import java-util-Arrays- import java-io-PrintWriter- import java-io-Fi.pdf
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import java.util.Arrays; import java.io.PrintWriter; import java.io.File; import java.util.Scanner; public class SortTimes { /* convert nanos to seconds */ static final double ONE_BILLION = 1_000_000_000.; /* sizes of input range should range up to 512k elements */ static int[] sizes = new int[32]; public static void main(String[] args) throws Exception { int sIdx = 0; long startTime; long endTime; double time; int size = 128*1024/sizes.length; /* let's make size smaller, comment out this next line and see what happens */ size = 16*1024/sizes.length; for( int i=0; i<sizes.length; i++ ) { sizes[i] = (i+1)*size; } Scanner inp = new Scanner(System.in); System.out.print("What is your last name? "); String lastName = inp.next(); PrintWriter pw = new PrintWriter(new File(lastName + "_sortTimes.csv")); pw.println( "Size, BubbleSort, MergeSort, QuickSort, JavaSort, BubbleSort (Sorted Array), MergeSort (Sorted Array), QuickSort (Sorted Array), JavaSort (Sorted Array)" ); while( sIdx < sizes.length ) { int[] A = new int[sizes[sIdx]]; int[] B = new int[sizes[sIdx]]; int[] C = new int[sizes[sIdx]]; int[] D = new int[sizes[sIdx]]; int[] E = new int[sizes[sIdx]]; for( int i=0; i<A.length; i++ ) { A[i] = (int)(Math.random()*sizes[sIdx]*2); B[i] = A[i]; C[i] = A[i]; D[i] = A[i]; E[i] = 1000; // Array E is used for testing algo on array that is already sorted. } System.out.println(); System.out.println("--------------------------------------------------"); System.out.println("Initial Values: "); printArray(A); pw.print(A.length + ", "); System.out.println( "Starting bubble sort ---------------- Size = " + B.length); startTime = System.nanoTime(); bubbleSort(A); endTime = System.nanoTime(); time = (endTime - startTime) / ONE_BILLION; System.out.println( "Bubble sort finished - time = " + time + " seconds" ); pw.print(time + ", "); printArray(A); System.out.println(); System.out.println( "Starting merge sort ---------------- Size = " + C.length); startTime = System.nanoTime(); mergeSort(B); endTime = System.nanoTime(); time = (endTime - startTime) / ONE_BILLION; System.out.println( "Merge sort finished - time = " + time + " seconds."); pw.print(time + ", "); printArray(B); System.out.println(); System.out.println( "Starting quick sort ---------------- Size = " + D.length); startTime = System.nanoTime(); quickSort(C); endTime = System.nanoTime(); time = (endTime - startTime) / ONE_BILLION; System.out.println( "quickSort finished - time = " + time + " seconds." ); pw.print(time + ", "); printArray(C); System.out.println(); System.out.println( "Starting java's Array sort ---------------- Size = " + D.length); startTime = System.nanoTime(); javaSort(D); endTime = System.nanoTime(); time = (endTime - startTime) / ONE_BILLION; System.out.println( "Java's Array sort finished - time = " + time + " seconds." ); pw.print(time + ", "); printArray(D); System.out.println(); System.out.println( "Starting bubble sort on sorted array ---------------- Size = " + B.length); star.
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![import java.util.Arrays;
import java.io.PrintWriter;
import java.io.File;
import java.util.Scanner;
public class SortTimes
{
/* convert nanos to seconds */
static final double ONE_BILLION = 1_000_000_000.;
/* sizes of input range should range up to 512k elements */
static int[] sizes = new int[32];
public static void main(String[] args) throws Exception
{
int sIdx = 0;
long startTime;
long endTime;
double time;
int size = 128*1024/sizes.length;
/* let's make size smaller, comment out this next line and see what happens */
size = 16*1024/sizes.length;
for( int i=0; i<sizes.length; i++ ) {
sizes[i] = (i+1)*size;
}
Scanner inp = new Scanner(System.in);](https://image.slidesharecdn.com/importjava-util-arrays-importjava-io-printwriter-importjava-io-fi-230314184234-14bb5d8f/85/import-java-util-Arrays-import-java-io-PrintWriter-import-java-io-Fi-pdf-1-320.jpg)
![System.out.print("What is your last name? ");
String lastName = inp.next();
PrintWriter pw = new PrintWriter(new File(lastName + "_sortTimes.csv"));
pw.println( "Size, BubbleSort, MergeSort, QuickSort, JavaSort, BubbleSort (Sorted Array),
MergeSort (Sorted Array), QuickSort (Sorted Array), JavaSort (Sorted Array)" );
while( sIdx < sizes.length )
{
int[] A = new int[sizes[sIdx]];
int[] B = new int[sizes[sIdx]];
int[] C = new int[sizes[sIdx]];
int[] D = new int[sizes[sIdx]];
int[] E = new int[sizes[sIdx]];
for( int i=0; i<A.length; i++ )
{
A[i] = (int)(Math.random()*sizes[sIdx]*2);
B[i] = A[i];
C[i] = A[i];
D[i] = A[i];
E[i] = 1000; // Array E is used for testing algo on array that is already sorted.
}
System.out.println();
System.out.println("--------------------------------------------------");
System.out.println("Initial Values: ");
printArray(A);](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
Recommended
please help finish sorting methods- import java-util-Arrays- import ja.pdf
please help finish sorting methods:
import java.util.Arrays;
import java.io.PrintWriter;
import java.io.File;
import java.util.Scanner;
public class SortTimes
{
/* convert nanos to seconds */
static final double ONE_BILLION = 1_000_000_000.;
/* sizes of input range should range up to 512k elements */
static int[] sizes = new int[32];
public static void main(String[] args) throws Exception
{
int sIdx = 0;
long startTime;
long endTime;
double time;
int size = 128*1024/sizes.length;
/* let's make size smaller, comment out this next line and see what happens */
size = 16*1024/sizes.length;
for( int i=0; i<sizes.length; i++ ) {
sizes[i] = (i+1)*size;
}
Scanner inp = new Scanner(System.in);
System.out.print("What is your last name? ");
String lastName = inp.next();
PrintWriter pw = new PrintWriter(new File(lastName + "_sortTimes.csv"));
pw.println( "Size, BubbleSort, MergeSort, QuickSort, JavaSort, BubbleSort (Sorted Array),
MergeSort (Sorted Array), QuickSort (Sorted Array), JavaSort (Sorted Array)" );
while( sIdx < sizes.length )
{
int[] A = new int[sizes[sIdx]];
int[] B = new int[sizes[sIdx]];
int[] C = new int[sizes[sIdx]];
int[] D = new int[sizes[sIdx]];
int[] E = new int[sizes[sIdx]];
for( int i=0; i<A.length; i++ )
{
A[i] = (int)(Math.random()*sizes[sIdx]*2);
B[i] = A[i];
C[i] = A[i];
D[i] = A[i];
E[i] = 1000; // Array E is used for testing algo on array that is already sorted.
}
System.out.println();
System.out.println("--------------------------------------------------");
System.out.println("Initial Values: ");
printArray(A);
pw.print(A.length + ", ");
System.out.println( "Starting bubble sort ---------------- Size = " + B.length);
startTime = System.nanoTime();
bubbleSort(A);
endTime = System.nanoTime();
time = (endTime - startTime) / ONE_BILLION;
System.out.println( "Bubble sort finished - time = " + time + " seconds" );
pw.print(time + ", ");
printArray(A);
System.out.println();
System.out.println( "Starting merge sort ---------------- Size = " + C.length);
startTime = System.nanoTime();
mergeSort(B);
endTime = System.nanoTime();
time = (endTime - startTime) / ONE_BILLION;
System.out.println( "Merge sort finished - time = " + time + " seconds.");
pw.print(time + ", ");
printArray(B);
System.out.println();
System.out.println( "Starting quick sort ---------------- Size = " + D.length);
startTime = System.nanoTime();
quickSort(C);
endTime = System.nanoTime();
time = (endTime - startTime) / ONE_BILLION;
System.out.println( "quickSort finished - time = " + time + " seconds." );
pw.print(time + ", ");
printArray(C);
System.out.println();
System.out.println( "Starting java's Array sort ---------------- Size = " + D.length);
startTime = System.nanoTime();
javaSort(D);
endTime = System.nanoTime();
time = (endTime - startTime) / ONE_BILLION;
System.out.println( "Java's Array sort finished - time = " + time + " seconds." );
pw.print(time + ", ");
printArray(D);
System.out.println();
System.out.println( "Starting bubble sort on sorted array -------.
I have already done the first 2 parts(factorial and fibonacci) also .pdf
I have already done the first 2 parts(factorial and fibonacci) also finding the time elapsed, need to
do the last 2 parts (palindromes and binary search) also finding the time elapsed on each version
(need to add code for that).
Assignments:
1) Write the recursive and nonrecursive version of
a. Int factorial(int n), (use nFactorial as the name for non-recursive version) (where n >= 0)
b. Int fibonacci(int n), (and nFibonacci as the name for non-recursive version)
c. boolean isPalindrome(String s, int low, int high), (and nIsPalindrome for the non-recursive
version)
d. int binarySearch(int[] v, int key, int low, int high), (and nBinarySearch for the non recursive
version)
2) Develop testing schemes as following:
a. For a and b: use at least 20 different values of n (from small to sufficiently large) and the time
function to find the time spent for different values of n.
b. For c, set up a string array of size 20 and initialize them with at least 10 palindromes. (already
made an array in the method)
c. For d, do the following: (need to make a new method for both versions and all the calls for
both )
i. Fill an integer array of 1000 multiples of 3 from 0 to 2997 (i.e., 0, 3, 6, 9,…,2997) and have
them properly sorted.
ii. Test your function and see if it works.
iii. Set up a test array with 20 random integers between 0 and 3000 to test the program.
Main method code:
public class factorials {
public static void main (String args[])
{
System.out.println(\"Recursive Fibonacci Method: \ \");
System.out.println(\" N\\t\\t\" + \"Fibonacci Sequence\\t\\t\" + \"Time Elapsed\");
for(int n = 0; n < 25; n++)
{
long startTime = System.nanoTime();
fibonacci(n);
long stopTime = System.nanoTime();
long elapsedTime = stopTime - startTime;
System.out.println( \" \" + n + \"\\t\\t\\t \" + fibonacci(n) + \"\\t\\t\\t \" + elapsedTime);
//System.out.println(/*\"Start Time: \" + startTime + \"\ Stop Time: \" + stopTime + */
//\"\ Elapsed Time: \" + elapsedTime);
}
System.out.println(\"\ \");
System.out.println(\"Iterative Fibonacci Method: \ \");
System.out.println(\" N\\t\\t\" + \"Fibonacci Sequence\\t\\t\" + \"Time Elapsed\");
for ( int n = 0; n < 25; n++)
{
long startTime = System.nanoTime();
nfactorial(n);
long stopTime = System.nanoTime();
long elapsedTime = stopTime - startTime;
System.out.println( \" \" + n + \"\\t\\t\\t \" + nfibonacci(n) + \"\\t\\t\\t \" + elapsedTime);
//System.out.println(/*\"Start Time: \" + startTime + \"\ Stop Time: \" + stopTime + */
// \"\ Elapsed Time: \" + elapsedTime);
}
System.out.println(\"\ \ \ \");
System.out.println(\"Recursive Factorial Method: \ \");
System.out.println(\" N\\t\\t\" + \"\\t Factorial\\t\\t\" + \"\\tTime Elapsed\");
for ( int n = 0; n < 25; n++)
{
long startTime = System.nanoTime();
nfactorial(n);
long stopTime = System.nanoTime();
long elapsedTime = stopTime - startTime;
System.out.println( \" \" + n + \"\\t\\t\\t \\t\" + factorial(n) + \"\\t\\t\\t\\t\\t\" +
elapsedTime);
//System.out.println(/*\"Start Time: .
Write a program that obtains the execution time of selection sort, bu.pdf
Write a java program called Question39 that does the following: Gets input for temperature
Utilizing a branching statement: If temperature is 76-100, call method outputHot passing the
temperature input as an argument. If temperature is 0-39, call method outputCold passing the
temperature input as an argument. If temperature is 40 to 75, call method outputJustRight
passing the temperature input as an argument. If temperature is outside these ranges, output
“Temperature outside range” to the screen. Be precise, import modules, include comments,
prologue, etc. as needed.
Solution
Question39.java
import java.util.Scanner;
public class Question39 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter temperature: \");
int temp = scan.nextInt();
if(temp >=76 && temp <=100){
outputHot(temp);
}
else if(temp >=0 && temp <=39){
outputCold(temp);
}
else if(temp >=40 && temp <=75){
outputJustRight(temp);
}
else{
System.out.println(\"Temperature outside range\");
}
}
public static void outputHot(int temp){
System.out.println(\"Temperature \"+temp+\" Hot \");
}
public static void outputCold(int temp){
System.out.println(\"Temperature \"+temp+\" Cold \");
}
public static void outputJustRight(int temp){
System.out.println(\"Temperature \"+temp+\" Just Right \");
}
}
Output:
Enter temperature:
45
Temperature 45 Just Right.
My code is below How do I write this code without usin.pdf
My code is below...... How do I write this code without using arrays? I also need to write it without
using the while (true) or using a break statement. I need to use a prime read and a sentinel value
to control the repetition.
import java.text.DecimalFormat;
import java.util.Scanner;
public class AuthorPay2 {
public static void main(String[] args) {
//Declaring variables
double largestAmount = 0.0, avgAmt = 0.0, totAmount = 0.0;
int totNoOfPayments = 0;
//Creating an integer array
int[] countPayLevels = new int[3];
//DecimalFormat class is used to format the output
DecimalFormat df = new DecimalFormat(".00");
/*
*Creating an Scanner class object which is used to get the inputs
*entered by the user
*/
Scanner sc = new Scanner(System.in);
while (true) {
//Getting the input entered by the user
System.out.print("Enter the length of the story: ");
int length = sc.nextInt();
if (length != -1) {
System.out.print("Enter the level of the author: ");
char level = sc.next().charAt(0);
double amount = calculateAmount(length, level);
if (largestAmount < amount) {
largestAmount = amount;
}
totAmount += amount;
totNoOfPayments++;
if (level == 'A' || length == 'a') {
countPayLevels[0]++;
} else if (level == 'B' || length == 'b') {
countPayLevels[1]++;
} else if (level == 'C' || length == 'c') {
countPayLevels[2]++;
}
System.out.println("Length of the story: " + length + " Amount to be paid; $" + df.format(amount));
} else {
avgAmt = totAmount / totNoOfPayments;
System.out.println("nTotal no of payments: " + totNoOfPayments);
System.out.println("No of Payments made at level 'A': " + countPayLevels[0]);
System.out.println("No of Payments made at level 'B': " + countPayLevels[1]);
System.out.println("No of Payments made at level 'C': " + countPayLevels[2]);
System.out.printf("The average amount paid for a story: %.2fn", avgAmt);
System.out.printf("The largest amount paid for a story: %.2fn", largestAmount);
break;
}
}
}
private static double calculateAmount(int length, char level) {
double amount = 0.0;
if (length < 7500) {
amount = 0.08 * length;
} else if (length >= 7500 && length < 8000) {
amount = 600;
} else if (length >= 8000 && length < 17500) {
amount = 0.075 * length;
} else if (length >= 17500 && length <= 190000) {
amount = 1313;
} else if (length > 19000) {
amount = 0.07 * length;
}
if (level == 'A' || length == 'a') {
amount = 1.75 * amount;
} else if (level == 'B' || length == 'b') {
amount = 1.25 * amount;
}
return amount;
}
}.
Problem1 java codeimport java.util.Scanner; Java code to pr.pdf
// Problem1 java code
import java.util.Scanner;
// Java code to print all possible strings of letter L and R
class Problem1 {
static void print(char set[], int length) {
int setLength = set.length;
printRecursion(set, \"\", setLength, length);
}
static void printRecursion(char set[], String prefixset, int setLength, int length) {
// Base case: length is 0
if (length == 0) {
System.out.println(prefixset);
return;
}
for (int i = 0; i < setLength; ++i) {
String newPrefixset = prefixset + set[i];
printRecursion(set, newPrefixset, setLength, length - 1);
}
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter length: \");
int length = scan.nextInt();
char set[] = {\'L\', \'R\'};
print(set, length);
}
}
/*
output:
Enter length:
3
LLL
LLR
LRL
LRR
RLL
RLR
RRL
RRR
*/
// Problem2 java code
import java.util.Scanner;
// Java code to print all possible strings of letter L and R
class Problem2 {
static void print(char set[], int length) {
int setLength = set.length;
printRecursion(set, \"\", setLength, length);
}
static void printRecursion(char set[], String prefixset, int setLength, int length) {
// Base case: length is 0
if (length == 0) {
System.out.print(prefixset + \" \");
return;
}
for (int i = 0; i < setLength; ++i) {
String newPrefixset = prefixset + set[i];
printRecursion(set, newPrefixset, setLength, length - 1);
}
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter length: \");
int length = scan.nextInt();
char set[] = {\'1\', \'3\', \'5\', \'7\', \'9\'};
print(set, length);
System.out.println();
}
}
/*
output:
Enter length:
3
111 113 115 117 119 131 133 135 137 139 151 153 155
157 159 171 173 175 177 179 191 193 195 197 199 311
313 315 317 319 331 333 335 337 339 351 353 355 357
359 371 373 375 377 379 391 393 395 397 399 511 513
515 517 519 531 533 535 537 539 551 553 555 557 559
571 573 575 577 579 591 593 595 597 599 711 713 715
717 719 731 733 735 737 739 751 753 755 757 759 771
773 775 777 779 791 793 795 797 799 911 913 915 917
919 931 933 935 937 939 951 953 955 957 959 971 973
975 977 979 991 993 995 997 999
*/
// Problem3 java code
import java.util.Scanner;
// Java code to multiply 2 numbers
class Problem3 {
public static int multiply(int a, int b)
{
int temp = b;
for (int i = 1; i < a; i++ ) {
b = b + temp;
}
return b;
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter a: \");
int a = scan.nextInt();
System.out.println(\"Enter b: \");
int b = scan.nextInt();
System.out.println(a + \"X\" + b + \" = \" + multiply(a,b));
}
}
/*
output:
Enter a:
7
Enter b:
6
7X6 = 42
*/
// Problem4 java code
import java.util.Scanner;
// Java code to find gcd 2 numbers
class Problem4 {
public static int gcd(int a, int b)
{
if (b!=0)
return gcd(b, a%b);
else
return a;
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter a: \");
int a = scan.ne.
Add an interactive command line to your C++ application
How to add an interactive shell (remote, too) to a C++ application by using my open-source C++14 library:
https://github.com/daniele77/cli
In the slide deck you can learn how to use it, how does it work, and find some thoughts about C++ design and patterns used by the library.
Recommended
please help finish sorting methods- import java-util-Arrays- import ja.pdf
please help finish sorting methods:
import java.util.Arrays;
import java.io.PrintWriter;
import java.io.File;
import java.util.Scanner;
public class SortTimes
{
/* convert nanos to seconds */
static final double ONE_BILLION = 1_000_000_000.;
/* sizes of input range should range up to 512k elements */
static int[] sizes = new int[32];
public static void main(String[] args) throws Exception
{
int sIdx = 0;
long startTime;
long endTime;
double time;
int size = 128*1024/sizes.length;
/* let's make size smaller, comment out this next line and see what happens */
size = 16*1024/sizes.length;
for( int i=0; i<sizes.length; i++ ) {
sizes[i] = (i+1)*size;
}
Scanner inp = new Scanner(System.in);
System.out.print("What is your last name? ");
String lastName = inp.next();
PrintWriter pw = new PrintWriter(new File(lastName + "_sortTimes.csv"));
pw.println( "Size, BubbleSort, MergeSort, QuickSort, JavaSort, BubbleSort (Sorted Array),
MergeSort (Sorted Array), QuickSort (Sorted Array), JavaSort (Sorted Array)" );
while( sIdx < sizes.length )
{
int[] A = new int[sizes[sIdx]];
int[] B = new int[sizes[sIdx]];
int[] C = new int[sizes[sIdx]];
int[] D = new int[sizes[sIdx]];
int[] E = new int[sizes[sIdx]];
for( int i=0; i<A.length; i++ )
{
A[i] = (int)(Math.random()*sizes[sIdx]*2);
B[i] = A[i];
C[i] = A[i];
D[i] = A[i];
E[i] = 1000; // Array E is used for testing algo on array that is already sorted.
}
System.out.println();
System.out.println("--------------------------------------------------");
System.out.println("Initial Values: ");
printArray(A);
pw.print(A.length + ", ");
System.out.println( "Starting bubble sort ---------------- Size = " + B.length);
startTime = System.nanoTime();
bubbleSort(A);
endTime = System.nanoTime();
time = (endTime - startTime) / ONE_BILLION;
System.out.println( "Bubble sort finished - time = " + time + " seconds" );
pw.print(time + ", ");
printArray(A);
System.out.println();
System.out.println( "Starting merge sort ---------------- Size = " + C.length);
startTime = System.nanoTime();
mergeSort(B);
endTime = System.nanoTime();
time = (endTime - startTime) / ONE_BILLION;
System.out.println( "Merge sort finished - time = " + time + " seconds.");
pw.print(time + ", ");
printArray(B);
System.out.println();
System.out.println( "Starting quick sort ---------------- Size = " + D.length);
startTime = System.nanoTime();
quickSort(C);
endTime = System.nanoTime();
time = (endTime - startTime) / ONE_BILLION;
System.out.println( "quickSort finished - time = " + time + " seconds." );
pw.print(time + ", ");
printArray(C);
System.out.println();
System.out.println( "Starting java's Array sort ---------------- Size = " + D.length);
startTime = System.nanoTime();
javaSort(D);
endTime = System.nanoTime();
time = (endTime - startTime) / ONE_BILLION;
System.out.println( "Java's Array sort finished - time = " + time + " seconds." );
pw.print(time + ", ");
printArray(D);
System.out.println();
System.out.println( "Starting bubble sort on sorted array -------.
I have already done the first 2 parts(factorial and fibonacci) also .pdf
I have already done the first 2 parts(factorial and fibonacci) also finding the time elapsed, need to
do the last 2 parts (palindromes and binary search) also finding the time elapsed on each version
(need to add code for that).
Assignments:
1) Write the recursive and nonrecursive version of
a. Int factorial(int n), (use nFactorial as the name for non-recursive version) (where n >= 0)
b. Int fibonacci(int n), (and nFibonacci as the name for non-recursive version)
c. boolean isPalindrome(String s, int low, int high), (and nIsPalindrome for the non-recursive
version)
d. int binarySearch(int[] v, int key, int low, int high), (and nBinarySearch for the non recursive
version)
2) Develop testing schemes as following:
a. For a and b: use at least 20 different values of n (from small to sufficiently large) and the time
function to find the time spent for different values of n.
b. For c, set up a string array of size 20 and initialize them with at least 10 palindromes. (already
made an array in the method)
c. For d, do the following: (need to make a new method for both versions and all the calls for
both )
i. Fill an integer array of 1000 multiples of 3 from 0 to 2997 (i.e., 0, 3, 6, 9,…,2997) and have
them properly sorted.
ii. Test your function and see if it works.
iii. Set up a test array with 20 random integers between 0 and 3000 to test the program.
Main method code:
public class factorials {
public static void main (String args[])
{
System.out.println(\"Recursive Fibonacci Method: \ \");
System.out.println(\" N\\t\\t\" + \"Fibonacci Sequence\\t\\t\" + \"Time Elapsed\");
for(int n = 0; n < 25; n++)
{
long startTime = System.nanoTime();
fibonacci(n);
long stopTime = System.nanoTime();
long elapsedTime = stopTime - startTime;
System.out.println( \" \" + n + \"\\t\\t\\t \" + fibonacci(n) + \"\\t\\t\\t \" + elapsedTime);
//System.out.println(/*\"Start Time: \" + startTime + \"\ Stop Time: \" + stopTime + */
//\"\ Elapsed Time: \" + elapsedTime);
}
System.out.println(\"\ \");
System.out.println(\"Iterative Fibonacci Method: \ \");
System.out.println(\" N\\t\\t\" + \"Fibonacci Sequence\\t\\t\" + \"Time Elapsed\");
for ( int n = 0; n < 25; n++)
{
long startTime = System.nanoTime();
nfactorial(n);
long stopTime = System.nanoTime();
long elapsedTime = stopTime - startTime;
System.out.println( \" \" + n + \"\\t\\t\\t \" + nfibonacci(n) + \"\\t\\t\\t \" + elapsedTime);
//System.out.println(/*\"Start Time: \" + startTime + \"\ Stop Time: \" + stopTime + */
// \"\ Elapsed Time: \" + elapsedTime);
}
System.out.println(\"\ \ \ \");
System.out.println(\"Recursive Factorial Method: \ \");
System.out.println(\" N\\t\\t\" + \"\\t Factorial\\t\\t\" + \"\\tTime Elapsed\");
for ( int n = 0; n < 25; n++)
{
long startTime = System.nanoTime();
nfactorial(n);
long stopTime = System.nanoTime();
long elapsedTime = stopTime - startTime;
System.out.println( \" \" + n + \"\\t\\t\\t \\t\" + factorial(n) + \"\\t\\t\\t\\t\\t\" +
elapsedTime);
//System.out.println(/*\"Start Time: .
Write a program that obtains the execution time of selection sort, bu.pdf
Write a java program called Question39 that does the following: Gets input for temperature
Utilizing a branching statement: If temperature is 76-100, call method outputHot passing the
temperature input as an argument. If temperature is 0-39, call method outputCold passing the
temperature input as an argument. If temperature is 40 to 75, call method outputJustRight
passing the temperature input as an argument. If temperature is outside these ranges, output
“Temperature outside range” to the screen. Be precise, import modules, include comments,
prologue, etc. as needed.
Solution
Question39.java
import java.util.Scanner;
public class Question39 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter temperature: \");
int temp = scan.nextInt();
if(temp >=76 && temp <=100){
outputHot(temp);
}
else if(temp >=0 && temp <=39){
outputCold(temp);
}
else if(temp >=40 && temp <=75){
outputJustRight(temp);
}
else{
System.out.println(\"Temperature outside range\");
}
}
public static void outputHot(int temp){
System.out.println(\"Temperature \"+temp+\" Hot \");
}
public static void outputCold(int temp){
System.out.println(\"Temperature \"+temp+\" Cold \");
}
public static void outputJustRight(int temp){
System.out.println(\"Temperature \"+temp+\" Just Right \");
}
}
Output:
Enter temperature:
45
Temperature 45 Just Right.
My code is below How do I write this code without usin.pdf
My code is below...... How do I write this code without using arrays? I also need to write it without
using the while (true) or using a break statement. I need to use a prime read and a sentinel value
to control the repetition.
import java.text.DecimalFormat;
import java.util.Scanner;
public class AuthorPay2 {
public static void main(String[] args) {
//Declaring variables
double largestAmount = 0.0, avgAmt = 0.0, totAmount = 0.0;
int totNoOfPayments = 0;
//Creating an integer array
int[] countPayLevels = new int[3];
//DecimalFormat class is used to format the output
DecimalFormat df = new DecimalFormat(".00");
/*
*Creating an Scanner class object which is used to get the inputs
*entered by the user
*/
Scanner sc = new Scanner(System.in);
while (true) {
//Getting the input entered by the user
System.out.print("Enter the length of the story: ");
int length = sc.nextInt();
if (length != -1) {
System.out.print("Enter the level of the author: ");
char level = sc.next().charAt(0);
double amount = calculateAmount(length, level);
if (largestAmount < amount) {
largestAmount = amount;
}
totAmount += amount;
totNoOfPayments++;
if (level == 'A' || length == 'a') {
countPayLevels[0]++;
} else if (level == 'B' || length == 'b') {
countPayLevels[1]++;
} else if (level == 'C' || length == 'c') {
countPayLevels[2]++;
}
System.out.println("Length of the story: " + length + " Amount to be paid; $" + df.format(amount));
} else {
avgAmt = totAmount / totNoOfPayments;
System.out.println("nTotal no of payments: " + totNoOfPayments);
System.out.println("No of Payments made at level 'A': " + countPayLevels[0]);
System.out.println("No of Payments made at level 'B': " + countPayLevels[1]);
System.out.println("No of Payments made at level 'C': " + countPayLevels[2]);
System.out.printf("The average amount paid for a story: %.2fn", avgAmt);
System.out.printf("The largest amount paid for a story: %.2fn", largestAmount);
break;
}
}
}
private static double calculateAmount(int length, char level) {
double amount = 0.0;
if (length < 7500) {
amount = 0.08 * length;
} else if (length >= 7500 && length < 8000) {
amount = 600;
} else if (length >= 8000 && length < 17500) {
amount = 0.075 * length;
} else if (length >= 17500 && length <= 190000) {
amount = 1313;
} else if (length > 19000) {
amount = 0.07 * length;
}
if (level == 'A' || length == 'a') {
amount = 1.75 * amount;
} else if (level == 'B' || length == 'b') {
amount = 1.25 * amount;
}
return amount;
}
}.
Problem1 java codeimport java.util.Scanner; Java code to pr.pdf
// Problem1 java code
import java.util.Scanner;
// Java code to print all possible strings of letter L and R
class Problem1 {
static void print(char set[], int length) {
int setLength = set.length;
printRecursion(set, \"\", setLength, length);
}
static void printRecursion(char set[], String prefixset, int setLength, int length) {
// Base case: length is 0
if (length == 0) {
System.out.println(prefixset);
return;
}
for (int i = 0; i < setLength; ++i) {
String newPrefixset = prefixset + set[i];
printRecursion(set, newPrefixset, setLength, length - 1);
}
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter length: \");
int length = scan.nextInt();
char set[] = {\'L\', \'R\'};
print(set, length);
}
}
/*
output:
Enter length:
3
LLL
LLR
LRL
LRR
RLL
RLR
RRL
RRR
*/
// Problem2 java code
import java.util.Scanner;
// Java code to print all possible strings of letter L and R
class Problem2 {
static void print(char set[], int length) {
int setLength = set.length;
printRecursion(set, \"\", setLength, length);
}
static void printRecursion(char set[], String prefixset, int setLength, int length) {
// Base case: length is 0
if (length == 0) {
System.out.print(prefixset + \" \");
return;
}
for (int i = 0; i < setLength; ++i) {
String newPrefixset = prefixset + set[i];
printRecursion(set, newPrefixset, setLength, length - 1);
}
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter length: \");
int length = scan.nextInt();
char set[] = {\'1\', \'3\', \'5\', \'7\', \'9\'};
print(set, length);
System.out.println();
}
}
/*
output:
Enter length:
3
111 113 115 117 119 131 133 135 137 139 151 153 155
157 159 171 173 175 177 179 191 193 195 197 199 311
313 315 317 319 331 333 335 337 339 351 353 355 357
359 371 373 375 377 379 391 393 395 397 399 511 513
515 517 519 531 533 535 537 539 551 553 555 557 559
571 573 575 577 579 591 593 595 597 599 711 713 715
717 719 731 733 735 737 739 751 753 755 757 759 771
773 775 777 779 791 793 795 797 799 911 913 915 917
919 931 933 935 937 939 951 953 955 957 959 971 973
975 977 979 991 993 995 997 999
*/
// Problem3 java code
import java.util.Scanner;
// Java code to multiply 2 numbers
class Problem3 {
public static int multiply(int a, int b)
{
int temp = b;
for (int i = 1; i < a; i++ ) {
b = b + temp;
}
return b;
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter a: \");
int a = scan.nextInt();
System.out.println(\"Enter b: \");
int b = scan.nextInt();
System.out.println(a + \"X\" + b + \" = \" + multiply(a,b));
}
}
/*
output:
Enter a:
7
Enter b:
6
7X6 = 42
*/
// Problem4 java code
import java.util.Scanner;
// Java code to find gcd 2 numbers
class Problem4 {
public static int gcd(int a, int b)
{
if (b!=0)
return gcd(b, a%b);
else
return a;
}
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter a: \");
int a = scan.ne.
Add an interactive command line to your C++ application
How to add an interactive shell (remote, too) to a C++ application by using my open-source C++14 library:
https://github.com/daniele77/cli
In the slide deck you can learn how to use it, how does it work, and find some thoughts about C++ design and patterns used by the library.
Hands on Session on Python
This is the collection of slides when conducting a Python Workshop held at Dayananda Sagar College of Engineering.
JAVA.Q4 Create a Time class. This class will represent a point in.pdf
JAVA.
Q4: Create a Time class. This class will represent a point in time, such as a departure time. It
should contain 2 constructors, 2 instance variables (hour and minute), and 10 methods (see
below). All methods but toString should be in terms of the 24 hour format. [30 points]
default constructor: Creates a Time object for 12:00AM.
overloaded constructor: Creates a Time object at a specific hour and minute.
getHour(): Returns an integer representing the hour of the Time object.
getMinute(): Returns an integer representing the minute of the Time object.
addHours(...): Updates the object by moving it forward a number of hours.
addMinute(...): Updates the object by moving it forward a number of minutes. (Hint: Be careful
that you don\'t allow minutes to be more than 59.)
addTime(...): Updates the object by moving it forward by the hour and minute from another
Time object.
getCopy(...): Returns a new Time object that has the same hour and minute of the existing Time
object.
isEarlierThan(...): Returns true if this Time object is earlier than another Time object.
isSameTime(...): Returns true if this Time object is the same as another Time object.
isLaterThan(...): Returns true if this Time object is later than another Time object.
toString(): Returns a string representing the Time object. Uses 12 hour AM/PM format and pads
minutes to be two digits. See the sample output for an example.
Q5: Create a Flight class that uses the Plane and Time class. This class will represent a flight
between two airports, using a specific Plane, and departing at a specific Time. It should contain a
constructor, 7 instance variables (plane, flight number, cost, departure, duration, source,
destination), and 9 methods (see below). [25 points]
overloaded constructor: Creates a Flight object that is setup up with a Plane, a flight number, a
cost, a departure Time, a duration time, a source Airport, and a destination Airport.
getPlane(): Returns the Plane that operates this flight.
getNumber(): Returns the flight number.
getCost(): Returns the flight cost.
getDestination(): Returns the destination Airport.
getDeparture(): Returns the departure Time.
getArrival(): Returns a Time object with the arrival time (computed from the departure time and
duration).
getSource(): Returns a Airport object for the departure location.
toOverviewString(): Returns a String representing an overview of the flight. Use NumberFormat
to display the price. See the sample output for an example.
toDetailedString(): Returns a String representing the flight\'s detail information. See the sample
output for an example.
Included below is an overall UML diagram that describes the three classes you will be
constructing. It provides a useful summary of all of the methods you are expected to implement,
and their corresponding types and visibility. Notice that one private method is listed here
(formatDigits in Time) that isn\'t mentioned above. This is a method that was in our solution, you
may not need.
FileName EX06_1java Programmer import ja.pdf
//FileName: EX06_1.java
//Programmer: ............
import java.util.Scanner;
import java.util.Arrays;
public class EX06_1
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
System.out.print("Enter a list of SAT scores: ");
String str = input.nextLine();
String[] items = str.split(" ");
double[] values = new double[items.length];
for (int i = 0; i<items.length; i++)
{
values[i] = Double.parseDouble(items[i]);
}
//count
System.out.print("Count: ");
int count = 0;
for (int i=0; i<values.length; i++)
{
count++;
}
System.out.println((count == values.length) ? values.length : count);
//sum
double sum = 0;
for (int i = 0; i<values.length; i++)
{
sum += values[i];
}
System.out.println("Sum: " + sum);
//mean
double mean = (sum/count);
System.out.println("Mean: " + mean);
//min and max
double min = values[0];
double max = values[0];
for (int i=1; i<values.length; i++)
{
if (values[i] < min)
{
min = values[i];
}
if (values[i] > max)
{
max = values[i];
}
}
System.out.println("Minimum: " + min);
System.out.println("Maximum: " + max);
System.out.println("Range: " + (max-min));
//sorting
Arrays.sort(values);
//median
int position = (count+1) / 2;
if (count%2 != 0)
{
System.out.println("Median: " + values[position-1]);
}
else
{
System.out.println("Median: " + (( values[position-1] + values[position]) / 2));
}
//Standard deviation
double sigmap = 0; //for population
double sigmas = 0; //for sample
for (int i=0; i<values.length; i++)
{
sigmap += (values[i] - mean)*(values[i] - mean);
}
sigmas = sigmap;
//sd for population
sigmap = sigmap / count;
System.out.println("Variance of population: " + sigmap);
double sdp = Math.sqrt(sigmap);
System.out.println("Standard Deviation of population: " + sdp);
//sd for sample
sigmas = sigmas / (count - 1);
System.out.println("Variance of sample: " + sigmas);
double sds = Math.sqrt(sigmas);
System.out.println("Standard Deviation of sample: " + sds);
//Skewness
double sigma = 0;
for (int i=0; i<values.length; i++)
{
sigma += (values[i] - mean)*(values[i] - mean)*(values[i] - mean);
}
double N = (double) count;
double sk = (Math.sqrt(N*(N-1)) / (N-2)) * (sigma / (sdp*sdp*sdp*N));
System.out.println("Skewness: " + sk);
//mode
int repeated = -1;
int[] frq = new int[values.length];
for (int i=0; i<values.length; i++)
{
int occ = 1;
for (int j=i+1; j<values.length; j++)
{
if (values[i] == values[j])
{
frq[j] = repeated;
occ++;
}
}
if (frq[i] != repeated)
{
frq[i] = occ;
}
}
int maxFrq = frq[0];
for (int i=0; i<frq.length; i++)
{
if (frq[i] > maxFrq)
{
maxFrq = frq[i];
}
}
System.out.print("Mode: ");
if (maxFrq == 1)
{
System.out.println("N/A");
}
else
{
for (int i=0; i<frq.length; i++)
{
if (frq[i] == maxFrq)
{
System.out.print(values[i] + " ");
}
}
System.out.print("n");
}
double standardError=sds/Math.sqrt(count);
System.out.println("Standard Error: "+standardError);
}
}
Programming Exercise #06_2:
Instruction:
1. Make a copy of your EX06_01.java file and rename it to EX06_2.java.
2. Open the EX06.
2 BytesC++ course_2014_c3_ function basics¶meters and overloading
This is my C++ course / basic , intermediate and a little advance
I have written the code but cannot complete the assignment please help.pdf
I have written the code but cannot complete the assignment please help me to complete.
Please don't just copy other s answers as your own.
// Insertion sort
/*
#include <ctime>
#include <iostream>
using namespace std;
void insertionSort(int arr[], int n) {
int i, key, j;
for (i = 1; i < n; i++) {
key = arr[i];
j = i - 1;
// Move elements of arr[0..i-1], that are greater than key, to one position ahead of their current
position
while (j >= 0 && arr[j] > key) {
arr[j + 1] = arr[j];
j = j - 1;
}
arr[j + 1] = key;
}
}
int main() {
int n;
cout << "Enter the size of the array: ";
cin >> n;
int arr[n];
cout << "Enter the elements of the array: ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int num = sizeof(arr) / sizeof(arr[0]);
clock_t start, end;
double timetaken;
start = clock();
insertionSort(arr, num);
end = clock();
cout << "Sorted array: \n";
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
cout << endl;
timetaken = ((double) (end - start)) / CLOCKS_PER_SEC;
cout << "Time taken : " << fixed << timetaken << "s" << endl;
return 0;
}
*/
// Shell Sort
/*
#include <ctime>
#include <iostream>
using namespace std;
int shellSort(int arr[], int n) {
for (int gap = n/2; gap > 0; gap /= 2) {
for (int i = gap; i < n; i += 1) {
int temp = arr[i];
int j;
for (j = i; j >= gap && arr[j - gap] > temp; j -= gap)
arr[j] = arr[j - gap];
arr[j] = temp;
}
}
return 0;
}
void printArray(int arr[], int n) {
for (int i=0; i<n; i++)
cout << arr[i] << " ";
}
int main() {
int n;
cout << "Enter the size of the array: ";
cin >> n;
int arr[n];
cout << "Enter the elements of the array: ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int num = sizeof(arr) / sizeof(arr[0]);
clock_t start, end;
double timetaken;
start = clock();
shellSort(arr, num);
end = clock();
cout << "\nArray after sorting: \n";
printArray(arr, n);
cout << endl;
timetaken = (double)(end - start) / CLOCKS_PER_SEC;
cout << "Time taken : " << fixed << timetaken << "s" << endl;
return 0;
}
*/
// MergeSort
/*
#include <ctime>
#include <iostream>
using namespace std;
void merge(int arr[], int l, int m, int r) {
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
int L[n1], R[n2];
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1 + j];
i = 0;
j = 0;
k = l;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
}
else {
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1) {
arr[k] = L[i];
i++;
k++;
}
while (j < n2) {
arr[k] = R[j];
j++;
k++;
}
}
void mergeSort(int arr[], int l, int r) {
if (l < r) {
int m = l + (r - l) / 2;
mergeSort(arr, l, m);
mergeSort(arr, m + 1, r);
merge(arr, l, m, r);
}
}
int main() {
int n;
cout << "Enter the size of the array: ";
cin >> n;
int arr[n];
cout << "Enter the elements of the array: ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int arr_size = sizeof(arr) / sizeof(arr[0]);
clock_t start;
clock_t end;
double timetaken;
start = clock();
mergeSort(arr, 0, arr_size - 1);
end = clock();
timetaken = (double)(end - start) / CLOCKS_PER_SEC;
.
Instruction1. Please read the two articles. (Kincheloe part 1 &.docx
Instruction:
1. Please read the two articles. (Kincheloe part 1 & 2)
2. Please choose some of the topics covered in each chapter, provide a brief summary (2-3 sentences) of those topics.
3. Then add your reflections, insights, or relevant experiences, etc. to help illustrate or expand upon the course.
4. This journal should be at least 400 words.
p5-start.cppp5-start.cpp/**
* @author Jane Student
* @cwid 123 45 678
* @class CSci 430, Spring 2018
* @ide Visual Studio Express 2010
* @date November 15, 2018
* @assg prog-04
*
* @description This program implements a simulation of process
* scheduling policies. In this program, we implement round-robin
* scheduling, where the time slice quantum can be specified as
* as a command line parameter. And we also implement shortest
* remaining time (SRT) scheduling policy
*/
#include<stdlib.h>
#include<iostream>
#include<iomanip>
#include<fstream>
#include<string>
#include<list>
usingnamespace std;
// global constants
// I won't test your round robin implementation with more than 20 processes
constint MAX_PROCESSES =20;
constint NO_PROCESS =0;
// Simple structure, holds all of the information about processes, their names
// arrival and service times, that we are to simulate.
typedefstruct
{
string processName;
int arrivalTime;
int serviceTime;
// holds running count of time slices for current time quantum, when
// serviceTime == quantum, time slice is up
int sliceTime;
// holds total number of time steps currently run, when == to
// serviceTime process is done
int totalTime;
// holds time when process finishes, used to calculate final stats,
// like T_r, T_r/T_s
int finishTime;
// a boolean flag, we will set this to true when the process is complete
bool finished;
}Process;
// Process table, holds table of information about processes we are simulating
typedefstruct
{
int numProcesses;
Process* process[MAX_PROCESSES];
}ProcessTable;
/** Create process table
* Allocate memory for a new process table. Load the process
* information from the simulation file into a table with the process
* information needed to perform the simulation. At the same time we
* initialize other information in process table for use in the
* simulation. Return the newly created ProcessTable
*
* @param processFilanem The name (char*) of the file to open and read
* the process information from.
* @param processTable This is actually a return parameter. This
* should be a pointer to an already allocated array of
* Process structure items. We will fill in this structure
* and return the process information.
*
* @returns ProcessTable* The newly allocated and initialized ProcessTable
* structure.
*/
ProcessTable* createProcessTable(char* processFilename)
{
ifstream simprocessfile(processFilename);
ProcessTable* processTable;
int pid;
string processName;
int arrivalTime;
int serviceTime;
// If we can't open file, abort and let ...
Java programs - bubble sort, iterator, linked list, hash set, reverse string,...
Java programs - bubble sort, iterator, linked list, hash set, reverse string,...Sunil Kumar Gunasekaran
VTU Design and Analysis of Algorithms(DAA) Lab Manual by Nithin, VVCE, Mysuru...
VTU Design and Analysis of Algorithms(DAA) Lab Manual by Nithin, VVCE, Mysuru...Nithin Kumar,VVCE, Mysuru
VTU Fourth Semester Design and Analysis of Algorithms Laboratory Manual As per CBCS Scheme by Nithin Kumar, VVCE, MysuruMilitary time and Standard time, JavaOne of the assignments given .pdf
Military time and Standard time, Java
One of the assignments given is to create a program that will essentially test Military time and
Standard time. The Guidlines to the problem are as follows:
Each class should be in a separate file. All related classes should be in a package. The main class
should not be in the package.
Create a Time class (abstract) that includes the following members:
* hour
* minutes
* seconds
* Include overloaded constructor and accessor/mutator functions
Create a MilTime class derived from Time that includes the following members
* milHours
- stores military time as an integer
- 1:00 PM would be stored as 1300
- 4:30 PM would be stored as 1630
* accessor for MilHour
- override frombaseclass
- return milHours
* Mutator for milHour
* Overloaded constructor
- accept milHours and seconds
- convert milHours to hours and minutes and store in class variables
* setTime
- same asconstructor
* getStandHr
- return hour instandardformat
- call base class accessor
Create a TimeClock class derived from MilTime that includes:
* another milHours and seconds variables
- these hours represent an ending time
- the variables inherited from MilTime represent a starting time
* overloaded constructor
- accept start time and end time
- usebaseclass constructor to storestarttime
* accessor and mutator for new class variables
* TimeDifference
- calculate the number of hours betweenstarttime and end time
Create a main program that uses a MilTIme object and a TimeClock object
* Test that all class member functions work properly
* Create the objects based on user input
The MilTime and TimeClock classes should validate that the data entered is within the proper
ranges
* seconds should be between 0 and 59 (inclusive)
* military time should be between 0 and 2359 (inclusive)
- the last two digits should never be between 60 and 99 (inclusive)
* numbers outside the range should be set to the closest valid number
=====================================================================
==================
I have created a total of 4 .java files so far; main.java, Time.java, MilTime.java, and
TimeClock.java
My I decided to work on this problem from top to bottom, starting with Time.java.
/* Time.java */
package demo
abstract public class Time
{
private int hour, minute, second;
// Default constructor.
public Time()
{
hour = 0;
minute = 0;
second = 0;
}
// Overloaded constructor.
public Time(int getHour, int getMinute, int getSecond)
{
hour = getHour;
minute = getMinute;
second = getSecond;
}
// Accessor/Mutator functions.
public int getHour()
{
return hour;
}
public int getMinute()
{
return minute;
}
public int getSecond()
{
return second;
}
public void setHour(int hour)
{
this.hour = hour;
}
public void setMinute(int minute)
{
this.minute = minute;
}
public void setSecond(int second)
{
this.second = second;
}
}
This class I believe I have correct according to the assignment guide. However after this I am
completely stuck.
/* MilTime.java */
package demo
publ.
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000 $35 , 000 , 000 $105 , 000 , 000.
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JAVA.Q4 Create a Time class. This class will represent a point in.pdf
JAVA.
Q4: Create a Time class. This class will represent a point in time, such as a departure time. It
should contain 2 constructors, 2 instance variables (hour and minute), and 10 methods (see
below). All methods but toString should be in terms of the 24 hour format. [30 points]
default constructor: Creates a Time object for 12:00AM.
overloaded constructor: Creates a Time object at a specific hour and minute.
getHour(): Returns an integer representing the hour of the Time object.
getMinute(): Returns an integer representing the minute of the Time object.
addHours(...): Updates the object by moving it forward a number of hours.
addMinute(...): Updates the object by moving it forward a number of minutes. (Hint: Be careful
that you don\'t allow minutes to be more than 59.)
addTime(...): Updates the object by moving it forward by the hour and minute from another
Time object.
getCopy(...): Returns a new Time object that has the same hour and minute of the existing Time
object.
isEarlierThan(...): Returns true if this Time object is earlier than another Time object.
isSameTime(...): Returns true if this Time object is the same as another Time object.
isLaterThan(...): Returns true if this Time object is later than another Time object.
toString(): Returns a string representing the Time object. Uses 12 hour AM/PM format and pads
minutes to be two digits. See the sample output for an example.
Q5: Create a Flight class that uses the Plane and Time class. This class will represent a flight
between two airports, using a specific Plane, and departing at a specific Time. It should contain a
constructor, 7 instance variables (plane, flight number, cost, departure, duration, source,
destination), and 9 methods (see below). [25 points]
overloaded constructor: Creates a Flight object that is setup up with a Plane, a flight number, a
cost, a departure Time, a duration time, a source Airport, and a destination Airport.
getPlane(): Returns the Plane that operates this flight.
getNumber(): Returns the flight number.
getCost(): Returns the flight cost.
getDestination(): Returns the destination Airport.
getDeparture(): Returns the departure Time.
getArrival(): Returns a Time object with the arrival time (computed from the departure time and
duration).
getSource(): Returns a Airport object for the departure location.
toOverviewString(): Returns a String representing an overview of the flight. Use NumberFormat
to display the price. See the sample output for an example.
toDetailedString(): Returns a String representing the flight\'s detail information. See the sample
output for an example.
Included below is an overall UML diagram that describes the three classes you will be
constructing. It provides a useful summary of all of the methods you are expected to implement,
and their corresponding types and visibility. Notice that one private method is listed here
(formatDigits in Time) that isn\'t mentioned above. This is a method that was in our solution, you
may not need.
FileName EX06_1java Programmer import ja.pdf
//FileName: EX06_1.java
//Programmer: ............
import java.util.Scanner;
import java.util.Arrays;
public class EX06_1
{
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
System.out.print("Enter a list of SAT scores: ");
String str = input.nextLine();
String[] items = str.split(" ");
double[] values = new double[items.length];
for (int i = 0; i<items.length; i++)
{
values[i] = Double.parseDouble(items[i]);
}
//count
System.out.print("Count: ");
int count = 0;
for (int i=0; i<values.length; i++)
{
count++;
}
System.out.println((count == values.length) ? values.length : count);
//sum
double sum = 0;
for (int i = 0; i<values.length; i++)
{
sum += values[i];
}
System.out.println("Sum: " + sum);
//mean
double mean = (sum/count);
System.out.println("Mean: " + mean);
//min and max
double min = values[0];
double max = values[0];
for (int i=1; i<values.length; i++)
{
if (values[i] < min)
{
min = values[i];
}
if (values[i] > max)
{
max = values[i];
}
}
System.out.println("Minimum: " + min);
System.out.println("Maximum: " + max);
System.out.println("Range: " + (max-min));
//sorting
Arrays.sort(values);
//median
int position = (count+1) / 2;
if (count%2 != 0)
{
System.out.println("Median: " + values[position-1]);
}
else
{
System.out.println("Median: " + (( values[position-1] + values[position]) / 2));
}
//Standard deviation
double sigmap = 0; //for population
double sigmas = 0; //for sample
for (int i=0; i<values.length; i++)
{
sigmap += (values[i] - mean)*(values[i] - mean);
}
sigmas = sigmap;
//sd for population
sigmap = sigmap / count;
System.out.println("Variance of population: " + sigmap);
double sdp = Math.sqrt(sigmap);
System.out.println("Standard Deviation of population: " + sdp);
//sd for sample
sigmas = sigmas / (count - 1);
System.out.println("Variance of sample: " + sigmas);
double sds = Math.sqrt(sigmas);
System.out.println("Standard Deviation of sample: " + sds);
//Skewness
double sigma = 0;
for (int i=0; i<values.length; i++)
{
sigma += (values[i] - mean)*(values[i] - mean)*(values[i] - mean);
}
double N = (double) count;
double sk = (Math.sqrt(N*(N-1)) / (N-2)) * (sigma / (sdp*sdp*sdp*N));
System.out.println("Skewness: " + sk);
//mode
int repeated = -1;
int[] frq = new int[values.length];
for (int i=0; i<values.length; i++)
{
int occ = 1;
for (int j=i+1; j<values.length; j++)
{
if (values[i] == values[j])
{
frq[j] = repeated;
occ++;
}
}
if (frq[i] != repeated)
{
frq[i] = occ;
}
}
int maxFrq = frq[0];
for (int i=0; i<frq.length; i++)
{
if (frq[i] > maxFrq)
{
maxFrq = frq[i];
}
}
System.out.print("Mode: ");
if (maxFrq == 1)
{
System.out.println("N/A");
}
else
{
for (int i=0; i<frq.length; i++)
{
if (frq[i] == maxFrq)
{
System.out.print(values[i] + " ");
}
}
System.out.print("n");
}
double standardError=sds/Math.sqrt(count);
System.out.println("Standard Error: "+standardError);
}
}
Programming Exercise #06_2:
Instruction:
1. Make a copy of your EX06_01.java file and rename it to EX06_2.java.
2. Open the EX06.
2 BytesC++ course_2014_c3_ function basics¶meters and overloading
This is my C++ course / basic , intermediate and a little advance
I have written the code but cannot complete the assignment please help.pdf
I have written the code but cannot complete the assignment please help me to complete.
Please don't just copy other s answers as your own.
// Insertion sort
/*
#include <ctime>
#include <iostream>
using namespace std;
void insertionSort(int arr[], int n) {
int i, key, j;
for (i = 1; i < n; i++) {
key = arr[i];
j = i - 1;
// Move elements of arr[0..i-1], that are greater than key, to one position ahead of their current
position
while (j >= 0 && arr[j] > key) {
arr[j + 1] = arr[j];
j = j - 1;
}
arr[j + 1] = key;
}
}
int main() {
int n;
cout << "Enter the size of the array: ";
cin >> n;
int arr[n];
cout << "Enter the elements of the array: ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int num = sizeof(arr) / sizeof(arr[0]);
clock_t start, end;
double timetaken;
start = clock();
insertionSort(arr, num);
end = clock();
cout << "Sorted array: \n";
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
cout << endl;
timetaken = ((double) (end - start)) / CLOCKS_PER_SEC;
cout << "Time taken : " << fixed << timetaken << "s" << endl;
return 0;
}
*/
// Shell Sort
/*
#include <ctime>
#include <iostream>
using namespace std;
int shellSort(int arr[], int n) {
for (int gap = n/2; gap > 0; gap /= 2) {
for (int i = gap; i < n; i += 1) {
int temp = arr[i];
int j;
for (j = i; j >= gap && arr[j - gap] > temp; j -= gap)
arr[j] = arr[j - gap];
arr[j] = temp;
}
}
return 0;
}
void printArray(int arr[], int n) {
for (int i=0; i<n; i++)
cout << arr[i] << " ";
}
int main() {
int n;
cout << "Enter the size of the array: ";
cin >> n;
int arr[n];
cout << "Enter the elements of the array: ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int num = sizeof(arr) / sizeof(arr[0]);
clock_t start, end;
double timetaken;
start = clock();
shellSort(arr, num);
end = clock();
cout << "\nArray after sorting: \n";
printArray(arr, n);
cout << endl;
timetaken = (double)(end - start) / CLOCKS_PER_SEC;
cout << "Time taken : " << fixed << timetaken << "s" << endl;
return 0;
}
*/
// MergeSort
/*
#include <ctime>
#include <iostream>
using namespace std;
void merge(int arr[], int l, int m, int r) {
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
int L[n1], R[n2];
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1 + j];
i = 0;
j = 0;
k = l;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
}
else {
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1) {
arr[k] = L[i];
i++;
k++;
}
while (j < n2) {
arr[k] = R[j];
j++;
k++;
}
}
void mergeSort(int arr[], int l, int r) {
if (l < r) {
int m = l + (r - l) / 2;
mergeSort(arr, l, m);
mergeSort(arr, m + 1, r);
merge(arr, l, m, r);
}
}
int main() {
int n;
cout << "Enter the size of the array: ";
cin >> n;
int arr[n];
cout << "Enter the elements of the array: ";
for (int i = 0; i < n; i++) {
cin >> arr[i];
}
int arr_size = sizeof(arr) / sizeof(arr[0]);
clock_t start;
clock_t end;
double timetaken;
start = clock();
mergeSort(arr, 0, arr_size - 1);
end = clock();
timetaken = (double)(end - start) / CLOCKS_PER_SEC;
.
Instruction1. Please read the two articles. (Kincheloe part 1 &.docx
Instruction:
1. Please read the two articles. (Kincheloe part 1 & 2)
2. Please choose some of the topics covered in each chapter, provide a brief summary (2-3 sentences) of those topics.
3. Then add your reflections, insights, or relevant experiences, etc. to help illustrate or expand upon the course.
4. This journal should be at least 400 words.
p5-start.cppp5-start.cpp/**
* @author Jane Student
* @cwid 123 45 678
* @class CSci 430, Spring 2018
* @ide Visual Studio Express 2010
* @date November 15, 2018
* @assg prog-04
*
* @description This program implements a simulation of process
* scheduling policies. In this program, we implement round-robin
* scheduling, where the time slice quantum can be specified as
* as a command line parameter. And we also implement shortest
* remaining time (SRT) scheduling policy
*/
#include<stdlib.h>
#include<iostream>
#include<iomanip>
#include<fstream>
#include<string>
#include<list>
usingnamespace std;
// global constants
// I won't test your round robin implementation with more than 20 processes
constint MAX_PROCESSES =20;
constint NO_PROCESS =0;
// Simple structure, holds all of the information about processes, their names
// arrival and service times, that we are to simulate.
typedefstruct
{
string processName;
int arrivalTime;
int serviceTime;
// holds running count of time slices for current time quantum, when
// serviceTime == quantum, time slice is up
int sliceTime;
// holds total number of time steps currently run, when == to
// serviceTime process is done
int totalTime;
// holds time when process finishes, used to calculate final stats,
// like T_r, T_r/T_s
int finishTime;
// a boolean flag, we will set this to true when the process is complete
bool finished;
}Process;
// Process table, holds table of information about processes we are simulating
typedefstruct
{
int numProcesses;
Process* process[MAX_PROCESSES];
}ProcessTable;
/** Create process table
* Allocate memory for a new process table. Load the process
* information from the simulation file into a table with the process
* information needed to perform the simulation. At the same time we
* initialize other information in process table for use in the
* simulation. Return the newly created ProcessTable
*
* @param processFilanem The name (char*) of the file to open and read
* the process information from.
* @param processTable This is actually a return parameter. This
* should be a pointer to an already allocated array of
* Process structure items. We will fill in this structure
* and return the process information.
*
* @returns ProcessTable* The newly allocated and initialized ProcessTable
* structure.
*/
ProcessTable* createProcessTable(char* processFilename)
{
ifstream simprocessfile(processFilename);
ProcessTable* processTable;
int pid;
string processName;
int arrivalTime;
int serviceTime;
// If we can't open file, abort and let ...
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VTU Fourth Semester Design and Analysis of Algorithms Laboratory Manual As per CBCS Scheme by Nithin Kumar, VVCE, MysuruMilitary time and Standard time, JavaOne of the assignments given .pdf
Military time and Standard time, Java
One of the assignments given is to create a program that will essentially test Military time and
Standard time. The Guidlines to the problem are as follows:
Each class should be in a separate file. All related classes should be in a package. The main class
should not be in the package.
Create a Time class (abstract) that includes the following members:
* hour
* minutes
* seconds
* Include overloaded constructor and accessor/mutator functions
Create a MilTime class derived from Time that includes the following members
* milHours
- stores military time as an integer
- 1:00 PM would be stored as 1300
- 4:30 PM would be stored as 1630
* accessor for MilHour
- override frombaseclass
- return milHours
* Mutator for milHour
* Overloaded constructor
- accept milHours and seconds
- convert milHours to hours and minutes and store in class variables
* setTime
- same asconstructor
* getStandHr
- return hour instandardformat
- call base class accessor
Create a TimeClock class derived from MilTime that includes:
* another milHours and seconds variables
- these hours represent an ending time
- the variables inherited from MilTime represent a starting time
* overloaded constructor
- accept start time and end time
- usebaseclass constructor to storestarttime
* accessor and mutator for new class variables
* TimeDifference
- calculate the number of hours betweenstarttime and end time
Create a main program that uses a MilTIme object and a TimeClock object
* Test that all class member functions work properly
* Create the objects based on user input
The MilTime and TimeClock classes should validate that the data entered is within the proper
ranges
* seconds should be between 0 and 59 (inclusive)
* military time should be between 0 and 2359 (inclusive)
- the last two digits should never be between 60 and 99 (inclusive)
* numbers outside the range should be set to the closest valid number
=====================================================================
==================
I have created a total of 4 .java files so far; main.java, Time.java, MilTime.java, and
TimeClock.java
My I decided to work on this problem from top to bottom, starting with Time.java.
/* Time.java */
package demo
abstract public class Time
{
private int hour, minute, second;
// Default constructor.
public Time()
{
hour = 0;
minute = 0;
second = 0;
}
// Overloaded constructor.
public Time(int getHour, int getMinute, int getSecond)
{
hour = getHour;
minute = getMinute;
second = getSecond;
}
// Accessor/Mutator functions.
public int getHour()
{
return hour;
}
public int getMinute()
{
return minute;
}
public int getSecond()
{
return second;
}
public void setHour(int hour)
{
this.hour = hour;
}
public void setMinute(int minute)
{
this.minute = minute;
}
public void setSecond(int second)
{
this.second = second;
}
}
This class I believe I have correct according to the assignment guide. However after this I am
completely stuck.
/* MilTime.java */
package demo
publ.
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2 BytesC++ course_2014_c3_ function basics¶meters and overloading
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I have written the code but cannot complete the assignment please help.pdf
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Instruction1. Please read the two articles. (Kincheloe part 1 &.docx
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Military time and Standard time, JavaOne of the assignments given .pdf
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metals to build the turbines B. Legal fees to defend the wind farm in court challenges from the
community C. Impact on the 'pe'ape'a (Native Hawaiian bat) population from interactions with
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electrical connections to HECO's utility grid F. Impact to the health and well-being of Kahuku
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a bill that sets a goal for 100% renewable energy production by any electric utility (e.g., HECO)
in the state by 2045. Assuming zero government intervention, which of the following would be
considered external costs? Steel and other metals to build the turbines B. Legal fees to defend the
wind farm in court challenges from the community C. Impact on the 'pe'ape'a (Native Hawaiian
bat) population from interactions with the turbines D. Loss of access to culturally-significant
land where the turbines sit E. Installation of electrical connections to HECO's utility grid F.
Impact to the health and well-being of Kahuku residents and students the agricultural land
owners H. Physical capital maintenance I. Low frequency vibration felt from the turbines'
operation J. Living with 20 of the 50 wind turbines on O 'ahu in their North Shore community
Based on your answers to the two questions above, A. the private cost is greater than the social
cost of the N Pua Makani wind project. B. the social cost is greater than the private cost of the N
Pua Makani wind project. C. the private cost is equal to the social cost of the N Pua Makani wind
project. D. it cannot be determined if either the private or social cost is larger for the N Pua
Makani wind project without additional information. Based only on the information given in this
assignment about the N Pua Makani wind farm, it can be concluded that it creates a externality.
A. Negative B. Positive.
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the Hague
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Responses:
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YouTube:- youtube.com/@mulesoftmysore
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- 1. import java.util.Arrays; import java.io.PrintWriter; import java.io.File; import java.util.Scanner; public class SortTimes { /* convert nanos to seconds */ static final double ONE_BILLION = 1_000_000_000.; /* sizes of input range should range up to 512k elements */ static int[] sizes = new int[32]; public static void main(String[] args) throws Exception { int sIdx = 0; long startTime; long endTime; double time; int size = 128*1024/sizes.length; /* let's make size smaller, comment out this next line and see what happens */ size = 16*1024/sizes.length; for( int i=0; i<sizes.length; i++ ) { sizes[i] = (i+1)*size; } Scanner inp = new Scanner(System.in);
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- 3. pw.print(A.length + ", "); System.out.println( "Starting bubble sort ---------------- Size = " + B.length); startTime = System.nanoTime(); bubbleSort(A); endTime = System.nanoTime(); time = (endTime - startTime) / ONE_BILLION; System.out.println( "Bubble sort finished - time = " + time + " seconds" ); pw.print(time + ", "); printArray(A); System.out.println(); System.out.println( "Starting merge sort ---------------- Size = " + C.length); startTime = System.nanoTime(); mergeSort(B); endTime = System.nanoTime(); time = (endTime - startTime) / ONE_BILLION; System.out.println( "Merge sort finished - time = " + time + " seconds."); pw.print(time + ", "); printArray(B); System.out.println(); System.out.println( "Starting quick sort ---------------- Size = " + D.length); startTime = System.nanoTime(); quickSort(C); endTime = System.nanoTime();
- 4. time = (endTime - startTime) / ONE_BILLION; System.out.println( "quickSort finished - time = " + time + " seconds." ); pw.print(time + ", "); printArray(C); System.out.println(); System.out.println( "Starting java's Array sort ---------------- Size = " + D.length); startTime = System.nanoTime(); javaSort(D); endTime = System.nanoTime(); time = (endTime - startTime) / ONE_BILLION; System.out.println( "Java's Array sort finished - time = " + time + " seconds." ); pw.print(time + ", "); printArray(D); System.out.println(); System.out.println( "Starting bubble sort on sorted array ---------------- Size = " + B.length); startTime = System.nanoTime(); bubbleSort(E); endTime = System.nanoTime(); time = (endTime - startTime) / ONE_BILLION; System.out.println( "Bubble sort finished - time = " + time + " seconds" ); pw.print(time + ", "); printArray(E); System.out.println();
- 5. System.out.println( "Starting merge sort on sorted array ---------------- Size = " + C.length); startTime = System.nanoTime(); mergeSort(E); endTime = System.nanoTime(); time = (endTime - startTime) / ONE_BILLION; System.out.println( "Merge sort finished - time = " + time + " seconds."); pw.print(time + ", "); printArray(E); System.out.println(); System.out.println( "Starting quick sort on sorted array ---------------- Size = " + D.length); startTime = System.nanoTime(); quickSort(E); endTime = System.nanoTime(); time = (endTime - startTime) / ONE_BILLION; System.out.println( "quickSort finished - time = " + time + " seconds." ); pw.print(time + ", "); printArray(E); System.out.println(); System.out.println( "Starting java's Array sorted array ---------------- Size = " + D.length); startTime = System.nanoTime(); javaSort(E); endTime = System.nanoTime(); time = (endTime - startTime) / ONE_BILLION;
- 6. System.out.println( "Java's Array sort finished - time = " + time + " seconds." ); pw.println(time); printArray(E); System.out.println(); sIdx = sIdx + 1; } pw.close(); } /* prints array or a portion of it if it's too large */ public static void printArray(int[] X) { if( X == null ) return; System.out.print("[ "); int i=0; for( ; i<15 && i<X.length; i++ ) { System.out.print( X[i] + " " ); } if( i < X.length ) System.out.print("... "); System.out.println("]"); } /* Finish Methods Below*/ public static void bubbleSort(int[] X)
- 7. { } public static void mergeSort(int[] X) { } private static void merge(int[] B, int[] C, int[] A ) { } /* this is simply the sort used by Arrays class */ public static void javaSort(int[] X) { } public static void quickSort(int[] X) { quickSort(X, 0, X.length-1); } private static void quickSort(int[] X, int s, int e) { // check our stopping condition // get pivot point from partition method // call quickSort on left portion and quickSort on right portion } private static int partition( int[]X, int s, int e )