 (the technical committee responsible for the
ECMAScript) works, how to participate, contribute, and how an idea can make it
through to the language, all step-by-step. Together with a sneak pick and
debate about some recent and interesting proposals in different stages.
(c) Fest.js Porto 2023
April 20, 2023
Porto, Portugal
https://fest.dev/events/js/porto-2023/
struct procedure { Date dateOfProcedure; int procedureID.pdf
struct procedure
{
Date dateOfProcedure;
int procedureID;
int procedureProviderID;
};
Data.h
#include
#include
class Date
{
friend ostream& operator<<( ostream &, const Date & );
// allows easy output to a ostream
public:
Date( int m = 1, int d = 1, int y = 1900 ); // constructor, note the default values
void setDate( int, int, int ); // set the date const
Date &operator+=( int ); // add days, modify object
bool leapYear( int) const; // is this a leap year?
bool endOfMonth( int ) const; // is this end of month?
int getMonth ( ) const;
int getDay ( ) const;
int getYear ( ) const;
string getMonthString( ) const;
private:
int month;
int day;
int year;
static const int days[]; // array of days per month static const string monthName[]; // array of
month names
void helpIncrement();
// utility function
}; #endif Data.cpp // Member function definitions for Date
class in separate
date.cpp
#include
#include \"date.h\"
#include // Initialize static members at file scope; // one class-wide copy.
const int Date::days[] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; const string
Date::monthName[] = { \"January\", \"February\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\", \"October\", \"November\", \"December\" };
// Date constructor
Date::Date(int m, int d, int y)
{
setDate(m, d, y); } // Set the date
void Date::setDate(int mm, int dd, int yy)
{
month = (mm >= 1 && mm <= 12) ? mm : 1; year = (yy >=1900&& yy <= 2100) ? yy
: 1900; // test for a leap year
if (month == 2 && leapYear(year)) day = (dd >= 1 && dd <= 29) ? dd : 1;
else
day =(dd >= 1 && dd <= days[month]) ? dd : 1;
} // Add a specific number of days to a date
const Date &Date::operator+=(int additionalDays)
{
for (int i = 0; i < additionalDays; i++) helpIncrement();
return *this; // enables cascading
} // If the year is leap year, return true; //
otherwise, return false
bool Date::leapYear(int testYear)
const
{
if (testYear % 400 == 0 || (testYear %100 != 0 && testYear % 4 == 0)) return true; // a
leap year
else
return false; // not a leap year
} // Determine if the day is the end of the month
bool Date::endOfMonth(int testDay)
const
{
if (month == 2 && leapYear(year))
return (testDay == 29); // last day of Feb. in leap year
else
return (testDay == days[month]);
} // Function to help increment the date
void Date::helpIncrement()
{
if (!endOfMonth(day))
{
// date is not at the end of the month
day++;
}
else if (month < 12)
{
// date is at the end of the month, but month < 12 day = 1; ++month;
}
else
{
// end of month and year: last day of the year
day = 1;
month = 1;
++year;
} }
// Overloaded output operator
ostream &operator<<(ostream& output, const Date &d)
{
output << d.monthName[d.month] << \' \' << d.day << \", \"<< d.year;
return output; // enables cascading
}
int Date::getMonth() const //
{
Public:
int month[20];
If(month[]!=”/0”)
Cout<<”Enter Month :-<>month;
Elseif (month[]<12)
{
Cout<>month[];
else
return month;
}
int Date::getDay() const //
{
Pu.
Java 8 Date and Time API
This presentation provides an overview of using the Java SE 8 Date & Time API. It covers how to:
1. Create and manage date-based and time-based events including a combination of date and time into a single object using LocalDate, LocalTime, LocalDateTime, Instant, Period, and Duration
2. Work with dates and times across timezones and manage changes resulting from daylight savings including format date and times values
3. Define and create and manage date-based and time-based events using Instant, Period, Duration, and TemporalUnit
딥러닝, 야 너도 할 수 있어(feat. PyTorch)
※다운로드하시면 더 선명한 자료를 보실 수 있습니다.
딥러닝을 처음 시작하시는 분, 특히 개발자에게 맞춰 필요한 내용을 준비하였습니다.
딥러닝의 수학적인 이해보다는 딥러닝을 활용하는 방법을 공유하고자 합니다.
목차
1. 딥러닝 권장 개발 환경
2. 코드 작성을 위한 약간의 이론
3. 가장 빠르게 그럴싸한 모델 만들기
대상
딥러닝에 관심이 있거나 딥러닝을 시작하려는 모든 개발자
※발표 영상은 제공하지 않습니다.
I am exploring the basic components of advanced broadband networks..pdf
I am exploring the basic components of advanced broadband networks.
Describe the three layered architecture:
Briefly, explain the difference between link state and distance vector routing
Describe how the sliding window protocol works
Explain the CIDR concept and its major benefit
Solution
The three layered architecture is a client server architecture.This architecture is used for
functional perspective logic,data access, data storage and user interface.
Three layer architecture is used for software design.
Three layers are as follows,
SLIDING WINDOW PROTOCOL:
It is one of the packet based data transmission protocol.Packets are send and its too hard to send
duplicate packets and to identify the missing packets.
In sliding window protocol we set a limit for number of packets to be transmitted at a time.
CIDR concept:
Classless Inter-domain routing is an alternate to the traditional subnetting. It is a way to allow the
more flexible way to allocate Internet protocol.
Benefit:
CIDR provides the better way to address since it prevents the wastage of addressing space since
it is a flexible way of addressing.Link state vectordistance vector routingLink state vector makes
all the nodes to know about the route.Distance vector doubting only the neighbouring node is
knownLink state information is flooded throughout the link state domain.Distance vector
routing information is not knownl.
How many different instances of the ls command are installed on .pdf
How many different instances of the \"ls\" command are installed on your linux machine, and
what is/are the location(s) path i.e. what is the complete path for each?
Solution
ls is one of the basic command in Linux. This command lists or displays files or directories.
ls command runs with different parameters or instances like
ls -l means long listing format
ls -a displays all files with hidden files
ls -lh show size of files
To find where a particular command is stored use whereis command e:g
whereis ls outputs Output:
ls: /bin/ls /usr/share/man/man1p/ls.1p.gz /usr/share/man/man1/ls.1.gz.
How are the Allen Bradley SLC 500 program files organizedSolutio.pdf
How are the Allen Bradley SLC 500 program files organized
Solution
Allen Bradley SLC 500 program files include System Functions, Reserved, User Program and
Subroutine Programs.
System Functions are of file 0, Reserved are of file 1, User Program are of file 2 and Subroutine
Programs are of file 3-255.
System Functions file contains various system related information and user programmed
information. User programmed information are processor type, I/O configuration, processor file
name and password.
Reserved file is reserved by the processor and user cannot access these files.
User program files contain user programmed instructions. These instructions define how the
controller should operate.
Subroutine programs are created by users and are activated based on subroutine instructions
which are in the main ladder program file..
Hi, I need help please, this is about KaseyaWich o the following .pdf
Hi, I need help please, this is about Kaseya:
Wich o the following pages has the ability to configure and apply the ATSE code:
____Monitor > Agent Monitoring > Event Log Alerts
____Monitor > Agent Monitoring > SNMP Traps Alert
____Monitor > Agent Monitoring > Alerts
____Monitor > Agent Monitoring > Monitor Log
____Monitor > SNMP Monitoring > Assign SNMP
____Monitor > Agent Monitoring > SNMP Log
Solution
The answer to this is the first one
Monitor > Agent Monitoring > Event Log Alerts has the ability to configure and apply the ATSE
code.
How did WWI and its aftermath provide African Americans with opportu.pdf
How did WWI and its aftermath provide African Americans with opportunities? Discuss W.E.B.
DuBois and the rise of the NAACP. Discuss Marcus Garvey and the rise of Garveyism. 5-7
sentences for answer
Solution
For most of the African American soldiers, serving in World War I had provided them a broaden
social , political, geographical and cultural acceptance in the society, As they received the
opportunity to travel the different locations of the country, and more than 200000 African-
American soldiers served in other countries to various part of the globe, had really improved the
life experience to these soldiers.
NAACP was founded on 12th Feb 1909 and it is considered to be the oldest, largest and most
widely recognized grassroots-based civil right organization in the United States. NAACP was
mainly constituted in response to the increasing horrific incidents of lynching and the race riots
which resulted in the year 1908 in Springfield. Following the racial violence against the Afro-
Americans, Mary White Ovington, and Oswald Garrison Villard, both the descendants of
abolitionists, William English Walling and Dr. Henry Moscowitz issued a call for a meeting to
discuss racial justice. Almost 60 individuals out of them 7 were Afro-American including W. E.
B. Du Bois, Ida B. Wells-Barnett, and Mary Church Terrell signed a declaration which was
released on the occasion of the centennial of Lincoln’s birth..
For problems 3 and 4, consider the following functions that implemen.pdf
For problems 3 and 4, consider the following functions that implement the dequeue operation for
a priority queue that is implemented with a heap.
int[] pQueue;
int length;
int dequeue()
{
int node = 1;
int value = pQueue[--length];
int maxValue = pQueue[node];
int location = sift(node * 2, value);
pQueue[location] = value;
return maxValue;
}
int sift(int node, int value)
{
if (node <= length)
{
if (node < length && pQueue[node] < pQueue[node + 1]) node++;
if (value < pQueue[node])
{
pQueue[node / 2] = pQueue[node];
return sift(node * 2, value);
}
}
return node / 2;
}
3. Write the recurrence equation and initial conditions that expresses the execution time cost for
the sift function in the above algorithm. Draw the recursion tree assuming that n = 8. Assume
that n represents the number of nodes in the subtree at each step of the sifting operation.
4. Determine the critical exponent for the recurrence equation in problem 3. Apply the Little
Master Theorem to solve that equation specifically stating which part of the theorem is applied to
arrive at the solution.
Solution
using System.Collections;
using System.Collections.Generic;
namespace PriorityQueueDemo
{
///
/// Priority queue based on binary heap in program,
/// Elements with minimum priority dequeued first in program
///
/// Type of priorities
/// Type of values
public class PriorityQueues : ICollection>
{
private List> _baseHeap;
private IComparer _comparers;
#region Constructor
///
/// Initializes a new instance of priority queue with default initial capacity and default
priority comparer in program
///
public PriorityQueues()
: this(Comparers.Default)
{
}
///
/// Initializes a new instance of priority queue with specified initial capacity and default
priority comparer in program
///
/// initial capacity
public PriorityQueues(int capacity)
: this(capacity, Comparers.Default)
{
}
///
/// Initializes a new instance of priority queue with specified initial capacity and specified
priority comparer in program
///
/// initial capacity
/// priority comparer
public PriorityQueues(int capacity, IComparers comparer)
{
if (comparer == null)
throw new ArgumentNullException();
_baseHeaps = new List>(capacity);
_comparers= comparer;
}
///
/// Initializes a new instance of priority queue with default initial capacity and specified
priority comparer in program
///
/// priority comparer
public PriorityQueues(IComparer comparers)
{
if (comparers == null)
throw new ArgumentNullException();
_baseHeap = new List>();
_comparer = comparers;
}
///
/// Initializes a new instance of priority queue with specified data and default priority
comparer in program
///
/// data to be inserted into priority queue in program
public PriorityQueues(IEnumerable> data)
: this(data, Comparers.Default)
{
}
///
/// Initializes a new instance of priority queue with specified data and specified priority
comparer in program
///
/// data to be inserted into priority queue
/// priority comparer in program
public PriorityQueue(IEnumer.
More Related Content
Similar to Need to make a Java program which calculates the number of days betw.pdf
Date.h#ifndef DateFormat#define DateFormat#includetime.h.pdf
//Date.h
#ifndef DateFormat
#define DateFormat
#include
#include
class Date {
public:
Date();
Date(int, int);
Date(int, int, int);
Date(char *, int, int);
voidsetMonth(int);
voidsetDay(int);
voidsetYear(int);
voidprintDateSlash(void) const;
voidprintDateMonth(void) const;
voidprintDateDay(void) const;
constchar *monthName(void) const;
intleapYear(void) const;
intdaysOfMonth(void) const;
voidchange(int);
intchange1(void) const;
voidchange2(constchar * const);
constchar *monthList(int) const;
intdays(int) const;
private:
int day;
int month;
int year;
};
class main {
public:
main();
virtual~main();
};
#endif
//Date.cpp
#include\"stdafx.h\"
#include
#include
#include
#include \"Date.h\"
Date::Date()
{
struct tm *ptr;
time_t t = time( 0 );
ptr = localtime( &t );
day = ptr->tm_mday;
month = 1 + ptr->tm_mon;
year = ptr->tm_year + 1900;
}
Date::Date( int ss, int ppp )
{
setYear( ppp );
change(ss);
}
Date::Date( int mm, int day1, int yy )
{
setYear( yy + 1900 );
setMonth( mm );
setDay(day1);
}
Date::Date( char *mPtr, int day1, int ppp )
{
setYear( ppp );
change2( mPtr );
setDay(day1);
}
voidDate::setDay( int d1 )
{ day = d1 >= 1 && d1 <= daysOfMonth() ? d1 : 1; }
voidDate::setMonth( int m ) { month = m >= 1 && m <= 12 ? m : 1; }
voidDate::setYear( int y ) { year = y >= 1900 && y <= 1999 ? y : 1900; }
voidDate::printDateSlash( void ) const
{
cout<< month << \'/\' << day << \'/\' << year << \'\ \'; }
voidDate::printDateMonth( void ) const
{
cout << monthName() << \' \' << day << \", \" << year << \'\ \'; }
voidDate::printDateDay( void ) const
{
cout << convert2() << \' \' << year << \'\ \'; }
constchar *Date::monthName( void ) const { return monthList( month - 1 ); }
intDate::daysOfMonth( void ) const
{ return leapYear() && month == 2 ? 29 : days( month ); }
intDate::leapYear( void ) const
{
if ( year % 400 == 0 || ( year % 4 == 0 && year % 100 != 0 ) )
return 1;
else
return 0;
}
voidDate::convert1( int ss )
{
int dayTotal = 0;
if ( ss < 1 || ss > 366 )
ss = 1;
setMonth( 1 );
for ( int m = 1; m < 13 && ( dayTotal + daysOfMonth() ) < ss; ++m ) {
dayTotal += daysOfMonth();
setMonth( m + 1 );
}
setDay( ss - dayTotal );
setMonth( m );
}
intDate::change1( void ) const
{
int ss = 0;
for ( int m = 1; m < month; ++m )
ss += days( m );
ss += day;
return ss;
}
voidDate::change2( constchar * const mPtr )
{
int flag = 0;
for ( int subscript = 0; subscript < 12; ++subscript )
if (!strcmp( mPtr, monthList( subscript ) ) ) {
setMonth( subscript + 1 );
flag = 1;
break;
}
if ( !flag )
setMonth( 1 );
}
constchar *Date::monthList( int mm ) const
{
char *months[] = { \"January\", \"February\", \"March\", \"April\", \"May\",
\"June\", \"July\", \"August\", \"September\", \"October\",
\"November\", \"December\" };
return months[ mm ];
}
intDate::days( int m ) const
{
constint monthDays[] = { 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 };
return monthDays[ m - 1 ];
}
//main.cpp
#include \"Date.h\"
#include
using sts::cout
using std::endl
#include
#include
#include \"p7_07..
Python and PostgreSQL: Let's Work Together! | PyConFr 2018 | Dimitri Fontaine
Python is often used to maintain application backends. When the backend should implement user oriented workflows, it may rely on a RDBMS component to take care of the system's integrity.
PostgreSQL is the world's most advanced open source relational database, and is very good at taking care of your system's integrity. PostgreSQL also comes with a ton of data processing power, and in many cases a simple enough SQL statement may replace hundreds of lines of code written in Python.
In this talk, we learn advanced SQL techniques and how to reason about which part of the backend code should be done in the database, and which parf of the backend code is so easier to write as a SQL query.
Date class that represents a date consisting of a year, month, and a.pdf
Date class that represents a date consisting of a year, month, and a day
import java.util.TimeZone;
public class Date {
// constants
private static final int JANUARY = 1;
private static final int FEBRUARY = 2;
private static final int DECEMBER = 12;
private static final int DAYS_PER_WEEK = 7;
private static final int DAYS_PER_YEAR = 365;
private static final int DAYS_PER_LEAP_YEAR = 366;
private static final String[] DAY_NAMES = {\"SUNDAY\", \"MONDAY\", \"TUESDAY\",
\"WEDNESDAY\", \"THURSDAY\", \"FRIDAY\", \"SATRDAY\"};
private static final int[] DAYS_PER_MONTH = { -1,
// 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31,
};
// fields
private int year;
private int month;
private int day;
/**
Constructs a new object representing the given year, month, and day.
@throws IllegalArgumentException if month is not between 1 and 12, or if day is not between 1
and the number of days in that month.
*/
public Date(int year, int month, int day) {
this.year = year;
this.month = month;
this.day = day;
if (month < 1 || month > 12 || day < 1 || day > getDaysInMonth()) {
throw new IllegalArgumentException(\"Invalid day or month: \" + toString());
}
}
/** Constructs a new object representing today\'s date. */
public Date() {
this(2006, july
, 22);
int daysSinceEpoch = (int) ((System.currentTimeMillis() +
TimeZone.getDefault().getRawOffset()) / 1000 / 60 / 60 / 24);
for (int i = 0; i < daysSinceEpoch; i++) {
nextDay();
}
}
/**
Adds the given number of days to this Date.
@throws IllegalArgumentException if days < 0.
*/
public void addDays(int days) {
while (days > getDaysInYear()) {
days -= getDaysInYear();
year++;
}
for (int i = 1; i <= days; i++) {
nextDay();
}
}
/**
Returns whether o refers to a Date object representing the same
year, month, and day as this one.
*/
public boolean equals(Object o) {
Date other = (Date) o;
return day == other.day &&
month == other.month &&
year == other.year;
}
/**
Returns the day represented by this Date object.
@return the day, between 1 and the number of days in this Date\'s month
(which varies from 28 to 31)
*/
public int getDay() {
return day;
}
/** Returns the number of days in the month represented by the current Date. */
public int getDaysInMonth() {
int result = DAYS_PER_MONTH[month];
if (month == FEBRUARY && isLeapYear()) {
result++;
}
return result;
}
public int getDaysInYear() {
if (isLeapYear()) {
return 366;
} else {
return 365;
}
}
/**
Returns a string such as \"Monday\" representing what day of the week
this Date fell on.
@return the day as either \"Sunday\", \"Monday\", \"Tuesday, \"Wednesday\", \"Thursday\",
\"Friday\", or \"Satrday\"
*/
public String getDayOfWeek() {
int index = 1;
Date temp = new Date(1753, JANUARY, 1);
while (temp != this) {
temp.nextDay();
index = (index + 1) % DAYS_PER_WEEK;
}
return DAY_NAMES[index];
}
/**
Returns the month represented by this Date object.
@return the month, between 1 and 12
*/
public int getMonth() {
return month;
}
/**
Returns the yea.
publicclass Date {privatestatic String DATE_SEPARATOR = ;pr.pdf
publicclass Date {
privatestatic String DATE_SEPARATOR = \"/\";
privatestaticintDAYS_PER_WEEK = 7;
//Attributes
privateint day;
privateint month;
privateint year;
/**
* Default Constructor
* Instantiates an object of type Date to 1/1/2000
*/
public Date() {
this.day = 1;
this.month = 1;
this.year = 2000;
}
/**
* Constructs a new date object to represent the given date.
* @param day
* @param month
* @param year
*/
public Date(int day, int month, int year) {
if(isValid(day, month, year)) {
this.day = day;
this.month = month;
this.year = year;
} else
System.out.println(\"Invalid Date.\");
}
/**
* Returns the day value of this date for example, for the date 2006/07/22, returns 22.
* @return
*/
publicint getDay() {
return day;
}
/**
* Returns the month value of this date ,for example, for the date 2006/07/22, returns 7.
* @return
*/
publicint getMonth() {
return month;
}
/**
* Returns the year value of this date, for example , the date 2006/07/22, returns 2006.
* @return
*/
publicint getYear() {
return year;
}
/**
* @param day the day to set
*/
publicvoid setDay(int day) {
this.day = day;
}
/**
* @param month the month to set
*/
publicvoid setMonth(int month) {
this.month = month;
}
/**
* @param year the year to set
*/
publicvoid setYear(int year) {
this.year = year;
}
/**
* Returns true if the year of this date is a leap year.
* A leap year occurs every 4 years , except for multiples of 100 that are not multiples of 400.
* For example, 1956,1844,1600,and 2000 are leap years, but 1983,2002,1700,and 1900 are not.
* @return
*/
publicboolean isLeapYear(){
if(((this.year % 400) == 0) || (((this.year % 4) == 0) && ((this.year % 100) != 0)))
returntrue;
else
returnfalse;
}
/**
* Checks if the date is valid
* @param day
* @param month
* @param year
* @return
*/
publicboolean isValid(int day, int month, int year) {
if((month < 1) || (12 < month))
returnfalse;
else {
if(((month == 1) || (month == 3) || (month == 5) || (month == 7) ||
(month == 8) || (month == 10) || (month == 12)) && ((day < 1) || (31 < day)))
returnfalse;
elseif(((month == 4) || (month == 6) || (month == 9) || (month == 11)) && ((day < 1) || (30 <
day)))
returnfalse;
elseif(month == 2) {
if(isLeapYear() && ((day < 1) || (29 < day)))
returnfalse;
elseif((day < 1) || (28 < day))
returnfalse;
}
}
returntrue;
}
/**
* Returns the maximum number of days in a month
* @return
*/
publicint maxMonthDays() {
if(this.month == 2) {
if(isLeapYear())
return 29;
else
return 28;
} elseif((this.month == 1) || (this.month == 3) || (this.month == 5) || (this.month == 7) ||
(this.month == 8) || (this.month == 10) || (this.month == 12))
return 31;
else
return 30;
}
/**
* Checks if this dat is same as other date
* @param other
* @return
*/
publicboolean isEqual(Date other) {
if((this.day == other.day) && (this.month == other.month) && (this.year == other.year))
returntrue;
else
returnfalse;
}
/**
* Moves this Date object forward in time by the given number of days .
* @param days
*/
publicvoid addDays(int days).
From the proposal to ECMAScript – Step by Step
Do ever wonder how ECMAScript evolves and it's available almost everywhere? We
gonna do a walkthrough of how TC39 (the technical committee responsible for the
ECMAScript) works, how to participate, contribute, and how an idea can make it
through to the language, all step-by-step. Together with a sneak pick and
debate about some recent and interesting proposals in different stages.
(c) Fest.js Porto 2023
April 20, 2023
Porto, Portugal
https://fest.dev/events/js/porto-2023/
struct procedure { Date dateOfProcedure; int procedureID.pdf
struct procedure
{
Date dateOfProcedure;
int procedureID;
int procedureProviderID;
};
Data.h
#include
#include
class Date
{
friend ostream& operator<<( ostream &, const Date & );
// allows easy output to a ostream
public:
Date( int m = 1, int d = 1, int y = 1900 ); // constructor, note the default values
void setDate( int, int, int ); // set the date const
Date &operator+=( int ); // add days, modify object
bool leapYear( int) const; // is this a leap year?
bool endOfMonth( int ) const; // is this end of month?
int getMonth ( ) const;
int getDay ( ) const;
int getYear ( ) const;
string getMonthString( ) const;
private:
int month;
int day;
int year;
static const int days[]; // array of days per month static const string monthName[]; // array of
month names
void helpIncrement();
// utility function
}; #endif Data.cpp // Member function definitions for Date
class in separate
date.cpp
#include
#include \"date.h\"
#include // Initialize static members at file scope; // one class-wide copy.
const int Date::days[] = { 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 }; const string
Date::monthName[] = { \"January\", \"February\", \"March\", \"April\", \"May\", \"June\",
\"July\", \"August\", \"September\", \"October\", \"November\", \"December\" };
// Date constructor
Date::Date(int m, int d, int y)
{
setDate(m, d, y); } // Set the date
void Date::setDate(int mm, int dd, int yy)
{
month = (mm >= 1 && mm <= 12) ? mm : 1; year = (yy >=1900&& yy <= 2100) ? yy
: 1900; // test for a leap year
if (month == 2 && leapYear(year)) day = (dd >= 1 && dd <= 29) ? dd : 1;
else
day =(dd >= 1 && dd <= days[month]) ? dd : 1;
} // Add a specific number of days to a date
const Date &Date::operator+=(int additionalDays)
{
for (int i = 0; i < additionalDays; i++) helpIncrement();
return *this; // enables cascading
} // If the year is leap year, return true; //
otherwise, return false
bool Date::leapYear(int testYear)
const
{
if (testYear % 400 == 0 || (testYear %100 != 0 && testYear % 4 == 0)) return true; // a
leap year
else
return false; // not a leap year
} // Determine if the day is the end of the month
bool Date::endOfMonth(int testDay)
const
{
if (month == 2 && leapYear(year))
return (testDay == 29); // last day of Feb. in leap year
else
return (testDay == days[month]);
} // Function to help increment the date
void Date::helpIncrement()
{
if (!endOfMonth(day))
{
// date is not at the end of the month
day++;
}
else if (month < 12)
{
// date is at the end of the month, but month < 12 day = 1; ++month;
}
else
{
// end of month and year: last day of the year
day = 1;
month = 1;
++year;
} }
// Overloaded output operator
ostream &operator<<(ostream& output, const Date &d)
{
output << d.monthName[d.month] << \' \' << d.day << \", \"<< d.year;
return output; // enables cascading
}
int Date::getMonth() const //
{
Public:
int month[20];
If(month[]!=”/0”)
Cout<<”Enter Month :-<>month;
Elseif (month[]<12)
{
Cout<>month[];
else
return month;
}
int Date::getDay() const //
{
Pu.
Java 8 Date and Time API
This presentation provides an overview of using the Java SE 8 Date & Time API. It covers how to:
1. Create and manage date-based and time-based events including a combination of date and time into a single object using LocalDate, LocalTime, LocalDateTime, Instant, Period, and Duration
2. Work with dates and times across timezones and manage changes resulting from daylight savings including format date and times values
3. Define and create and manage date-based and time-based events using Instant, Period, Duration, and TemporalUnit
딥러닝, 야 너도 할 수 있어(feat. PyTorch)
※다운로드하시면 더 선명한 자료를 보실 수 있습니다.
딥러닝을 처음 시작하시는 분, 특히 개발자에게 맞춰 필요한 내용을 준비하였습니다.
딥러닝의 수학적인 이해보다는 딥러닝을 활용하는 방법을 공유하고자 합니다.
목차
1. 딥러닝 권장 개발 환경
2. 코드 작성을 위한 약간의 이론
3. 가장 빠르게 그럴싸한 모델 만들기
대상
딥러닝에 관심이 있거나 딥러닝을 시작하려는 모든 개발자
※발표 영상은 제공하지 않습니다.
Similar to Need to make a Java program which calculates the number of days betw.pdf (16)
Date.h#ifndef DateFormat#define DateFormat#includetime.h.pdf
Date.h#ifndef DateFormat#define DateFormat#includetime.h.pdf
Python and PostgreSQL: Let's Work Together! | PyConFr 2018 | Dimitri Fontaine
Python and PostgreSQL: Let's Work Together! | PyConFr 2018 | Dimitri Fontaine
Date class that represents a date consisting of a year, month, and a.pdf
Date class that represents a date consisting of a year, month, and a.pdf
publicclass Date {privatestatic String DATE_SEPARATOR = ;pr.pdf
publicclass Date {privatestatic String DATE_SEPARATOR = ;pr.pdf
struct procedure { Date dateOfProcedure; int procedureID.pdf
struct procedure { Date dateOfProcedure; int procedureID.pdf
Date and Timestamp Types In Snowflake (By Faysal Shaarani)
Date and Timestamp Types In Snowflake (By Faysal Shaarani)
More from anjandavid
I am exploring the basic components of advanced broadband networks..pdf
I am exploring the basic components of advanced broadband networks.
Describe the three layered architecture:
Briefly, explain the difference between link state and distance vector routing
Describe how the sliding window protocol works
Explain the CIDR concept and its major benefit
Solution
The three layered architecture is a client server architecture.This architecture is used for
functional perspective logic,data access, data storage and user interface.
Three layer architecture is used for software design.
Three layers are as follows,
SLIDING WINDOW PROTOCOL:
It is one of the packet based data transmission protocol.Packets are send and its too hard to send
duplicate packets and to identify the missing packets.
In sliding window protocol we set a limit for number of packets to be transmitted at a time.
CIDR concept:
Classless Inter-domain routing is an alternate to the traditional subnetting. It is a way to allow the
more flexible way to allocate Internet protocol.
Benefit:
CIDR provides the better way to address since it prevents the wastage of addressing space since
it is a flexible way of addressing.Link state vectordistance vector routingLink state vector makes
all the nodes to know about the route.Distance vector doubting only the neighbouring node is
knownLink state information is flooded throughout the link state domain.Distance vector
routing information is not knownl.
How many different instances of the ls command are installed on .pdf
How many different instances of the \"ls\" command are installed on your linux machine, and
what is/are the location(s) path i.e. what is the complete path for each?
Solution
ls is one of the basic command in Linux. This command lists or displays files or directories.
ls command runs with different parameters or instances like
ls -l means long listing format
ls -a displays all files with hidden files
ls -lh show size of files
To find where a particular command is stored use whereis command e:g
whereis ls outputs Output:
ls: /bin/ls /usr/share/man/man1p/ls.1p.gz /usr/share/man/man1/ls.1.gz.
How are the Allen Bradley SLC 500 program files organizedSolutio.pdf
How are the Allen Bradley SLC 500 program files organized
Solution
Allen Bradley SLC 500 program files include System Functions, Reserved, User Program and
Subroutine Programs.
System Functions are of file 0, Reserved are of file 1, User Program are of file 2 and Subroutine
Programs are of file 3-255.
System Functions file contains various system related information and user programmed
information. User programmed information are processor type, I/O configuration, processor file
name and password.
Reserved file is reserved by the processor and user cannot access these files.
User program files contain user programmed instructions. These instructions define how the
controller should operate.
Subroutine programs are created by users and are activated based on subroutine instructions
which are in the main ladder program file..
Hi, I need help please, this is about KaseyaWich o the following .pdf
Hi, I need help please, this is about Kaseya:
Wich o the following pages has the ability to configure and apply the ATSE code:
____Monitor > Agent Monitoring > Event Log Alerts
____Monitor > Agent Monitoring > SNMP Traps Alert
____Monitor > Agent Monitoring > Alerts
____Monitor > Agent Monitoring > Monitor Log
____Monitor > SNMP Monitoring > Assign SNMP
____Monitor > Agent Monitoring > SNMP Log
Solution
The answer to this is the first one
Monitor > Agent Monitoring > Event Log Alerts has the ability to configure and apply the ATSE
code.
How did WWI and its aftermath provide African Americans with opportu.pdf
How did WWI and its aftermath provide African Americans with opportunities? Discuss W.E.B.
DuBois and the rise of the NAACP. Discuss Marcus Garvey and the rise of Garveyism. 5-7
sentences for answer
Solution
For most of the African American soldiers, serving in World War I had provided them a broaden
social , political, geographical and cultural acceptance in the society, As they received the
opportunity to travel the different locations of the country, and more than 200000 African-
American soldiers served in other countries to various part of the globe, had really improved the
life experience to these soldiers.
NAACP was founded on 12th Feb 1909 and it is considered to be the oldest, largest and most
widely recognized grassroots-based civil right organization in the United States. NAACP was
mainly constituted in response to the increasing horrific incidents of lynching and the race riots
which resulted in the year 1908 in Springfield. Following the racial violence against the Afro-
Americans, Mary White Ovington, and Oswald Garrison Villard, both the descendants of
abolitionists, William English Walling and Dr. Henry Moscowitz issued a call for a meeting to
discuss racial justice. Almost 60 individuals out of them 7 were Afro-American including W. E.
B. Du Bois, Ida B. Wells-Barnett, and Mary Church Terrell signed a declaration which was
released on the occasion of the centennial of Lincoln’s birth..
For problems 3 and 4, consider the following functions that implemen.pdf
For problems 3 and 4, consider the following functions that implement the dequeue operation for
a priority queue that is implemented with a heap.
int[] pQueue;
int length;
int dequeue()
{
int node = 1;
int value = pQueue[--length];
int maxValue = pQueue[node];
int location = sift(node * 2, value);
pQueue[location] = value;
return maxValue;
}
int sift(int node, int value)
{
if (node <= length)
{
if (node < length && pQueue[node] < pQueue[node + 1]) node++;
if (value < pQueue[node])
{
pQueue[node / 2] = pQueue[node];
return sift(node * 2, value);
}
}
return node / 2;
}
3. Write the recurrence equation and initial conditions that expresses the execution time cost for
the sift function in the above algorithm. Draw the recursion tree assuming that n = 8. Assume
that n represents the number of nodes in the subtree at each step of the sifting operation.
4. Determine the critical exponent for the recurrence equation in problem 3. Apply the Little
Master Theorem to solve that equation specifically stating which part of the theorem is applied to
arrive at the solution.
Solution
using System.Collections;
using System.Collections.Generic;
namespace PriorityQueueDemo
{
///
/// Priority queue based on binary heap in program,
/// Elements with minimum priority dequeued first in program
///
/// Type of priorities
/// Type of values
public class PriorityQueues : ICollection>
{
private List> _baseHeap;
private IComparer _comparers;
#region Constructor
///
/// Initializes a new instance of priority queue with default initial capacity and default
priority comparer in program
///
public PriorityQueues()
: this(Comparers.Default)
{
}
///
/// Initializes a new instance of priority queue with specified initial capacity and default
priority comparer in program
///
/// initial capacity
public PriorityQueues(int capacity)
: this(capacity, Comparers.Default)
{
}
///
/// Initializes a new instance of priority queue with specified initial capacity and specified
priority comparer in program
///
/// initial capacity
/// priority comparer
public PriorityQueues(int capacity, IComparers comparer)
{
if (comparer == null)
throw new ArgumentNullException();
_baseHeaps = new List>(capacity);
_comparers= comparer;
}
///
/// Initializes a new instance of priority queue with default initial capacity and specified
priority comparer in program
///
/// priority comparer
public PriorityQueues(IComparer comparers)
{
if (comparers == null)
throw new ArgumentNullException();
_baseHeap = new List>();
_comparer = comparers;
}
///
/// Initializes a new instance of priority queue with specified data and default priority
comparer in program
///
/// data to be inserted into priority queue in program
public PriorityQueues(IEnumerable> data)
: this(data, Comparers.Default)
{
}
///
/// Initializes a new instance of priority queue with specified data and specified priority
comparer in program
///
/// data to be inserted into priority queue
/// priority comparer in program
public PriorityQueue(IEnumer.
Give an example of a system. What are the Components, Attributes,.pdf
Give an example of a system. What are the Components, Attributes, and Relationship of the
system?
Give an example of a system. What are the Components, Attributes, and Relationship of the
system?
Solution
An arrangement of point by point techniques, procedures and schedules made to do a particular
action, play out an obligation, or take care of an issue.
And no more fundamental level, frameworks are partitioned into two classifications:
(1) Closed systems
theoretical systems that don\'t associate with the environment and are not impacted by its
surroundings.Only the components inside the system are noteworthy.
Example: a sealed jug - nothing enters or leaves the jug, however whatever is inside can interact.
(2) Open systems
real-world systems whose limits permit exchanges of energy, material and data with the external
environment or system in which they exist.
Examples: an organization - regardless of whether there are separate divisions in a single
association, the specialists share information and connect with each other once a day. Diverse
systems procedures, (for example, dynamics and systems thinking) characterize systems
differently.
Component # 1. System:
Each piece of the greater entire is a sub-system. These parts make the entire association. Each
sub-system is a piece of the bigger system which, thusly, is sub system of a still bigger
framework. For instance, division is a sub-arrangement of the association which is a sub-
arrangement of the business, which, additionally is a sub-arrangement of the national economy
which is a sub-arrangement of the world economy.
Part # 2. Synergy:
The aggregate of parts is not as much as that of the entirety. On the off chance that each division
works autonomously, add up to yield would be not as much as what is created by them together.
Synergy defines relationships among all parts of the association e.g., if production and marketing
departments have autonomous sub-departments to give them fund or work, it will be less
proficient than a system where both (production and marketing department) are associated with
one back or staff branch of the association all in all.
In this manner, system approach does not simply discussion of the parts and their sub-parts yet
additionally their course of action. All the parts and sub-parts are arranged in such a manner that
output of the whole (accomplished through coordination among sub systems) is more than the
aggregate of the yield of individual parts.
Component # 3. Open and closed system:
System can be open or closed. Open system effectively cooperates with environment. It gets
contributions as raw material, work, capital, administrative and specialized skill from the earth
and offers yields (goods and services) to the society. The Administration (framing the policies
and imposing taxes) and contenders additionally connect with business associations.
A closed system has no or very little interaction with the environment. For all intents and
purposes, all as.
Discuss the attribution process and attribution errorsSolution.pdf
Discuss the attribution process and attribution errors
Solution
Attribution is the process of inferring the causes of events or behaviors, this is something we do
many times during the day without us noticing. For example, if you fail to enter into the college
you wanted you might blame the committee that accepts or declines candidates that they didn’t
put attention to your grades or your extra work, or when you have a bad score on a test you
might blame the teacher for not explaining correctly.
Attribution errors happen usually when it comes to other people, we tend to attribute causes to
characteristics of people ignoring external variables, and for example when we blame someone
else for a problem over which they didn’t have control.
References: Jaspars, J., Fincham, F.D., & Hewstone, M. (1983). Attribution Theory and
Research: Conceptual Developmental and Social Dimensions. Academic Press..
E, an individual, received $40,000 of non-eligible dividends from Ca.pdf
E, an individual, received $40,000 of non-eligible dividends from Canadian corporations in the
current year. What amount is included in E’s net income for tax purposes for the current year?
Solution
When an individual receives non-eligible dividend from a Canadian Corporation, it is grossed up
to 116%. Thus, the amount included in E\'s net income for tax purposes for current year would
be =116%*$40000= $46400.
Describe one (1) example in which laws granting freedom of the press.pdf
Describe one (1) example in which laws granting freedom of the press resulted in the exposure of
political or business corruption.
Solution
hi,
below is the answer to ypur question
Freedom of the press or freedom of the media is the freedom of communication and expression
through mediums including various electronic media and published materials. While such
freedom mostly implies the absence of interference from an overreaching state, its preservation
may be sought through constitutional or other legal protections.
The freedom of the press has been found to be a potentally good mechanism as external control
oncorruption, particularly because it seems to be effective against both extortive and collusive
corruption.
As it has been explained before, the media and in particular the freedom of the press, have a role
inreducing corruption. The tangible way they act can take a variety of forms. Reporting, for
example,may induce the public bodies to start formal investigation into claims of corruption. In
addition,new sources disclose what public anti-corruption bodies have found, reinforcing in this
way thelegitimacy of these bodies. So, when journalism shows the faults and the corruption
involved in thedifferent bodies of the state, corruption starts to be investigated. Moreover, if the
public pressurethat follows leads to a reform of these bodies, the long-run effectiveness of the
media to play as acounterbalance against corruption is reinforced.Additionally, intangible effects
may also result. These can be defined as those controls oncorruption, which are generated from a
broader social environment of increased political pluralism,sharpened public discussion and an
expanded sense of accountability among political bodies andinstitutions, which derive from
independent media.
The Telegraph exposes a massive culture of expenses abuse by MPs
Giving politicians any sort of control over the media is a frightening prospect especially when
you consider its role in exposing their mistakes, abuses of power and, in some cases, corruption.
A look back over the last 60 years yields any number of examples of politicians who have
misbehaved and who have been brought to book by the dogged and noble efforts of reporters and
their editors.
In 1963, Secretary of State for War John Profumo had an affair with Christine Keeler, the
reputed mistress of an alleged Soviet spy.
He lied to the House of Commons about it and when the newspapers revealed all, he resigned.
MP Neil Hamilton was investigated by the Guardian and in 1994 an article claimed he had
received cash in a brown envelope from Mohamed Al-Fayed, the owner of Harrods, to ask
questions on behalf of Al-Fayed in the Commons. He claimed his innocence but was forced to
resign.
Neil Hamilton MP was caught up in the \'cash for questions\' scandal
Former Tory cabinet minister Jonathan Aitken was investigated by the Guardian and World In
Action for corrupt deals with leading Saudi Arabians.
He said he would fight the .
Briefly discuss 3–5 key trends in the modern health care operation.pdf
Briefly discuss 3–5 key trends in the modern health care operational environment that may
have an impact on the effective leadership and management of a hospital or health care
organization.
Solution
Trend #1 : Usage of HIT stands for Health Information Technology platforms. Electronic Health
Records or EHR\'s are digitized patient records & medical history that can be updated on a real
time basis and reviewed. These are a good example of the integration of information technology
with health care.
EHR Interoperability
Wireless devices are known to interfere with the smooth operation of health care systems.
Wireless devices are known to interfere with pacemaker, MRI\'s . Xrays , CAT scans etc.
However with HIT and EHR\'s introducing interoperability into their systems. its important to
transition health care technology into a wireless world. In healthcare, interoperability is the
ability of different information technology systems and software applications to communicate,
exchange data, and use the information that has been exchanged.
It\'s impact on the effective leadership and management of a hospital or health care organization
Health care organisations will need to address the following changes in order to keep up with the
trend in the health care industry
EHR downtime : Since EHR\'s are basically hardware and software, like all systems they may
face outages, system down times and hacking (as seen during the recent attack by the ransom
ware virus on UK hospitals where they couldn\'t access patient records).
These are complex systems and require strong enterprise grade IT support. With EHR\'s
becoming increasingly complex with a vast number of variables, there have been a significant
increase in outages and downtime resulting in hospitals not having access to patient records, real
time diagnosis tools etc.
Miscommunication of Data between between different components of EHR : As i mentioned
before EHR comprises of both hardware and software. Software systems are often linked to
hardware systems that interface with patients such as X ray, MRI, life support systems etc.
Sometimes, the manufacturer of the hardware and software may not be the same and this could
cause a certain degree of discrepancies. In other cases even when the manufacturer of the
hardware and software components were the same , there was a fair degree of miscommunication
between various components.
Alarm Fatigue
Alarm fatigue is a scenario where health care workers may normalise to alerts or alarms
triggered by EHR\'s. This desensitisation is often caused due to either false alarms or a high
frequency of low priority alarms being triggered by EHRs. When EHR\'s trigger alarms, health
care workers may brush it off as a low priority alarm when it might be quite the opposite. This
removes the sense of urgency causing real emergencies to be neglected. This could lead to
serious deterioration in patient\'s health and in some cases death.
While EHR\'s can do many things and capt.
A student obtained the following data Mass of water in calorimeter 3.pdf
A student obtained the following data: Mass of water in calorimeter 31.633 g Mass of metal
Temperature of metal Initial T of water in calorimeter Highest T obtained 75.224 g 98.6 C 23.2
C 47.4 C what is T of the metal in C-round off to correct number of significant digits Answer
Solution
Change in temperature of metal ,dT= initial temperature of metal- highest temperature of the
system
dT= 98.6 - 47.4 = 51.2 oC.
All answers must be in your own words.What is importance of the Wa.pdf
All answers must be in your own words.
What is importance of the Walnut Street Jail?
Solution
Walnut street jail in Philadelphia, Pennsylvania was first opened in 1775, which brought the
concepts of prison architecture and prison administration in the modern American history. The
building was designed in such a way that it could hold groups of prisoners in a room. The cells &
corridors were connected in such a way that prisoners could not communicate with each other.
Thus it was the first prison especially designed & managed for prison activities. Initially it served
as a holding facility for accused till they were convicted. After conviction offenders faced harsh
punishments including the stocks, whippings, brandings, and beatings.
But during a period of reforms humanitarians in Philadelphia & Quakers came to realize that
cruelty & harsh punishment are of little help to deter from crime. They successfully advocated
for combating crime through solitary confinement, isolation & labor. Quakers specially believed
in penitence & self-examination to achieve salvation. Therefore a new era of isolating prisoners
started in walnut street jail consider to be first of such an experiment in the US prison system.
All prisoners were sentenced to labor where they were given tasks of weaving, nail making
shoemaking & carding wool.
Due to the efforts of Quakers a part of walnut street jail was converted into the penitentiary
house for serious offenders which is consider first of such a facility in the United States. This
penitentiary house was built due to the humanitarian efforts of Quakers who thought cruelty &
harsh punishment are of little help to deter from crime. Thus such prisoners were sentenced to
solitary confinement in individual cells. This system later became famous as ‘Pennsylvanian
system’.
Another importance of Wall nut street jail is that first manned balloon flight was organized in the
prison yard of this jail..
Can someone please fix my code for a hashtable frequencey counter I.pdf
Can someone please fix my code for a hashtable frequencey counter: Implement a Java program
that counts word frequencies in a text file. Use a hashtable to store the data the words are the
keys, and their frequencies are the values. The output of the program is the complete list of all
the words and their frequencies in descending order of frequencies when two words have the
same frequency, output them by alphabetical order. Each line of output consists of a word, a tab,
and the frequency. A sample input is \"The Tragedy of Hamlet, Prince of Denmark\", in the file
Hamlet When processing the text, you should keep just the words, and discard all punctuation
marks, digits, and so on. You also need to turn all upper case letters to lower cases.
Solution
Your code is fine except for apostrophe was also being replacced by space because you were
replaceing everything other than alphabets with space. I put apostrophe in exclusion list and then
processed that seperately. Check the code below
import java.io.File;
import java.io.FileNotFoundException;
import java.util.*;
import java.util.ArrayList;
/**
* Created by abdul on 2/6/2017.
*/
public class WordCounter {
public static String finalWord(String str) {
String processedWord = str.replaceAll(\"[^a-zA-Z\']\", \" \").toLowerCase();
processedWord = processedWord.replaceAll(\"[\']\", \"\"); // replace \' with empty string
return processedWord;
}
private Hashtable hash = new Hashtable();
public void fileInput() throws FileNotFoundException {
File text = new File(\"E:/Shake.txt\");
String word;
int count = 1;
Scanner in = new Scanner(text);
ArrayList list = new ArrayList<>();
while (in.hasNextLine()) {
String line = in.nextLine();
line = finalWord(line);
StringTokenizer st = new StringTokenizer(line);
while (st.hasMoreTokens()) {
word = st.nextToken();
if (hash.containsKey(word)) {
hash.put(word, hash.get(word) + 1);
//int count = (Integer) hash.get(word);
//hash.put(word, count + 1);
count++;
} else {
hash.put(word, 1);
count = 1;
}
}
}
Map map = new TreeMap(hash);
//System.out.println(map);
Set set = map.entrySet();
Iterator i = set.iterator();
while(i.hasNext()){
Map.Entry me = (Map.Entry)i.next();
System.out.print(me.getKey() + \" \");
System.out.println(me.getValue());
}
}
public static void main (String[]args) throws FileNotFoundException {
WordCounter abc = new WordCounter();
abc.fileInput();
}
}.
A polycationic mRNA contains two or more promoter sequences. True Fal.pdf
A polycationic mRNA contains two or more promoter sequences. True False
Solution
Answer-False
Description-operon is a functioning unit of genomic DNA containing a cluster of genes under the
control of a single promoter.The genes are transcribed together into an mRNA strand and
translated together in the cytoplasm, to create monocistronic mRNAs that are translated
separately.
several strands of mRNA that each encode a single gene productoperon contains one or more
structural genes which are generally transcribed into one polycistronic mRNA-a single mRNA
molecule that codes for more than one protein. the structural genes lies a promoter sequence
which provides a site for RNA polymerase to bind and initiate transcription.Polycistronic mRNA
is a mRNA that encodes several proteins and is characteristic of many bacterial and chloroplast
mRNAs. Polycistronic mRNAs consist of a leader sequence which precedes the first gene. The
gene is followed by an intercistronic region and then another gene. A trailer sequence follows the
last gene in the mRNA. Examples of a polycistronic transcripts are found in the chloroplast.
Why are electrons shared in molecular compoundsWhy are electron.pdf
Why are electrons shared in molecular compounds?
Why are electrons shared in molecular compounds?
Solution
Electrons are shared because the atoms and elements don’t exist or aren’t stable in their crude
form. They need to complete their o outermost shell to become stable. They need to complete the
octet rule.
Any substance in nature prefers a low energy form and is thermodynamically stable at that lower
energy, hence atoms need to follow the octet rule. In doing so they form molecular compounds,
when they share their electrons with each other, hence completing and satisfying each other.
Sharing leads to formation of a Covalent bond between the 2 atoms.
Sharing of atoms mostly happens in non metals , metalloids and other electronegative atoms..
Which of these isare true of UDPa. It provides reliability, flow-c.pdf
Which of these is/are true of UDP?a. It provides reliability, flow-control, and error-recovery
functions.
Solution
Answer:
b. It functions basically like a transport-layer equivalent of IP.
c. Its simplicity means its headers contain fewer bytes and consume less network overhead than
TCP
UDP is a cnnection less transport layer protocol that provides no reliability, flow-control, and
error-recovery functions. Its simplicity means its headers contain fewer bytes and consume less
network overhead than TCP. It functions basically like a transport-layer equivalent of IP..
Which liquid would BaCl Which liquid would BaCl 4. Which liquid.pdf
Which liquid would BaCl
Which liquid would BaCl 4. Which liquid would BaCl2 more likely be soluble? CoH6, CHCl3,
CH3CH2OH
Solution
BaCl2 is an ionic solid. Ionic compounds tend to dissolve in polar solvents. Amont the given set
of solvents ethanol is the pmost polar one. So, BaCl2 is more likely to dissolve in CH3CH2OH..
What is employee involvement What are some of the benefits of invol.pdf
What is employee involvement? What are some of the benefits of involving employees in the
organization?
Solution
Employee involvement is defined as employee\'s commitment and participation towards
achieving organizational goals and objectives.
Below are some of the ways by which employee involvement can be increased:
Below are some of the benefits of improved employee involvement:.
What are the pros and cons of technological leader versus technologi.pdf
What are the pros and cons of technological leader versus technological follower as a functional
strategy? Do you believe a follower can become a leader? Why or why not? Explain your
answer
Solution
Technological leaders are the people who predict the changes in techology in early, who fosters
technological innovation, who understands the technology life cycle clearly, and use all these for
benefit of the business. These people takes initiation for commercialization of these
technological advances, they link all these with business, manage research & development in
technology area and understand technological revolutions.
Some of the times, due to early predictions, these people may not able to concentrate on the
running technologies and may focus on next trends of technologies. It does not help them to
achieve the present and current goals. This kind of attitude will be become a burden to the
organization in many ways. If every one be like this and do like this, the regular activities of the
business entity may not go further.
The followers may not always go with the leaders in this area because of many limitations.
Leaders have a freedome to select their activities and to plan their actions, where the followers
have some duties and obligations to deliver and perform.
In my openion, a follower never becomes a leader. Leadership is a seperate category of peole and
they have some raw heridity from their birth only. Leadership is not a skill where can be learned
from others or by experience. It is a characteristic which should be inculcate from their
innersense. These people have some unique, special, and different ideology than others, and
make them to become leaders.
Some of the people says that leaders can be made but not born. It is completely wrong, a simple
example: Martin Luther King is the first person who fought against the racism in US. But before
him and after him many lakhs of people born in the community, why not all these people raise
their voice in a strategic and purposeful way? Similarly Mahatma Gandhi, Nelson Mandela,
Mother Theressa, all these people born with some special characteristics, when the time comes
they show the abilities. But no others done this. What i mean to say is, leaders born with some
raw heridity related to leadership, when the time comes they only react and stand first, the
remaining, who trust the leaders, become followers. And a follower never becomes a leader..
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- 1. Need to make a Java program which calculates the number of days between two dates. I am NOT able to use anything too advanced yet such as import.java.util.Calender, import.java.util.Date; or import.java.util.GregorianCalender; Here are a couple of examples of what is should look like when complete. I am having troubles getting my calculations to come out correctly. I’ve also written the code I have so far. EXAMPLE “Enter first date now: Enter Month: 1 Enter Day: 1 Enter Year: 1900 Enter the second date: Enter Month: 1 Enter Day: 1 Enter Year: 2400 The first date is: 1/1/1900 The other date is: 1/1/2400 The total number of days between the dates is 182621” AND “Enter first date now: Enter Month: 2 Enter Day: 23 Enter Year: 2016 Enter the second date: Enter Month: 11 Enter Day: 30 Enter Year: 1955 The first date is: 2/23/2016 The other date is: 11/30/1955 The total number of days between the dates is 22000” Here is what I have so far: Import java.util.Scanner; Public class dates { Public static void main ( String args [] ) { int monthOne; int monthTwo;
- 2. int dayOne; int dayTwo; intyearOne; intyearTwo; Scanner get = new Scanner (System.in ); System.out.println ( “Enter first date now: “ ); do { System.out.print ( “Enter Month: “); monthOne = get.nextInt(); System.out.print ( “Enter Day: “ ); dayOne = get.nextInt(); System.out.print ( “Enter Year : “ ); yearOne = get.nextInt(); if (monthOne < 1 || monthOne > 12 ) { System.out.printf ( “%d is not a valid month ”, monthOne ); } If ( (!(yearOne % 4 == 0 && yearOne & 100 !=0 || yearOne % 400 == 0) && (monthOne == 2 && dayOne >= 29)) || ((yearOne % 4 == 0 && yearOne % 100 != 0 || yearOne % 400 == 0) && (monthOne == 2 && dayOne >= 30))) { System.out.printf ( “%d is not a valid day. ”, dayOne ); } If ( (monthOne == 4 && dayOne > 30) || (monthOne ==6 && dayOne > 30) || (monthOne == 9 && dayOne > 30) || (monthOne == 11 && dayOne > 30) || (dayOne < 1 || dayOne > 31)) { System.out.printf ( “%d is not a valid day. ”, dayOne ); If ( yearOne < 1900 || yearOne > 2400 ) { System.out.printf ( “%d is not a valid year. ”, yearOne ); } } while (( (!(yearOne % 4 == 0 && yearOne % 100 !=0 || yearOne % 400 == 0) && (monthOne == 2 && dayOne >= 29)) || ((yearOne % 4 == 0 && yearOne % 100 != 0 || yearOne % 400 == 0) && (monthOne == 2 && dayOne >= 30 )) || (monthOne == 4 && dayOne > 30) || (monthOne == 6 && dayOne > 30) || (monthOne == 9 && dayOne > 30) || (monthOne == 11 && dayOne > 30) || (monthOne < 1 || monthOne > 12) || (dayOne < 1 || dayOne > 31) || (yearOne < 1900 || yearOne > 2400) )); System.out.printf ( “ Enter the second date: ” ); do { System.out.print ( “Enter Month: “); monthTwo = get.nextInt(); System.out.print ( “Enter Day: “ ); dayTwo = get.nextInt();
- 3. System.out.print ( “Enter Year : “ ); yearTwo = get.nextInt(); if (monthTwo < 1 || monthTwo > 12 ) { System.out.printf ( “%d is not a valid month ”, monthTwo ); } If ( (!(yearTwo % 4 == 0 && yearTwo & 100 !=0 || yearTwo % 400 == 0) && (monthTwo == 2 && dayTwo >= 29)) || ((yearTwo % 4 == 0 && yearTwo % 100 != 0 || yearTwo % 400 == 0) && (monthTwo == 2 && dayTwo >= 30))) { System.out.printf ( “%d is not a valid day. ”, dayTwo ); } If ( (monthTwo == 4 && dayTwo > 30) || (monthTwo ==6 && dayTwo > 30) || (monthTwo == 9 && dayTwo > 30) || (monthTwo == 11 && dayTwo > 30) || (dayTwo < 1 || dayTwo > 31)) { System.out.printf ( “%d is not a valid day. ”, dayTwo ); If ( yearTwo < 1900 || yearTwo > 2400 ) { System.out.printf ( “%d is not a valid year. ”, yearTwo ); } } while (( (!(yearTwo % 4 == 0 && yearTwo % 100 !=0 || yearTwo % 400 == 0) && (monthTwo == 2 && dayTwo >= 29)) || ((yearTwo % 4 == 0 && yearTwo % 100 != 0 || yearTwo % 400 == 0) && (monthTwo == 2 && dayTwo >= 30 )) || (monthTwo == 4 && dayTwo > 30) || (monthTwo == 6 && dayTwo > 30) || (monthTwo == 9 && dayTwo > 30) || (monthTwo == 11 && dayTwo > 30) || (monthTwo < 1 || monthTwo > 12) || (dayTwo < 1 || dayTwo > 31) || (yearTwo < 1900 || yearTwo > 2400) )); System.out.printf ( “ The first date is: %d/%d/%d ”, monthOne, dayOne, yearOne ); System.out.printf ( “ The other date is: %d/%d/%d ”, monthTwo, dayTwo, yearTwo ); int totalDayOne = 0; int numDaysJan = dayOne; int numDaysFeb = 31 + dayOne; int numDaysMar = 31 + 28 + dayOne; int numDaysApr = 31 + 28 + 31 + dayOne; int numDaysMay = 31 + 28 + 31 + 30 + dayOne; int numDaysJun = 31 + 28 + 31 + 30 + 31 + dayOne; int numDaysJul = 31 + 28 + 31 + 30 + 31 + 30 + dayOne; int numDaysAug = 31 + 28 + 31 + 30 + 31 + 30 + 31 + dayOne; int numDaysSep = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + dayOne; int numDaysOct = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + dayOne; int numDaysNov = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + dayOne; int numDaysDec = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + dayOne; int totalDayTwo = 0; int numDaysJanTwo = dayTwo;
- 4. int numDaysFebTwo = 31 + dayTwo; int numDaysMarTwo = 31 + 28 + dayTwo; int numDaysAprTwo = 31 + 28 + 31 + dayTwo; int numDaysMayTwo = 31 + 28 + 31 + 30 + dayTwo; int numDaysJunTwo = 31 + 28 + 31 + 30 + 31 + dayTwo; int numDaysJulTwo = 31 + 28 + 31 + 30 + 31 + 30 + dayTwo; int numDaysAugTwo = 31 + 28 + 31 + 30 + 31 + 30 + 31 + dayTwo; int numDaysSepTwo = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + dayTwo; int numDaysOctTwo = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + dayTwo; int numDaysNovTwo = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + dayTwo; int numDaysDecTwo = 31 + 28 + 31 + 30 + 31 + 30 + 31 + 31 + 30 + 31 + 30 + dayTwo; if ( monthOne == 1 ) { totalDayOne += numDaysJan; } if ( monthOne == 2 ) { totalDayOne += numDaysFeb; } if ( monthOne == 3 ) { totalDayOne += numDaysMar; } if ( monthOne == 4 ) { totalDayOne += numDaysApr; } If ( monthOne == 5 ) { totalDayOne += numDaysMay; } if (monthOne == 6 ) { totalDayOne += numDaysJun; } if ( monthOne == 7 ) { totalDayOne += numDaysJul; } if ( monthOne == 8 ) { totalDayOne += numDaysAug; } if ( monthOne == 9 ) {
- 5. totalDayOne += numDaysSep; } if ( monthOne == 10 ) { totalDayOne += numDaysOct; } if (monthOne == 11 ) { totalDayOne += numDaysNov; } if (monthOne == 12 ) { totalDayOne += numDaysDec; } if ( monthTwo == 1 ) { totalDayTwo += numDaysJanTwo; } if ( monthTwo == 2 ) { totalDayTwo += numDaysFebTwo; } if ( monthTwo == 3 ) { totalDayTwo += numDaysMarTwo; } if ( monthTwo == 4 ) { totalDayTwo += numDaysAprTwo; } If ( monthTwo == 5 ) { totalDayTwo += numDaysMayTwo; } if (monthTwo == 6 ) { totalDayTwo += numDaysJunTwo; } if ( monthTwo == 7 ) { totalDayTwo += numDaysJulTwo; } if ( monthTwo == 8 ) { totalDayTwo += numDaysAugTwo; } if ( monthTwo == 9 ) {
- 6. totalDayTwo += numDaysSepTwo; } if ( monthTwo == 10 ) { totalDayTwo += numDaysOctTwo; } if (monthTwo == 11 ) { totalDayTwo += numDaysNovTwo; } if (monthTwo == 12 ) { totalDayTwo += numDaysDecTwo; } int daysBetween = totalDayOne – totalDayTwo; if ( totalDayOne < totalDayTwo ) { daysBetween = totalDayTwo – totalDayOne; } int daysTotal = 0; while ( yearOne < yearTwo ) { yearOne++; if ( yearOne % 4 == 0 && yearOne % 100 != 0 || yearOne % 400 == 0) daysTotal += 366; else daysTotal += 365; } while ( yearTwo < yearOne ) { yearTwo++; if ( yearTwo % 4 == 0 && yearTwo % 100 != 0 || yearTwo % 400 == 0) daysTotal += 366; else daysTotal += 365; } System.out.printf ( “ The total number of days between the dates is %d ”, daysTotal + daysBetween); } } Solution
- 7. import java.util.Calendar; import java.util.Date; import java.util.Scanner; public class NewDateDifference { public static void main(String[] args) { System.out.print("Insert first date : "); Scanner s = new Scanner(System.in); String[] eingabe1 = new String[3]; while (s.hasNext()) { int i = 0; insert1[i] = s.next(); if (!s.hasNext()) { s.close(); break; } i++; } System.out.print("Insert second date : "); Scanner t = new Scanner(System.in); String[] insert2 = new String[3]; while (t.hasNext()) { int i = 0; insert2[i] = t.next(); if (!t.hasNext()) { t.close(); break; } i++; } Calendar cal = Calendar.getInstance(); cal.set(Calendar.DAY_OF_MONTH, Integer.parseInt(insert1[0])); cal.set(Calendar.MONTH, Integer.parseInt(insert1[1])); cal.set(Calendar.YEAR, Integer.parseInt(insert1[2])); Date firstDate = cal.getTime(); cal.set(Calendar.DAY_OF_MONTH, Integer.parseInt(insert2[0]));
- 8. cal.set(Calendar.MONTH, Integer.parseInt(insert2[1])); cal.set(Calendar.YEAR, Integer.parseInt(insert2[2])); Date secondDate = cal.getTime(); long diff = secondDate.getTime() - firstDate.getTime(); System.out.println ("Days: " + diff / 1000 / 60 / 60 / 24); } }