Write a java program that reads a number N as input and displays the.pdf
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The program takes a number N as input, then uses stacks to print N lines of output. It prints the numbers from N to 1, with each number centered and surrounded by spaces that decrease by 2 each line. It uses methods to generate the outer and inner spaces for each line based on the line number and input N.
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![Write a java program that reads a number N as input and displays the following output using
stacks with parameters. Example o Input: Enter N: o Output: 5
Solution
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Stacks {
public static void main(String[] args) throws IOException {
// used variables
BufferedReader reader;
String input,outerSpaces,innerSpaces;
Integer number;
// taking input
System.out.print("Enter N : ");
reader = new BufferedReader(new InputStreamReader(System.in));
input = reader.readLine();
number = Integer.parseInt(input);
// printing upper stack
for(int i=1;i0;i--){
outerSpaces = getOuterSpaces(i);
innerSpaces = getInnerSpaces(number, i);
System.out.print(outerSpaces);
System.out.print(i);
System.out.print(innerSpaces);
if(i != number){
System.out.print(i);
}
System.out.print(outerSpaces);
System.out.println();
}
}](https://image.slidesharecdn.com/writeajavaprogramthatreadsanumbernasinputanddisplaysthe-230708181102-6b272f75/85/Write-a-java-program-that-reads-a-number-N-as-input-and-displays-the-pdf-1-320.jpg)
Recommended
Accepts number and base and returns result in integerimport jav.pdf
// Accepts number and base and returns result in integer
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Base {
public static void main(String[] args) throws IOException {
System.out.println(\"Enter number: \");
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String st=br.readLine();
System.out.println(\"Enter base\");
String st1=br.readLine();
int n=Integer.parseInt(st);
int b=Integer.parseInt(st1);
int sum=0;
int i=0;
while(n!=0)
{
sum=(int) (sum+(n%10)*Math.pow(b,i++));
n=n/10;
}
System.out.println(sum);
}
}
Solution
// Accepts number and base and returns result in integer
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Base {
public static void main(String[] args) throws IOException {
System.out.println(\"Enter number: \");
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String st=br.readLine();
System.out.println(\"Enter base\");
String st1=br.readLine();
int n=Integer.parseInt(st);
int b=Integer.parseInt(st1);
int sum=0;
int i=0;
while(n!=0)
{
sum=(int) (sum+(n%10)*Math.pow(b,i++));
n=n/10;
}
System.out.println(sum);
}
}.
Below is my code for a line editor import java.io.BufferedReader;.pdf
Below is my code for a line editor:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Main {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("Input file: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
lines.add(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}
The output should look like this:
EDIT: textin.txt (user inputs textin.txt)
1> The first line
2>
3> And another line
4> I 3
3> The second line
4> One more line
5> L
1> The first line
2>
3> The second line
4> One more line
5> And another line //This is now line 5; not 3;
5> D 2 // line 5, since L was issued from line 5;
4> L //line 4, since one line was deleted;
1> The first line
2> The second line
3> One more line
4> And another line
5> E
Here is a screenshot of the errors I am getting and need help fixing them: FileEdit [Java
Application] /Users/caseysharp/.p2/pool/plugins/org.eclipse.justj.openjdk.hotspot.jre.full.mac
EDIT: textin.txt 1> The First line 1> 2> And another line 3> I 3 3> The second line Exception in
thread "main" java..
Here is my code for a linefile editor import java.io.BufferedRea.pdf
Here is my code for a line/file editor:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class FileEdit {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("EDIT: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
if (count >= lines.size()) {
for(int i = lines.size(); i <= count; i++) {
lines.add(null);
}
}
lines.set(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}
Here is my output:
EDIT: textin.txt
1> The First Line
1>
2> And another line
3> I 3
3> The second line
4> One more line
5> L
1> null
2> null
3> The second line
4> One more line
5> D 2
3> L
1> null
2> The second line
3> One more line
4> E
How can I keep it from printing "null" after ">" and it just be blank. I need the output to look like
this:
1> The first line
2>
3> And another line
4> I 3
3> The second line
4> One more line
5> L
1> The first line
2>
3> The second line
4> One more line
5> And another line
5> D 2
4> L
1> The first line
2> The second line
3> One more line
4> And another line
5> E.
CODEimport java.util.; public class test { public static voi.pdf
CODE:
import java.util.*;
public class test {
public static void main(String args[]) {
int num, sum=0;
Scanner input=new Scanner(System.in);
System.out.print(\"Input num : \");
num=input.nextInt();
while(num != 13) {
sum=sum+num;
System.out.print(\"Input num : \");
num=input.nextInt();
}
sum=sum+num;
System.out.println(\"Sum = \"+sum);
}
}
save the above file as test.java
Output:
javac test.java
java test
Input num :
1
Input num :
2
Input num :
3
Input num :
4
Input num :
5
Input num :
6
Input num :
7
Input num :
8
Input num :
9
Input num :
0
Input num :
13
Sum = 58
Solution
CODE:
import java.util.*;
public class test {
public static void main(String args[]) {
int num, sum=0;
Scanner input=new Scanner(System.in);
System.out.print(\"Input num : \");
num=input.nextInt();
while(num != 13) {
sum=sum+num;
System.out.print(\"Input num : \");
num=input.nextInt();
}
sum=sum+num;
System.out.println(\"Sum = \"+sum);
}
}
save the above file as test.java
Output:
javac test.java
java test
Input num :
1
Input num :
2
Input num :
3
Input num :
4
Input num :
5
Input num :
6
Input num :
7
Input num :
8
Input num :
9
Input num :
0
Input num :
13
Sum = 58.
Recommended
Accepts number and base and returns result in integerimport jav.pdf
// Accepts number and base and returns result in integer
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Base {
public static void main(String[] args) throws IOException {
System.out.println(\"Enter number: \");
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String st=br.readLine();
System.out.println(\"Enter base\");
String st1=br.readLine();
int n=Integer.parseInt(st);
int b=Integer.parseInt(st1);
int sum=0;
int i=0;
while(n!=0)
{
sum=(int) (sum+(n%10)*Math.pow(b,i++));
n=n/10;
}
System.out.println(sum);
}
}
Solution
// Accepts number and base and returns result in integer
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Base {
public static void main(String[] args) throws IOException {
System.out.println(\"Enter number: \");
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String st=br.readLine();
System.out.println(\"Enter base\");
String st1=br.readLine();
int n=Integer.parseInt(st);
int b=Integer.parseInt(st1);
int sum=0;
int i=0;
while(n!=0)
{
sum=(int) (sum+(n%10)*Math.pow(b,i++));
n=n/10;
}
System.out.println(sum);
}
}.
Below is my code for a line editor import java.io.BufferedReader;.pdf
Below is my code for a line editor:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Main {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("Input file: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
lines.add(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}
The output should look like this:
EDIT: textin.txt (user inputs textin.txt)
1> The first line
2>
3> And another line
4> I 3
3> The second line
4> One more line
5> L
1> The first line
2>
3> The second line
4> One more line
5> And another line //This is now line 5; not 3;
5> D 2 // line 5, since L was issued from line 5;
4> L //line 4, since one line was deleted;
1> The first line
2> The second line
3> One more line
4> And another line
5> E
Here is a screenshot of the errors I am getting and need help fixing them: FileEdit [Java
Application] /Users/caseysharp/.p2/pool/plugins/org.eclipse.justj.openjdk.hotspot.jre.full.mac
EDIT: textin.txt 1> The First line 1> 2> And another line 3> I 3 3> The second line Exception in
thread "main" java..
Here is my code for a linefile editor import java.io.BufferedRea.pdf
Here is my code for a line/file editor:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class FileEdit {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("EDIT: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
if (count >= lines.size()) {
for(int i = lines.size(); i <= count; i++) {
lines.add(null);
}
}
lines.set(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}
Here is my output:
EDIT: textin.txt
1> The First Line
1>
2> And another line
3> I 3
3> The second line
4> One more line
5> L
1> null
2> null
3> The second line
4> One more line
5> D 2
3> L
1> null
2> The second line
3> One more line
4> E
How can I keep it from printing "null" after ">" and it just be blank. I need the output to look like
this:
1> The first line
2>
3> And another line
4> I 3
3> The second line
4> One more line
5> L
1> The first line
2>
3> The second line
4> One more line
5> And another line
5> D 2
4> L
1> The first line
2> The second line
3> One more line
4> And another line
5> E.
CODEimport java.util.; public class test { public static voi.pdf
CODE:
import java.util.*;
public class test {
public static void main(String args[]) {
int num, sum=0;
Scanner input=new Scanner(System.in);
System.out.print(\"Input num : \");
num=input.nextInt();
while(num != 13) {
sum=sum+num;
System.out.print(\"Input num : \");
num=input.nextInt();
}
sum=sum+num;
System.out.println(\"Sum = \"+sum);
}
}
save the above file as test.java
Output:
javac test.java
java test
Input num :
1
Input num :
2
Input num :
3
Input num :
4
Input num :
5
Input num :
6
Input num :
7
Input num :
8
Input num :
9
Input num :
0
Input num :
13
Sum = 58
Solution
CODE:
import java.util.*;
public class test {
public static void main(String args[]) {
int num, sum=0;
Scanner input=new Scanner(System.in);
System.out.print(\"Input num : \");
num=input.nextInt();
while(num != 13) {
sum=sum+num;
System.out.print(\"Input num : \");
num=input.nextInt();
}
sum=sum+num;
System.out.println(\"Sum = \"+sum);
}
}
save the above file as test.java
Output:
javac test.java
java test
Input num :
1
Input num :
2
Input num :
3
Input num :
4
Input num :
5
Input num :
6
Input num :
7
Input num :
8
Input num :
9
Input num :
0
Input num :
13
Sum = 58.
import java.io.BufferedReader;import java.io.File;import java.io.pdf
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class FileEdit {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("EDIT: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
if (count >= lines.size()) {
for(int i = lines.size(); i <= count; i++) {
lines.add(null);
}
}
lines.set(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}
Here is my output:
EDIT: textin.txt
1> The First Line
1>
2> And another line
3> I 3
3> The second line
4> One more line
5> L
1> null
2> null
3> The second line
4> One more line
5> D 2
3> L
1> null
2> The second line
3> One more line
4> E
How can I keep it from printing "null" after ">" and it just be blank. I need the output to look like
this:
1> The first line
2>
3> And another line
4> I 3
3> The second line
4> One more line
5> L
1> The first line
2>
3> The second line
4> One more line
5> And another line
5> D 2
4> L
1> The first line
2> The second line
3> One more line
4> And another line
5> E.
In JavaWrite a program that reads a text file that contains a gra.pdf
In Java:
Write a program that reads a text file that contains a grade (for instance, 87) on each line.
Calculate and print the average of the grades.
Solution
import java.io.BufferedReader;
import java.io.FileReader; // importing filereader
import java.io.IOException; // importing file io exceptions
public class grade {
public static void main(String[] args) {
BufferedReader br = null;
double count=0;
int noOfgrades=0;
try {
String sLine;
br = new BufferedReader(new FileReader(\"sample.txt\")); // reading file using buffer
reader
while ((sLine = br.readLine()) != null) { // rading line by line
noOfgrades=noOfgrades+1; // getting number of grades
int num=Integer.parseInt(sLine); // convert string into integer
count=count+num; // adding each grade to count
}
System.out.println(\"count:\\t\"+count);
System.out.println(\"noOfgrades:\\t\"+noOfgrades);
System.out.println(\"Average of the grades:\\t\"+(count/noOfgrades));
} catch (IOException e) {
e.printStackTrace(); // catch exception
} finally {
try {
if (br != null)br.close(); // closing buffer
} catch (IOException ex) {
ex.printStackTrace();// printing exceptions
}
}
}
}
/********** out file******
fileName:sample.txt
file contains \"
23
34
56
7
88
89
52
34
65
\"
*/
/************output*************
count: 448.0
noOfgrades: 9
Average of the grades: 49.77
*/.
Please help me with a UML class diagram for the following code im.pdf
Please help me with a UML class diagram for the following code:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class FileEdit {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("EDIT: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
if (count >= lines.size()) {
for(int i = lines.size(); i <= count; i++) {
lines.add(null);
}
}
lines.set(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}.
mport java.io.; class Student { int rollno; String name; .pdf
mport java.io.*;
class Student
{
int rollno;
String name;
int number_of_subjects;
int mark[];
Student(int roll,String stud_name,int noofsub)throws IOException
{
rollno=roll;
name=stud_name;
number_of_subjects= noofsub;
getMarks(noofsub);
}
public void getMarks(int nosub ) throws IOException
{
mark=new int[nosub];
BufferedReader br= new BufferedReader (new InputStreamReader(System.in));
for (int i=0; i “);
mark[i]=Integer.parseInt(br.readLine());
System.out.println(“”);
}
}
public void calculateMarks()
{
double percentage=0;
String grade;
int tmarks=0;
for (int i=0;i “+rollno);
System.out.println(“Name Of Student is :=> “+name);
System.out.println(“Number Of Subject :=> “+number_of_subjects);
System.out.println(“Percentage Is :=> “+percentage);
if (percentage>=70)
System.out.println(“Grade Is First Class With Distinction “);
else if(percentage>=60 && percentage<70)
System.out.println(“Grade Is First Class”);
else if(percentage>=50 && percentage<60)
System.out.println(“Grade Is Second Class”);
else if(percentage>=40 && percentage<50)
System.out.println(“Grade Is Pass Class”);
else
System.out.println(“You Are Fail”);
}
}
class StudentDemo
{
public static void main(String args[])throws IOException
{
int rno,no,nostud;
String name;
BufferedReader br= new BufferedReader (new InputStreamReader(System.in));
System.out.println(“Enter How many Students:=> “);
nostud=Integer.parseInt(br.readLine());
Student s[]=new Student[nostud];
for(int i=0;i “);
rno=Integer.parseInt(br.readLine());
System.out.println(“Enter Name:=> “);
name=br.readLine();
System.out.println(“Enter No of Subject:=> “);
no=Integer.parseInt(br.readLine());
s[i]=new Student(rno,name,no);
}
for(int i=0;i
Solution
mport java.io.*;
class Student
{
int rollno;
String name;
int number_of_subjects;
int mark[];
Student(int roll,String stud_name,int noofsub)throws IOException
{
rollno=roll;
name=stud_name;
number_of_subjects= noofsub;
getMarks(noofsub);
}
public void getMarks(int nosub ) throws IOException
{
mark=new int[nosub];
BufferedReader br= new BufferedReader (new InputStreamReader(System.in));
for (int i=0; i “);
mark[i]=Integer.parseInt(br.readLine());
System.out.println(“”);
}
}
public void calculateMarks()
{
double percentage=0;
String grade;
int tmarks=0;
for (int i=0;i “+rollno);
System.out.println(“Name Of Student is :=> “+name);
System.out.println(“Number Of Subject :=> “+number_of_subjects);
System.out.println(“Percentage Is :=> “+percentage);
if (percentage>=70)
System.out.println(“Grade Is First Class With Distinction “);
else if(percentage>=60 && percentage<70)
System.out.println(“Grade Is First Class”);
else if(percentage>=50 && percentage<60)
System.out.println(“Grade Is Second Class”);
else if(percentage>=40 && percentage<50)
System.out.println(“Grade Is Pass Class”);
else
System.out.println(“You Are Fail”);
}
}
class StudentDemo
{
public static void main(String args[])throws IOException
{
int rno,no,nostud;
String name;
BufferedReader br= new BufferedReader (new InputStre.
import java.io.BufferedReader; import java.io.IOException; impor.pdf
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Hw3_4
{
public static void main(String arg[]) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println(\"Enter the Name Of Employee\");
String str = br.readLine();
str= str.toUpperCase();
str = str.replaceAll(\"\\\\s\",\"\"); //for removing space
char[] c=str.toCharArray();//code is used for converting the String into Array for one-by-one
matching
int lp=0;
int count=0;
for(int i=0;i
Solution
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Hw3_4
{
public static void main(String arg[]) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println(\"Enter the Name Of Employee\");
String str = br.readLine();
str= str.toUpperCase();
str = str.replaceAll(\"\\\\s\",\"\"); //for removing space
char[] c=str.toCharArray();//code is used for converting the String into Array for one-by-one
matching
int lp=0;
int count=0;
for(int i=0;i.
import java.util.;public class Program{public static void.pdf
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println(x);
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain
using the most basic codes.
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println(x);
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain
using the most basic codes.
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println(x);
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain
using the most basic codes.
Solution
Hi,
I have modified the code. it is working as expected now. Highlighted the code changes below.
Issue here is with you have written this statement twice System.out.println(x); which causes the
issue.
Program.java
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println();
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Output:
Enter amount
10
Amount is: 10
X
XX
XXX
XXXX
XXXXX
XXXXXX
XXXXXXX
XXXXXXXX
XXXXXXXXX
XXXXXXXXXX.
JAVA Q2- Write a program that reads strings from the user and writes t.docx
JAVA
Q2: Write a program that reads strings from the user and writes them to an output file called userStrings.txt. Stop processing when the user enters the string \"DONE\". Do not write the sentinel string (\"DONE\") to the output file. [15 points] Hint: For a complete example of file IO, please see the attached TestData program.
Sample Output:
A string.
Another string.
Yet more text...
DONE
Solution
Answer
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Scanner;
class WriteStringFile {
public static void main(String[] args) {
File file = new File(\"D:/userFile.txt\");
String data;
Scanner input = new Scanner(System.in);
data = input.readLine();
try (FileOutputStream fop = new FileOutputStream(file)) {
// if file doesn\'t exists, then create it
if (!file.exists()) {
file.createNewFile();
}
if (data != \"DONE\") {
// get the content in bytes
byte[] contentInBytes = content.getBytes();
fop.write(contentInBytes);
fop.flush();
fop.close();
} else {
System.out.println(\"User Enter the value Done.. So unable to write into file\");
}
} catch (IOException e) {
e.printStackTrace();
}
}
}
.
InsertionSorting.javaimport java.util.Arrays; import java.util.S.pdf
InsertionSorting.java
import java.util.Arrays;
import java.util.Scanner;
public class InsertionSorting {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Please enter number of letters:\");
int n = scan.nextInt();
char letters[] = new char[n];
for(int i=0; i -1) && ( charArray [i] > key ) ) {
charArray [i+1] = charArray [i];
i--;
}
charArray[i+1] = key;
}
}
}
Output:
Please enter number of letters:
5
Enter letter:
f
Enter letter:
s
Enter letter:
w
Enter letter:
a
Enter letter:
b
Before Sorting: [f, s, w, a, b]
After insertion Sorting: [a, b, f, s, w]
BubbleSorting.java
import java.util.Arrays;
import java.util.Scanner;
public class BubbleSorting {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Please enter number of letters:\");
int n = scan.nextInt();
char letters[] = new char[n];
for(int i=0; i array [1] swap it.
* 3. Compare array[1] & array[2]
* 4. If array[1] > array[2] swap it.
* ...
* 5. Compare array[n-1] & array[n]
* 6. if [n-1] > array[n] then swap it.
*
* After this step we will have largest element at the last index.
*
* Repeat the same steps for array[1] to array[n-1]
*
*/
int n = charArray.length;
char temp;
for(int i=0; i < n; i++){
for(int j=1; j < (n-i); j++){
if(charArray[j-1] > charArray[j]){
//swap the elements!
temp = charArray[j-1];
charArray[j-1] = charArray[j];
charArray[j] = temp;
}
}
}
}
}
Output:
Please enter number of letters:
5
Enter letter:
f
Enter letter:
d
Enter letter:
w
Enter letter:
a
Enter letter:
b
Before Sorting: [f, d, w, a, b]
After Bubble Sorting: [a, b, d, f, w]
Solution
InsertionSorting.java
import java.util.Arrays;
import java.util.Scanner;
public class InsertionSorting {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Please enter number of letters:\");
int n = scan.nextInt();
char letters[] = new char[n];
for(int i=0; i -1) && ( charArray [i] > key ) ) {
charArray [i+1] = charArray [i];
i--;
}
charArray[i+1] = key;
}
}
}
Output:
Please enter number of letters:
5
Enter letter:
f
Enter letter:
s
Enter letter:
w
Enter letter:
a
Enter letter:
b
Before Sorting: [f, s, w, a, b]
After insertion Sorting: [a, b, f, s, w]
BubbleSorting.java
import java.util.Arrays;
import java.util.Scanner;
public class BubbleSorting {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Please enter number of letters:\");
int n = scan.nextInt();
char letters[] = new char[n];
for(int i=0; i array [1] swap it.
* 3. Compare array[1] & array[2]
* 4. If array[1] > array[2] swap it.
* ...
* 5. Compare array[n-1] & array[n]
* 6. if [n-1] > array[n] then swap it.
*
* After this step we will have largest elem.
Write a program that will count the number of characters- words- and l.docx
Write a program that will count the number of characters, words, and lines in a file. Words are separated by whitespace characters. The file name should be passed as a command-line argument, as shown;
c:\\exercise>java Exercise12_13 Loan.java
File Loan.java has
1919 characters
210 words
71 lines
c:\\exercise>
Solution
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
import java.util.StringTokenizer;
public class ass1
{
public static void main(String[] args) throws FileNotFoundException
{
//Â Â Scanner console = new Scanner(System.in);
System.out.println(args[0]);
//System.out.println(args[1]);
//System.out.println(args[2]);
String inputFile=args[0];
File file = new File(inputFile);
Scanner in = new Scanner(file);
int words = 0;
int lines = 0;
int chars = 0;
while(in.hasNextLine()) {
lines++;
String line = in.nextLine();
chars += line.length();
words += new StringTokenizer(line, \" \").countTokens();
}
System.out.println(\"characters are \"+chars);
System.out.println(\"words are \"+words);
System.out.println(\"lines are \"+lines);
}
}
.
Javaa. The mean of a list of numbers is its arithmetic average. T.pdf
Java:
a. The mean of a list of numbers is its arithmetic average. The median of a list is its middle value
when the values are placed in order. For example, if a list contains 1, 4, 7, 8, and 10, then the
mean is 6 and the median is 7. Write an application that allows you to enter five integers and
displays the values, their mean, and their median.
Save the file as MeanMedian.java.
b. Revise the MeanMedian class so that the user can enter any number of values up to 20. If the
list has an even number of values, the median is the numeric average of the values in the two
middle positions. Save the file as
MeanMedian2.java
Solution
Ans a):
import java.util.*;
import java.lang.*;
import java.io.*;
class MeanMedian
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner scan=new Scanner(System.in);
int arr[]=new int[5];
System.out.println(\"Enter the integers :\"); //asking the user to enter 5 integers
for(int i=0;i<5;i++)
arr[i]=scan.nextInt();
int sum=0;
for(int i=0;i<5;i++)
sum=sum+arr[i];
int mean=sum/5; //calculating mean
int median=arr[2]; //calculating median
System.out.println(\"Integers entered by the user \");
for(int i=0;i<5;i++)
System.out.print(arr[i]+\" \"); //printing values entered by user
System.out.println();
System.out.println(\"Mean = \"+mean); //printing mean
System.out.println(\"Median = \"+median); //printing median
}
}
******OUTPUT******
Enter the integers :
1 4 7 8 10
Integers entered by the user
1 4 7 8 10
Mean = 6
Median = 7
******OUTPUT******
Ansb):
import java.util.*;
import java.lang.*;
import java.io.*;
class MeanMedian //class MeanMedian
{
public static void main (String[] args) throws java.lang.Exception
{
int sum=0; //variable to store the sum of values
int mean=0; //variable to store the mean of values
int median=0; //variable to store the median of values
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter the number of values upto you want to enter,it must not be greater
than 20 :\"); //asking the user for no.of values
int n=scan.nextInt(); //store size in n
int arr[]=new int[n]; //array to store integers
System.out.println(\"Enter the integers\"); //asking the user to enter the integers
for(int i=0;i.
(JAVA NetBeans) Write a Java program able to perform selection sort-So.docx
(JAVA NetBeans) Write a Java program able to perform selection sort.
Solution
import java.util.Scanner;
/**
* @author
*
*/
public class SelectionSort {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int arr[] = setArray();
System.out.println(\"\ Before Sorting:\");
printArray(arr);
arr = selectionSort(arr);
System.out.println(\"\ After Sorting:\");
printArray(arr);
}
/**
* reads and stores the data in the one-dimensional array
*/
/**
* @return
*/
public static int[] setArray() {
int arr[] = new int[12];
System.out.println(\"Enter 12 integers.\");
Scanner scanner = new Scanner(System.in);
for (int i = 0; i < arr.length; i++) {
arr[i] = scanner.nextInt();
}
return arr;
}
/**
* prints the data in the one-dimensional array
*/
public static void printArray(int[] arr) {
System.out.println();
for (int i = 0; i < arr.length; i++) {
System.out.print(arr[i] + \" \");
}
System.out.println();
}
/**
* method will sort the array in ascending order
*/
public static int[] selectionSort(int[] arr) {
for (int i = 0; i < arr.length - 1; i++) {
int index = i;
for (int j = i + 1; j < arr.length; j++)
if (arr[j] < arr[index])
index = j;
int smallerNumber = arr[index];
arr[index] = arr[i];
arr[i] = smallerNumber;
}
return arr;
}
}
OUTPUT:
Enter 12 integers.
4 23 65 34 84 37 12 17 24 36 82 51
Before Sorting:
4 23 65 34 84 37 12 17 24 36 82 51
After Sorting:
4 12 17 23 24 34 36 37 51 65 82 84
.
IN JAVA Write a program to create a binary data file named RandomInt.pdf
IN JAVA Write a program to create a binary data file named RandomInt.dat that consists of 10
random integer numbers using DataOutputStream. Use DataInputStream to read the integers
from RandomInt.dat just created and print them on the console in the same program.
Solution
DataInputStreamTest.java
import java.io.DataInputStream;
import java.io.DataOutputStream;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Random;
public class DataInputStreamTest {
public static void main(String[] args) {
String strFilePath = \"D:\\\\RandomInt.dat\";
Random r = new Random();
try
{
FileOutputStream outputStr = new FileOutputStream(strFilePath);
DataOutputStream outputStrData = new DataOutputStream(outputStr);
for(int i=0; i<10; i++){
outputStrData.writeInt(r.nextInt(100)+1);
outputStrData.writeChar(\'\\t\');
}
outputStrData.flush();
outputStrData.close();
outputStr.close();
FileInputStream inputStr = new FileInputStream(strFilePath);
DataInputStream inputStrData = new DataInputStream(inputStr);
int count = 0;
while(count<10){
int i = inputStrData.readInt();
inputStrData.readChar();
count++;
System.out.println( i);
}
outputStrData.flush();
inputStrData.close();
outputStrData.close();
}
catch(FileNotFoundException fe)
{
System.out.println(\"FileNotFoundException : \" + fe);
}
catch(IOException ioe)
{
System.out.println(\"IOException : \" + ioe);
}
}
}
Output:
65
8
95
69
1
77
16
97
85
1.
I studied analysis in Japanese textbook. Though I know many theorems.pdf
I studied analysis in Japanese textbook. Though I know many theorems in analysis, I don\'t know
the names of the theorems in English. The theorem in the following is called \"Koubetusekibun
teiri\" in Japanese. If you know the name of the theorem in English, let me know. Note that it is
not of Lebesgue integral, but of RIemann integral.
Solution
Riemann dominated uniform convergence theorem.
I am in need of two separate files that I can save independently.pdf
I am in need of two \"separate\" files that I can save independently of each other written from
beginning to end. A crontab and an associated bash file that will delete all user\'s trash every
night at 10 p.m. I need to be able to attach both files to an email.
Solution
Using rsync to make a backup
->Basics
->Using the --delete flag
->Be lazy: use cron
Incremental backups with rsync
->Review of hard links
->Using cp -al
->Putting it all together
Isolating the backup from the rest of the system
->The easy (bad) way
->Keep it on a separate partition
->Keep that partition on a separate disk
->Keep that disk on a separate machine
Making the backup as read-only as possible
->Bad: mount/unmount
->Better: mount read-only most of the time
->Tempting but it doesn\'t seem to work: the 2.4 kernel\'s mount --bind
->My solution: using NFS on localhost
Extensions: hourly, daily, and weekly snapshots
->Keep an extra script for each level
->Run it all with cron
Known bugs and problems
->Maintaining Permissions and Owners in the snapshots
->mv updates timestamp bug
->Windows-related problems.
More Related Content
Similar to Write a java program that reads a number N as input and displays the.pdf
import java.io.BufferedReader;import java.io.File;import java.io.pdf
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class FileEdit {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("EDIT: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
if (count >= lines.size()) {
for(int i = lines.size(); i <= count; i++) {
lines.add(null);
}
}
lines.set(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}
Here is my output:
EDIT: textin.txt
1> The First Line
1>
2> And another line
3> I 3
3> The second line
4> One more line
5> L
1> null
2> null
3> The second line
4> One more line
5> D 2
3> L
1> null
2> The second line
3> One more line
4> E
How can I keep it from printing "null" after ">" and it just be blank. I need the output to look like
this:
1> The first line
2>
3> And another line
4> I 3
3> The second line
4> One more line
5> L
1> The first line
2>
3> The second line
4> One more line
5> And another line
5> D 2
4> L
1> The first line
2> The second line
3> One more line
4> And another line
5> E.
In JavaWrite a program that reads a text file that contains a gra.pdf
In Java:
Write a program that reads a text file that contains a grade (for instance, 87) on each line.
Calculate and print the average of the grades.
Solution
import java.io.BufferedReader;
import java.io.FileReader; // importing filereader
import java.io.IOException; // importing file io exceptions
public class grade {
public static void main(String[] args) {
BufferedReader br = null;
double count=0;
int noOfgrades=0;
try {
String sLine;
br = new BufferedReader(new FileReader(\"sample.txt\")); // reading file using buffer
reader
while ((sLine = br.readLine()) != null) { // rading line by line
noOfgrades=noOfgrades+1; // getting number of grades
int num=Integer.parseInt(sLine); // convert string into integer
count=count+num; // adding each grade to count
}
System.out.println(\"count:\\t\"+count);
System.out.println(\"noOfgrades:\\t\"+noOfgrades);
System.out.println(\"Average of the grades:\\t\"+(count/noOfgrades));
} catch (IOException e) {
e.printStackTrace(); // catch exception
} finally {
try {
if (br != null)br.close(); // closing buffer
} catch (IOException ex) {
ex.printStackTrace();// printing exceptions
}
}
}
}
/********** out file******
fileName:sample.txt
file contains \"
23
34
56
7
88
89
52
34
65
\"
*/
/************output*************
count: 448.0
noOfgrades: 9
Average of the grades: 49.77
*/.
Please help me with a UML class diagram for the following code im.pdf
Please help me with a UML class diagram for the following code:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class FileEdit {
public static List lines = new ArrayList();
public static void main(String[] args) throws IOException
{
Scanner s = new Scanner(System.in);
System.out.print("EDIT: ");
String fileName = s.nextLine();
BufferedReader br = new BufferedReader(new FileReader(new File(fileName)));
String str="";
int count=0;
while((str=br.readLine())!=null)
{
lines.add(count, str);
System.out.println((++count) + "> "+ str);
}
String text="";
String command = "";
System.out.print((++count)+"> ");
command=s.nextLine();
String commandArr[];
int flag;
while(!command.equals("E"))
{
flag=0;
commandArr = command.split("\\s");
if(commandArr[0].equals("I")) //Insertion
{
while(true)
{
if(commandArr.length==1&&flag!=2)
{
System.out.print((count)+"> ");
text = s.nextLine();
insertLine(text,count-1);
}
else if(commandArr.length==2&&flag!=2)
{
System.out.print((commandArr[1])+"> ");
count=Integer.parseInt(commandArr[1]);
text = s.nextLine();
insertLine(text,Integer.parseInt(commandArr[1])-1);
}
else
insertLine(text,count-1);
System.out.print((++count)+"> ");
command=s.nextLine();
commandArr = command.split("\\s");
if(commandArr[0].equals("I")||commandArr[0].equals("L")||commandArr[0].equals("D")||comm
andArr[0].equals("E"))
{
flag=1;
break;
}
else
{
flag=2;
text=command;
}
}
}
else if(commandArr[0].equals("L")) //Listing
{
printList();
}
else if(commandArr[0].equals("D")) //Deletion
{
if(commandArr.length==1)
{
deleteLine(count-1,count-1);
}
else if(commandArr.length==2)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[1])-1);
}
else if(commandArr.length==3)
{
deleteLine(Integer.parseInt(commandArr[1])-1,Integer.parseInt(commandArr[2])-1);
}
count=lines.size();
}
if(flag!=1)
{
System.out.print((count++)+"> ");
command = s.nextLine();
}
}
}
public static void insertLine(String text,int count)
{
if (count >= lines.size()) {
for(int i = lines.size(); i <= count; i++) {
lines.add(null);
}
}
lines.set(count,text);
}
public static void deleteLine(int start,int end)
{
for(int i=start;i<=end;i++)
lines.remove(i);
}
public static void printList()
{
for(int i=0;i "+lines.get(i));
}
}.
mport java.io.; class Student { int rollno; String name; .pdf
mport java.io.*;
class Student
{
int rollno;
String name;
int number_of_subjects;
int mark[];
Student(int roll,String stud_name,int noofsub)throws IOException
{
rollno=roll;
name=stud_name;
number_of_subjects= noofsub;
getMarks(noofsub);
}
public void getMarks(int nosub ) throws IOException
{
mark=new int[nosub];
BufferedReader br= new BufferedReader (new InputStreamReader(System.in));
for (int i=0; i “);
mark[i]=Integer.parseInt(br.readLine());
System.out.println(“”);
}
}
public void calculateMarks()
{
double percentage=0;
String grade;
int tmarks=0;
for (int i=0;i “+rollno);
System.out.println(“Name Of Student is :=> “+name);
System.out.println(“Number Of Subject :=> “+number_of_subjects);
System.out.println(“Percentage Is :=> “+percentage);
if (percentage>=70)
System.out.println(“Grade Is First Class With Distinction “);
else if(percentage>=60 && percentage<70)
System.out.println(“Grade Is First Class”);
else if(percentage>=50 && percentage<60)
System.out.println(“Grade Is Second Class”);
else if(percentage>=40 && percentage<50)
System.out.println(“Grade Is Pass Class”);
else
System.out.println(“You Are Fail”);
}
}
class StudentDemo
{
public static void main(String args[])throws IOException
{
int rno,no,nostud;
String name;
BufferedReader br= new BufferedReader (new InputStreamReader(System.in));
System.out.println(“Enter How many Students:=> “);
nostud=Integer.parseInt(br.readLine());
Student s[]=new Student[nostud];
for(int i=0;i “);
rno=Integer.parseInt(br.readLine());
System.out.println(“Enter Name:=> “);
name=br.readLine();
System.out.println(“Enter No of Subject:=> “);
no=Integer.parseInt(br.readLine());
s[i]=new Student(rno,name,no);
}
for(int i=0;i
Solution
mport java.io.*;
class Student
{
int rollno;
String name;
int number_of_subjects;
int mark[];
Student(int roll,String stud_name,int noofsub)throws IOException
{
rollno=roll;
name=stud_name;
number_of_subjects= noofsub;
getMarks(noofsub);
}
public void getMarks(int nosub ) throws IOException
{
mark=new int[nosub];
BufferedReader br= new BufferedReader (new InputStreamReader(System.in));
for (int i=0; i “);
mark[i]=Integer.parseInt(br.readLine());
System.out.println(“”);
}
}
public void calculateMarks()
{
double percentage=0;
String grade;
int tmarks=0;
for (int i=0;i “+rollno);
System.out.println(“Name Of Student is :=> “+name);
System.out.println(“Number Of Subject :=> “+number_of_subjects);
System.out.println(“Percentage Is :=> “+percentage);
if (percentage>=70)
System.out.println(“Grade Is First Class With Distinction “);
else if(percentage>=60 && percentage<70)
System.out.println(“Grade Is First Class”);
else if(percentage>=50 && percentage<60)
System.out.println(“Grade Is Second Class”);
else if(percentage>=40 && percentage<50)
System.out.println(“Grade Is Pass Class”);
else
System.out.println(“You Are Fail”);
}
}
class StudentDemo
{
public static void main(String args[])throws IOException
{
int rno,no,nostud;
String name;
BufferedReader br= new BufferedReader (new InputStre.
import java.io.BufferedReader; import java.io.IOException; impor.pdf
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Hw3_4
{
public static void main(String arg[]) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println(\"Enter the Name Of Employee\");
String str = br.readLine();
str= str.toUpperCase();
str = str.replaceAll(\"\\\\s\",\"\"); //for removing space
char[] c=str.toCharArray();//code is used for converting the String into Array for one-by-one
matching
int lp=0;
int count=0;
for(int i=0;i
Solution
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Hw3_4
{
public static void main(String arg[]) throws IOException
{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println(\"Enter the Name Of Employee\");
String str = br.readLine();
str= str.toUpperCase();
str = str.replaceAll(\"\\\\s\",\"\"); //for removing space
char[] c=str.toCharArray();//code is used for converting the String into Array for one-by-one
matching
int lp=0;
int count=0;
for(int i=0;i.
import java.util.;public class Program{public static void.pdf
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println(x);
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain
using the most basic codes.
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println(x);
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain
using the most basic codes.
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println(x);
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain
using the most basic codes.
Solution
Hi,
I have modified the code. it is working as expected now. Highlighted the code changes below.
Issue here is with you have written this statement twice System.out.println(x); which causes the
issue.
Program.java
import java.util.*;
public class Program
{
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
String x=\"X\";
int amount;
System.out.println(\"Enter amount\");
amount=input.nextInt();
System.out.println(\"Amount is: \"+amount);
for(int i=1; i<=amount; i++)
{
System.out.println();
for(int j=1; j<=i; j++)
{
System.out.print(x);
}
}
}
}
Output:
Enter amount
10
Amount is: 10
X
XX
XXX
XXXX
XXXXX
XXXXXX
XXXXXXX
XXXXXXXX
XXXXXXXXX
XXXXXXXXXX.
JAVA Q2- Write a program that reads strings from the user and writes t.docx
JAVA
Q2: Write a program that reads strings from the user and writes them to an output file called userStrings.txt. Stop processing when the user enters the string \"DONE\". Do not write the sentinel string (\"DONE\") to the output file. [15 points] Hint: For a complete example of file IO, please see the attached TestData program.
Sample Output:
A string.
Another string.
Yet more text...
DONE
Solution
Answer
import java.io.File;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Scanner;
class WriteStringFile {
public static void main(String[] args) {
File file = new File(\"D:/userFile.txt\");
String data;
Scanner input = new Scanner(System.in);
data = input.readLine();
try (FileOutputStream fop = new FileOutputStream(file)) {
// if file doesn\'t exists, then create it
if (!file.exists()) {
file.createNewFile();
}
if (data != \"DONE\") {
// get the content in bytes
byte[] contentInBytes = content.getBytes();
fop.write(contentInBytes);
fop.flush();
fop.close();
} else {
System.out.println(\"User Enter the value Done.. So unable to write into file\");
}
} catch (IOException e) {
e.printStackTrace();
}
}
}
.
InsertionSorting.javaimport java.util.Arrays; import java.util.S.pdf
InsertionSorting.java
import java.util.Arrays;
import java.util.Scanner;
public class InsertionSorting {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Please enter number of letters:\");
int n = scan.nextInt();
char letters[] = new char[n];
for(int i=0; i -1) && ( charArray [i] > key ) ) {
charArray [i+1] = charArray [i];
i--;
}
charArray[i+1] = key;
}
}
}
Output:
Please enter number of letters:
5
Enter letter:
f
Enter letter:
s
Enter letter:
w
Enter letter:
a
Enter letter:
b
Before Sorting: [f, s, w, a, b]
After insertion Sorting: [a, b, f, s, w]
BubbleSorting.java
import java.util.Arrays;
import java.util.Scanner;
public class BubbleSorting {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Please enter number of letters:\");
int n = scan.nextInt();
char letters[] = new char[n];
for(int i=0; i array [1] swap it.
* 3. Compare array[1] & array[2]
* 4. If array[1] > array[2] swap it.
* ...
* 5. Compare array[n-1] & array[n]
* 6. if [n-1] > array[n] then swap it.
*
* After this step we will have largest element at the last index.
*
* Repeat the same steps for array[1] to array[n-1]
*
*/
int n = charArray.length;
char temp;
for(int i=0; i < n; i++){
for(int j=1; j < (n-i); j++){
if(charArray[j-1] > charArray[j]){
//swap the elements!
temp = charArray[j-1];
charArray[j-1] = charArray[j];
charArray[j] = temp;
}
}
}
}
}
Output:
Please enter number of letters:
5
Enter letter:
f
Enter letter:
d
Enter letter:
w
Enter letter:
a
Enter letter:
b
Before Sorting: [f, d, w, a, b]
After Bubble Sorting: [a, b, d, f, w]
Solution
InsertionSorting.java
import java.util.Arrays;
import java.util.Scanner;
public class InsertionSorting {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Please enter number of letters:\");
int n = scan.nextInt();
char letters[] = new char[n];
for(int i=0; i -1) && ( charArray [i] > key ) ) {
charArray [i+1] = charArray [i];
i--;
}
charArray[i+1] = key;
}
}
}
Output:
Please enter number of letters:
5
Enter letter:
f
Enter letter:
s
Enter letter:
w
Enter letter:
a
Enter letter:
b
Before Sorting: [f, s, w, a, b]
After insertion Sorting: [a, b, f, s, w]
BubbleSorting.java
import java.util.Arrays;
import java.util.Scanner;
public class BubbleSorting {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Please enter number of letters:\");
int n = scan.nextInt();
char letters[] = new char[n];
for(int i=0; i array [1] swap it.
* 3. Compare array[1] & array[2]
* 4. If array[1] > array[2] swap it.
* ...
* 5. Compare array[n-1] & array[n]
* 6. if [n-1] > array[n] then swap it.
*
* After this step we will have largest elem.
Write a program that will count the number of characters- words- and l.docx
Write a program that will count the number of characters, words, and lines in a file. Words are separated by whitespace characters. The file name should be passed as a command-line argument, as shown;
c:\\exercise>java Exercise12_13 Loan.java
File Loan.java has
1919 characters
210 words
71 lines
c:\\exercise>
Solution
import java.io.File;
import java.io.FileNotFoundException;
import java.util.Scanner;
import java.util.StringTokenizer;
public class ass1
{
public static void main(String[] args) throws FileNotFoundException
{
//Â Â Scanner console = new Scanner(System.in);
System.out.println(args[0]);
//System.out.println(args[1]);
//System.out.println(args[2]);
String inputFile=args[0];
File file = new File(inputFile);
Scanner in = new Scanner(file);
int words = 0;
int lines = 0;
int chars = 0;
while(in.hasNextLine()) {
lines++;
String line = in.nextLine();
chars += line.length();
words += new StringTokenizer(line, \" \").countTokens();
}
System.out.println(\"characters are \"+chars);
System.out.println(\"words are \"+words);
System.out.println(\"lines are \"+lines);
}
}
.
Javaa. The mean of a list of numbers is its arithmetic average. T.pdf
Java:
a. The mean of a list of numbers is its arithmetic average. The median of a list is its middle value
when the values are placed in order. For example, if a list contains 1, 4, 7, 8, and 10, then the
mean is 6 and the median is 7. Write an application that allows you to enter five integers and
displays the values, their mean, and their median.
Save the file as MeanMedian.java.
b. Revise the MeanMedian class so that the user can enter any number of values up to 20. If the
list has an even number of values, the median is the numeric average of the values in the two
middle positions. Save the file as
MeanMedian2.java
Solution
Ans a):
import java.util.*;
import java.lang.*;
import java.io.*;
class MeanMedian
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner scan=new Scanner(System.in);
int arr[]=new int[5];
System.out.println(\"Enter the integers :\"); //asking the user to enter 5 integers
for(int i=0;i<5;i++)
arr[i]=scan.nextInt();
int sum=0;
for(int i=0;i<5;i++)
sum=sum+arr[i];
int mean=sum/5; //calculating mean
int median=arr[2]; //calculating median
System.out.println(\"Integers entered by the user \");
for(int i=0;i<5;i++)
System.out.print(arr[i]+\" \"); //printing values entered by user
System.out.println();
System.out.println(\"Mean = \"+mean); //printing mean
System.out.println(\"Median = \"+median); //printing median
}
}
******OUTPUT******
Enter the integers :
1 4 7 8 10
Integers entered by the user
1 4 7 8 10
Mean = 6
Median = 7
******OUTPUT******
Ansb):
import java.util.*;
import java.lang.*;
import java.io.*;
class MeanMedian //class MeanMedian
{
public static void main (String[] args) throws java.lang.Exception
{
int sum=0; //variable to store the sum of values
int mean=0; //variable to store the mean of values
int median=0; //variable to store the median of values
Scanner scan=new Scanner(System.in);
System.out.println(\"Enter the number of values upto you want to enter,it must not be greater
than 20 :\"); //asking the user for no.of values
int n=scan.nextInt(); //store size in n
int arr[]=new int[n]; //array to store integers
System.out.println(\"Enter the integers\"); //asking the user to enter the integers
for(int i=0;i.
(JAVA NetBeans) Write a Java program able to perform selection sort-So.docx
(JAVA NetBeans) Write a Java program able to perform selection sort.
Solution
import java.util.Scanner;
/**
* @author
*
*/
public class SelectionSort {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int arr[] = setArray();
System.out.println(\"\ Before Sorting:\");
printArray(arr);
arr = selectionSort(arr);
System.out.println(\"\ After Sorting:\");
printArray(arr);
}
/**
* reads and stores the data in the one-dimensional array
*/
/**
* @return
*/
public static int[] setArray() {
int arr[] = new int[12];
System.out.println(\"Enter 12 integers.\");
Scanner scanner = new Scanner(System.in);
for (int i = 0; i < arr.length; i++) {
arr[i] = scanner.nextInt();
}
return arr;
}
/**
* prints the data in the one-dimensional array
*/
public static void printArray(int[] arr) {
System.out.println();
for (int i = 0; i < arr.length; i++) {
System.out.print(arr[i] + \" \");
}
System.out.println();
}
/**
* method will sort the array in ascending order
*/
public static int[] selectionSort(int[] arr) {
for (int i = 0; i < arr.length - 1; i++) {
int index = i;
for (int j = i + 1; j < arr.length; j++)
if (arr[j] < arr[index])
index = j;
int smallerNumber = arr[index];
arr[index] = arr[i];
arr[i] = smallerNumber;
}
return arr;
}
}
OUTPUT:
Enter 12 integers.
4 23 65 34 84 37 12 17 24 36 82 51
Before Sorting:
4 23 65 34 84 37 12 17 24 36 82 51
After Sorting:
4 12 17 23 24 34 36 37 51 65 82 84
.
IN JAVA Write a program to create a binary data file named RandomInt.pdf
IN JAVA Write a program to create a binary data file named RandomInt.dat that consists of 10
random integer numbers using DataOutputStream. Use DataInputStream to read the integers
from RandomInt.dat just created and print them on the console in the same program.
Solution
DataInputStreamTest.java
import java.io.DataInputStream;
import java.io.DataOutputStream;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Random;
public class DataInputStreamTest {
public static void main(String[] args) {
String strFilePath = \"D:\\\\RandomInt.dat\";
Random r = new Random();
try
{
FileOutputStream outputStr = new FileOutputStream(strFilePath);
DataOutputStream outputStrData = new DataOutputStream(outputStr);
for(int i=0; i<10; i++){
outputStrData.writeInt(r.nextInt(100)+1);
outputStrData.writeChar(\'\\t\');
}
outputStrData.flush();
outputStrData.close();
outputStr.close();
FileInputStream inputStr = new FileInputStream(strFilePath);
DataInputStream inputStrData = new DataInputStream(inputStr);
int count = 0;
while(count<10){
int i = inputStrData.readInt();
inputStrData.readChar();
count++;
System.out.println( i);
}
outputStrData.flush();
inputStrData.close();
outputStrData.close();
}
catch(FileNotFoundException fe)
{
System.out.println(\"FileNotFoundException : \" + fe);
}
catch(IOException ioe)
{
System.out.println(\"IOException : \" + ioe);
}
}
}
Output:
65
8
95
69
1
77
16
97
85
1.
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More from brijeshagarwa329898l
I studied analysis in Japanese textbook. Though I know many theorems.pdf
I studied analysis in Japanese textbook. Though I know many theorems in analysis, I don\'t know
the names of the theorems in English. The theorem in the following is called \"Koubetusekibun
teiri\" in Japanese. If you know the name of the theorem in English, let me know. Note that it is
not of Lebesgue integral, but of RIemann integral.
Solution
Riemann dominated uniform convergence theorem.
I am in need of two separate files that I can save independently.pdf
I am in need of two \"separate\" files that I can save independently of each other written from
beginning to end. A crontab and an associated bash file that will delete all user\'s trash every
night at 10 p.m. I need to be able to attach both files to an email.
Solution
Using rsync to make a backup
->Basics
->Using the --delete flag
->Be lazy: use cron
Incremental backups with rsync
->Review of hard links
->Using cp -al
->Putting it all together
Isolating the backup from the rest of the system
->The easy (bad) way
->Keep it on a separate partition
->Keep that partition on a separate disk
->Keep that disk on a separate machine
Making the backup as read-only as possible
->Bad: mount/unmount
->Better: mount read-only most of the time
->Tempting but it doesn\'t seem to work: the 2.4 kernel\'s mount --bind
->My solution: using NFS on localhost
Extensions: hourly, daily, and weekly snapshots
->Keep an extra script for each level
->Run it all with cron
Known bugs and problems
->Maintaining Permissions and Owners in the snapshots
->mv updates timestamp bug
->Windows-related problems.
Heat transfer QuestionGiven a too hot to taste spoonful of soup.pdf
Heat transfer Question:
Given a too hot to taste spoonful of soup. Will the soup temperature decrease faster (as compared
to forced convection), if it is allowed to cool solely under the effects of natural convection? yes ,
or no / why, or why not?
a) yes/ since the thermal conductivity of the soup is greater at higher temperatures
b) yes/ since the viscosity of the soup is lower at higher temperatures
c) no/ since the convective heat transfer coefficient of the air increases as the air velocity is
increased
d) no /since the specific heat of the air is greater as the air velocity increases
e) could be either yes or no depending upon the flow regime, as characterized by the Reynolds
Number b yes isme the wiscsity of the o.is lower at righe temera res cl no s1ncethe convastve he
transter costticient st the alr increases as the ar velccityis ncrease d; rolaih sa.ilit heal eathz air
acctvincoases
Solution
e)could be either yes or no depending upon the flow regime, as characterized by the reynolds
number.
Explain what it is that is used with a 9S12 processor to match a giv.pdf
Explain what it is that is used with a 9S12 processor to match a given interrupt request with its
corresponding ISR.
Solution
Each ISR have the some vector location in the memory. The interrupts are generated by the
hardware or software. The software interrupts are generated by programming, hardware
interrupts are generated by external triggerring pulses. When the interrupt request came by
software or hardware the PC (Program Counter) automatically loaded the vector address of the
matched Interrupt request ISR. The vector address have 8bytes of memory only in that 4 bytes
for base address and 4 bytes for offset of the ISR.
The on-chip resources of the 9s12 (freescale) MCU are controlled and enhanced by numerous
interrupts. These ISR-controlled resources include:
A 16-channel 10-bit A/D converter,
Pulse-Width Modulated (PWM) outputs on PortP,
Enhanced Capture/Timer (ECT) I/O on PortT,
A watchdog timer (COP),
A clock monitor,
Maskable and nonmaskable external interrupt pins,
Real-Time Interrupt (RTI),
Dual serial communications ports,
A high speed serial peripheral interface (SPI)
An inter-IC (IIC) serial bus, and,.
Duroc Jersey pigs are typically red, but a sandy variation is also s.pdf
Duroc Jersey pigs are typically red, but a sandy variation is also seen. When two different
varieties of true-breeding sandy pigs were crossed to each other, they produced F1 offspring that
were red. When these F1 offspring were crossed to each other, they produced red, sandy, and
white pigs in a 9:6:1 ratio. Explain this pattern of inheritance in terms of number of genes, and
any allele or gene interactions.
Solution
The dihybrid ratio for Mendelian pattern of inheritance often shows some deviations owing to
epistasis. One of these ratio include the pattern of 9:6:1 ratio rather than normal 9:3:3:1. This
ratio can be attributed to presence of duplicate genes with cumulative effects. According to the
above example, it can be seen that parents are pure breeding red and sandy in color. However,
their offsprings appear in the form of red, sandy and peculiar white pigs in the ratio of 9:6:1
respectively.
Explanation: The data and observation suggest that the genes responsible for red color and sandy
color are duplicate in nature and cumulatively, they provide sandy color to the pigs. However,
absence of any of these alleles in either homozygous or heterozygous condition fails to generate
a coat color and a colorless white pig is reared. This also suggests that deviation of Mendelian
pattern of inheritance occures in this case owing to specific interactive behaviour of these two
genes for red and sandy coat colors. Failure of interaction of these two alleles results in white pig
whereas presence of both the alleles results in inter-allelic interaction and unexpectedly, a very
large number of sandy pigs are observed. Cumulatively, it can be deciphered that both the genes
are dominant over each other. These alleles produce a different phenotype when present alone
but produce a peculiar phenotype when present together due to inter-allelic interaction.
Importantly, it is not a case of simply additive behaviour of alleles because if it had been, a new
phenotypic coat color would have been obtained. However, since the F1 progeny resumes the
color similar to their parents, it represents a case of epistatic duplicate genes..
Discuss the economic impact of technological sophistication.Solu.pdf
Discuss the economic impact of technological sophistication.
Solution
Economic impact of technological sophistication
In past few decades, a rapid rowing literature has emerged, arguing that a key aspect of economic
development and the growth process is an increase in the sophistication” of a countrys
production,
In export sophistication over the past years in low income countries and middle income countries
. In particular, it describes differences in the performance of different geographical regions, as
well as between natural-resource–rich and other economies.
Sophistication In information technology
In IT sector , There was a immense change and rapid growth of technology. From past two
decades, Internet was commercialized nearly 18 years ago and now we are moving a bit faster
than those old days
Advancements in fiber optic s have also played a vital role in IT revolution. Fiber optics enables
data rates, including voices in digital form, to be converted into tiny pulses of light and then
transmitted at high speeds through glass fibers wrapped into large capacity telecommunication
cables. Hundreds of thousands of miles of these cables were installed over the past ten years,
boosting the speed and capacity of telecommunications networks.
Sophistication In Government sector:
The IT sector emerged some rapid changes in the government sector such as e-government, can
enhance the efficiency and effectiveness of government services. E-government can also help
achieve other important goals of good governance, such as transparency and reliability. Access
to official information is critical to ensuring that governments are accountable to citizens—that
they are responsive to citizens and that they are doing what they are obligated to do under law.
Sophistication InExports and manufacturing Industry
China had become relatively sophisticated in manufacturing, whereas India produced relatively
sophisticated services, even controlling for income levels. composition of service exports moved
away from traditional activities to modern activities like business services and computer
services. As a result, service export sophistication grew relatively fast in India.
Sophistication In Education :
IT improves educational opportunities by enabling educators and students to overcome barriers
of distance and by enhancing the content of instructional materials. The use of IT to deliver
lessons or training from instructors in one location to students in another is called \"distance
learning.\" Distance learning has been around for a long time. For many years people have
listened to recordings of classroom lectures or other educational presentations, and millions of
people have watched educational programming on public televisions channels. Most colleges and
universities across the United States offer some online course offerings. In 2016, the latest
development is the rise of Coursera, which offers free online courses from elite universities, such
as MIT and Harvard.
Describe in detail the specimen of a potato with a drop of iodine in.pdf
Describe in detail the specimen of a potato with a drop of iodine in comparison with a potatoe
with just a drop of water under the microscope.
Solution
Answer:
Starch (a carbohydrate) can be identified by a positive iodine test. Few drops of iodine solution
are added to a raw or boiled sample of the substance suspected of containing starch. If starch is
present, the substance will turn blue-black in the area where the iodine solution touches it.
Substances with very small amounts of starch will turn faint blue or greenish instead of blue-
black. Many materials of biological origin have high starch content. These materials can be
identified with the iodine test.
The sample specimen with potato and a drop of iodine will turn blue black concluding for the
presence of large amount of starch.
The sample specimen with potato and a drop of water will not undergo a change in colour..
define two modes of speciation and provide examples of eachSolut.pdf
define two modes of speciation and provide examples of each
Solution
Speciation:
A species is a group of organisms capable of interbreed and produce fertile offspring. Speciation
is a process of evolution in biological population. Speciation gives distinct population over time.
There are various mode of speciation as follows:
1. Allopatric speciation: Any geographic, climatic or anthropogenic barriers in population split
the population in two or more. These populations evolve in own way to accumulate mutations
leads to genetic divergence.
Example: Isthmus of Panama
2. Peripatricspeciation: When a migrated population grow in a different ecology and adapts to
that environment causes to form a sub-species is called Peripatric speciation speciation.
Examples: organisms at oceanic island
3. Sympatric Speciation: When two or more species arises from a single parent species known
as sympatric speciation within same geographic environment.
Example: Nicaragua crater lake cichlid fishes evolves by Sympatric Speciation..
computer math Use the following to answer questions 1-2 In the q.pdf
computer math Use the following to answer questions 1-2: In the questions below means \"r +2y
or, where are integers, Determine the truth value af the statement, P(ry) and y A) True B) False
2. P(1,-1) A) True B) False
Solution
2) True
Given that
p(x,y) is x + 2y = xy
p(1,-1) is 1 + 2.(-1) = 1.(-1)
1 - 2 = -1
-1 = -1
Hence ,
p(1,-1) is true.
CASe Stic ato wad duun SolutionAggregates used in this method a.pdf
CASe Stic ato wad duun
Solution
Aggregates used in this method are chemically inert. Visually they look in reddish yellow colour.
Size of these aggregates vary from 4.75mm to 0.075mm. They are less permeable as compared to
coarse aggregates. Some of them are collapsible soils with dispersed structures..
Write MatLAb code that would produce the following printSolution.pdf
Write MatLAb code that would produce the following print:
Solution
rgb = imread(\'coloredChips.png\'); figure imshow(rgb) d = imdistline; delete(d);
gray_image = rgb2gray(rgb); imshow(gray_image); [centers, radii] = imfindcircles(rgb,[20
25],\'ObjectPolarity\',\'dark\') [centers, radii] = imfindcircles(rgb,[20
25],\'ObjectPolarity\',\'dark\', ... \'Sensitivity\',0.9) imshow(rgb); h = viscircles(centers,radii);
[centers, radii] = imfindcircles(rgb,[20 25],\'ObjectPolarity\',\'dark\', ... \'Sensitivity\',0.92);
length(centers) delete(h); % Delete previously drawn circles h = viscircles(centers,radii);.
Why do rays run radially in a woody stemHow would wood produced i.pdf
Why do rays run radially in a woody stem?
How would wood produced in a dry year differ from wood produced in a wet year?
What is dendrochronology?
Solution
Dear student,
Wood produced in dry year:
Is just brought on by serpula lacrymans and is the most genuine type of parasitic rot in a
building. It can spread onto and crush a great part of the timber. Signs for dry decay:
Particular mushroom smell
White contagious development with yellow and lilac tinges
Profound splits show up inside the wood
Both delicate and hardwoods can be harmed
Wood produced in wet year:
Happens all the more every now and again, yet is less genuine; rot is normally bound to the
range where timber has gotten to be and stays wet. Indications of wet decay :
Mutilation, discolouration, non-abrasiveness and breaking
Loss of quality to the timber
Noticeable contagious development – this may now and again happen
Smell – there might be a soggy smelly smell
Flare-ups of dry and wet decay begin in comparative ways. The develop fruiting assortments of
wood-annihilating organisms that create amid an assault deliver a large number of tiny spores
and these are generally scattered via air streams. In the event that they fall on untreated moist
wood they will develop by pushing out an empty tube called a hypha which develops and
branches to frame a mass of hyphal strings called mycelium.
The principle contrasts between dry year and wet year are the level of advancement of mycelium
on the wood surface and the capacity of the organism to spread into different timbers by means
of nearby brick work. It is imperative that the two sorts of rot be recognized since they require
distinctive treatment.
Part3
Dendrochronology is the investigation of environmental change as recorded by tree development
rings. Every year, trees include a layer of development between the more established wood and
the bark. This layer, or ring as found in cross segment, can be wide, recording a wet season, or
thin, recording a dry developing season. Since the rings are essentially recording a decent
developing season or a terrible developing season, they are in a roundabout way recording more
than just dampness. They likewise record temperature and overcast cover as they effect tree
development too. This record of yearly summer data is imperative when you consider that
specific sorts of trees develop gradually over several years, and subsequently contain a record of
the same number of years of atmosphere and environmental change..
Why is the DNA fragment amplified with Phusion DNA into pGEM-T plasmi.pdf
Why is the DNA fragment amplified with Phusion DNA into pGEM-T plasmids? The colorless
E. coli colonies in our experiment contain the recombinant pGEM-T/GFPuv plasmid. Some of
the colonies emit green fluorescence whereas others don\'t when they are irradiated with long
wave UV. light. Explain Why.
Solution
1) Phusion DNA Polymerase enzyme is a proof-reading polymerase, with extreme specificity.
The enzyme provides robust amplification. Since it is a proof-reading enzyme, Phusion DNA
Polymerase creates blunt ends in the DNA fragment, because of which it cannot be directly
cloned into the plasmid.
2) GFPuv ,the green fluorescent protein, exhibits structural polymorphism. Onlyproper
conformationproduces fluorescence. Some colonies emitted fluorescence, because the GFPuv in
them is in proper conformation while in colonies which did not emit fluorescence, the protein
was not in a proper conformation..
Which of these is a function of the skeletal system communication s.pdf
Which of these is a function of the skeletal system? communication system creation of the
structural framework within soft organs of the body storage site for calcium removal of old or
damaged blood cells
Solution
Answer:
storage site of calcium
Explanation:
The followings are the function of the skeletal system:
1. Support: Provides structural support for the entire body
2. Protection: Surrounds soft tissue- ex. ribs and sternum protect the heart and lungs, - skull
protects the brain
3. Movement: Skeletal muscle is attached to bone so it pulls on the bone when it contracts
4. Mineral homeostasis: Stores calcium and phosphorus--minerals are released into the blood
when needed.
5. Blood cell production: Red bone marrow produces red blood cells, white blood cells and other
blood elements
6. Storage: Storage of minerals and lipids(fats)---yellow marrow stores fat --(found in long
bones).
what is transport layer what are the typical attacks in transport l.pdf
what is transport layer? what are the typical attacks in transport layer? what are the controls that
are employed in the layer to minimize the attack or vulnerability that leads to the attack? cite
references
Solution
Transport Layer:-
In computer networking, the transport layer is a conceptual division of methods in the layered
architecture of protocols in the network stack in the Internet Protocol Suite and the Open
Systems Interconnection (OSI). The protocols of the layerprovide host-to-host communication
services for applications.
Typical attacks in transport layer:-
1. SESSION HIJACKING: Session Hijacking is commonly known as TCP session Hijacking is a
way of taking over a secure/ unsecure web user session by secretly obtaining user’s session ID
and pretending to be the authorized user for accessing the data. How it works and types: Session
hijacking works by taking advantage of the fact that most communications are protected (by
providing credentials) at session setup, but not thereafter. These attacks generally fall into three
categories: Man-in-the-middle (MITM), Blind Hijack, and Session Theft. In MITM attacks, an
attacker intercepts all communications between two hosts. With communications between a
client and server now flowing through the attacker, he or she is free to modify their content.
Protocols that rely on the exchange of public keys to protect communications are often the target
of these types of attacks. In blind hijacking, an attacker injects data such as malicious commands
into intercepted communications between two hosts commands like “net.exe local group
administrators /add Evil Attacker”. This is called blind hijacking because the attacker can only
inject data into the communications stream; he or she cannot see the response to that data (such
as “The command completed successfully.”) Essentially, the blind hijack attacker is shooting
data in the dark, but as you will see shortly, this method of hijacking is still very effective. In a
session theft attack, the attacker neither intercepts nor injects data into existing communications
between two hosts. Instead, the attacker creates new sessions or uses old ones. This type of
session hijacking is most common at the application level, especially Web applications. Main
features are: -URL (Uniform resource locator) -Cookies -Session ID The cookies stores the
previous records of the users and the URL logs can give the current visited site, a hacker take
benefits from it and hacks user’s session ID through it, after doing that it pretends to be the
authorized user and accesses the data. A cookie usually is a piece of text sent by a server to the
web client and sent back unchanged by the client, each time it access the data. 1.1 Methods used
to perform session hijacking: 1.1.1IP Spoofing: It basically means taking identity of someone
else to perform some task, in this the attacker pretends to be the authorized user and access some
confidential information, not only this, the.
Which monetary aggregate would you expect the Federal Reserve to hav.pdf
Which monetary aggregate would you expect the Federal Reserve to have the greatest control.
Explain.
Solution
Fed controls the money supply in the economy, by expanding money supply during recession
and contracting money supply during uncontrolled expansion (that leads to inflation).
In this context, Fed has highest control over M0 which is notes and coins in circulation. That\'s
because M0 is the most liquid form of monetary aggregate which Fed can directly control.
Starting from M1, other modes of savings are added to the aggregate, reducing their liquidity,
and Fed has no direct control over how individuals or firms save their money..
wavelet decomposition process code in matlabSolution WAVE.pdf
wavelet decomposition process code in matlab???
Solution
WAVELET DECOMPOSITION PROCESS :-
function [coef,scf]=mywavedec2(x,N, wname)
coef={cA_N;cV_N;cH_N;cD_N;cV_N-1;cH_N-1;cD_N-1;… …;cV_1;cH_1;cD_1}
[Lo_D,Hi_D] = wfilters(wname,\'d\');
lf=length(Lo_D); imshow(x);title(\'Original Image\');
[r,c]=size(x);
xlabel([\'Size : \',num2str(r),\'*\',num2str(c)]);
xd=double(x);
[rx,cx]=size(x);
[o2sa,f1sa,rsx]=sizcoef([rx,cx],lf,N); a=[o2sa,f1sa,rsx]
coef=[]; scf=[rx,cx];
for i=1:N
[cA,cV,cH,cD]=mydwt2(xd,Lo_D,Hi_D);
xd=cA;
outmp={cV;cH;cD};
scf=[size(cV);scf];
coef=[outmp;coef];
end
coef=[cA;coef];
scf=[size(cA);scf];
plotcoef(N,wname,coef);
subplot(N+1,3,2);
yt=uint8(coef{1});
[yrow,ycol]=size(yt);
imshow(yt);title( [\'Approximation A\',num2str(N)]);
xlabel([\'Size : \',num2str(yrow),\'*\',num2str(ycol)]);
titllist={[\'Vertical Detail V\'];[\'Horizontal Detail H\'];
[\'Diagonal Detail D\']};
pn=2;
for pr=1:N
for pc=1:3
subplot(N+1,3,pn+2);
yt=[ ];
yt=uint8(coef{pn});
[yrow,ycol]=size(yt);
imshow(yt);title([ titllist{pc},num2str(N-pr+1)]);
xlabel([\'Size : \',num2str(yrow),\'*\',num2str(ycol)]);
if mod(pn+2,3)==1
ylabel([\'Level \',num2str(N-pr+1)]);
end
pn=pn+1;
end
end
function [o2sa,f1sa,rsx]=sizcoef(sx,lf,N)^4
o2sa=sx;
f1sa=[0 0];
osx=sx;
for i=1:N
ot=osx+lf-1;
sa=floor(ot/2);
o2sa=[o2sa;ot];
f1sa=[f1sa;sa];
osx=sa;
end
rsx=zeros(N+1,2);
rsx(1,:)=sx;
addln=0;
for j=2:N+1
aln=floor((addln+lf-1)/2);
tsx=f1sa(j,:)-aln;
rsx(j,:)=tsx;
addln=aln;
end
function y=downspl(x)
N=length(x);
M=floor(N/2);
i=1:M;
y(i)=x(2*i).
What happens if a red hot inoculation loop plunges into the nutrient .pdf
What happens if a red hot inoculation loop plunges into the nutrient media? Name the type(s) of
culture used to make your cultures of two unknown specimens. For inoculating into a liquid
media, what is the most appropriate transfer (inoculation) tool? For inoculating into a solid
media, what is the most appropriate transfer (inoculation) tool? Describe & sketch the
appearance of growth on nutrient agar slant.
Solution
1. what happens when red hot inoculation loop plunges into nutrient media ?
ANSWER - An aerosol may be created, releasing bacteria in to the air .an aerosol release within
the laboratory can pose an infection risk, which may not be immediately recognized and may
affect all users of the laboratory .
2. name the culture used to make your cultures of two unknown specimens ?
ANSWER - USE OF SELECTIVE AND DIFFERENTIAL MEDIA
Differential medium types are those that distinguish microorganisms from one another based on
growth characteristics evident when grown on specific medium types. Organisms with differing
growth characteristics typically show visible differences in growth when placed on differential
media. Examples include blood agar, Eosin Methylene Blue (EMB) agar, Mannitol Salt agar, and
MacConkey agar.
Selective medium types are formulated to support the growth of one group of organisms, but
inhibit the growth of another. These media contain antimicrobials, dyes, or alcohol to inhibit the
growth of the organisms not targeted for study. Selective medium types include EMB agar,
Mannitol Salt agar, MacConkey agar, and Phenylethyl Alcohol (PEA) agar.
3. For inoculating in to liquid media what is the most appropriate transfer tool ?
ANSWER - Inoculating Loop , as the loop can hold a drop of liquid .
4. For inoculating in to solid media what is the most appropriate transfer tool ?
ANSWER - same , inoculating loop .
5. Describe the appearance of growth on nutrient agar slant ?
ANSWER - Cultures are often transferred to agar slants, in addition to broth tubes and agar
plates. An agar slant is a test tube containing agar, in which the solid agar forms a slant in the test
tube. When inoculating an agar slant, draw the loop containing the inoculum very lightly over
the surface in a zigzag formation while being careful not to break the surface. A needle can be
used instead of a loop to inoculate an agar slant by stabbing the needle containing the inoculum
into the agar ..
What are speaking notes Have you used speaking notes before What wo.pdf
What are speaking notes? Have you used speaking notes before? What works? What doesn\'t?
Solution
Speaker Notes – Meaning:
Speaker Notes is a tool / feature offered by power point presentation to allow presenter to add
reminder on what the presenter presents when a presentation is done using power point software.
Speaker Notes will be available on presenter\'s screen only and will not be visible on the
presentation shown to the viewers and can be added to all slides of the power point presentation
by the presenter. (Person who prepares and presents power point presentation).
Speaker Notes - How to insert?
Speaker Notes will appear at the bottom of each slide of the power point presentation if added by
the presenter. To insert Speaker Notes in a slide, presenter must click on \"Notes\" tab that
appears at the bottom of respective slide and add required notes on the pop-window that appears
at the bottom of respective slide on the screen. After completing the notes, presenter can click
anywhere on the slide other than on pop-up window to complete the notes and continue using
other slides. Notes added will appear constantly without being hidden always for presenter\'s
easy reference.
What works? What doesn\'t?
I\'ve made a power point presentation at my college during a seminar. Speaker Notes on power
point presentation done by me was very much helpful whenever I needed sudden references to
improve the presentation\'s quality. However, as the notes were appearing as low font and each
slide contained different notes applicable to respective slide, my presentation was affected due to
oversight in one or two instances. That is, referring to another slide\'s notes visually but tried
delivering current slide\'s presentation as a virtual image and the presentation was collapsed for a
while..
There are five components of internal control1) Control environme.pdf
There are five components of internal control:
1) Control environment
2) Risk assessment
3) Control activities
4) Information and communication system
5) Monitoring processes.
In your own words explain the purpose of internal controls. Then select one of the components
shown above and explain the role it plays in protecting the assets of the business. Many of these
activities can be observed in the workplace both inside and outside of the accounting department.
Please share any work experiences you may have had with these activities or any that you have
observed in other businesses, even if they are not in the accounting department. Be sure to
provide a reference for any material used from a source.
Solution
Internal Control Purpose -
It support business by provide an idea to employees and other personnel’s regarding the aspect,
objective of business. Due to the internal control the stake holders will be get facilitated that the
business transactions are recorded properly and there is no any under reporting or none reporting
of transactions. The consumers also preferred those organizations which have proper internal
control, as it helps consumers in understanding organisation.
Control Activities
The control activities led down policies and procedures to be followed by personnel’s so that
there will not be any misuse of assets of the organization. There should be proper backup as per
control over any event of loss of data due to any unfortunate event. The control activities ensure
that the applications and network are safeguarded from any threat as part of internal control.
Internal control ensure that control activities must have business continuity plan..
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- 1. Write a java program that reads a number N as input and displays the following output using stacks with parameters. Example o Input: Enter N: o Output: 5 Solution import java.io.BufferedReader; import java.io.IOException; import java.io.InputStreamReader; public class Stacks { public static void main(String[] args) throws IOException { // used variables BufferedReader reader; String input,outerSpaces,innerSpaces; Integer number; // taking input System.out.print("Enter N : "); reader = new BufferedReader(new InputStreamReader(System.in)); input = reader.readLine(); number = Integer.parseInt(input); // printing upper stack for(int i=1;i0;i--){ outerSpaces = getOuterSpaces(i); innerSpaces = getInnerSpaces(number, i); System.out.print(outerSpaces); System.out.print(i); System.out.print(innerSpaces); if(i != number){ System.out.print(i); } System.out.print(outerSpaces); System.out.println(); } }
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