Source code for word count

The Entities in java are slightly different. I adore them being included (finally).

float bigNum = 123456.789012;

String pattern = String.format( "%10.2f is my magic number%n", bigNum);

System.out.println( pattern);

or

System.out.println(

String.format( "%10.3f", bigNum) );

the "decimal place" is the rounding.

lead off with a zero to get leading zeroes

String.format("%03.2f", bigNum));

if I want evenly spaced columns, I have to construct an Object[] array

array[0] = salaryAmt;

array[1] = hourly Rate;

array[2] = "Cowboy Bob";

System.out.println(

"%5.2f,%3.2d,%s%n, array);

Here you will get more information about java string format.

coding in C++

#include<iostream.h>
#include<conio.h>
#define max 50

void main()
{
clrscr();
int arr[max],size=0;
int i=0,j=0,temp;
do
{
cout<<"Enter the size of array(less than or equal to 50)";
cin>>size;
} while(size>50);

for(i=0;i<size;i++){
cout<<"Enter a number";
cin>>arr[i];
}


for( i=1;i<size;i++){
for(j=0;j<size;j++){
if(arr[j+1]>arr[j]){
temp = arr[j];
arr[j] = arr[j+1];
arr[j+1] = temp;

}
}
}
cout<<”Array in descending order”;
for(i=0;i<size;i++){
cout<<arr[i]<<"\n";
}
getch();
return 0;
}

get the text as string , store the possible errors as array check the string against the array and display result

NEED A NEW KEY THE ONE ON THE BOX WONT WORK

the best thing to do here is to separate the steps for calculating the variance into different functions. To compute the variance the first thing to do is to compute the mean or the average of the numbers in the array. so your first function will loop through each of the numbers and divide it by the length of your array. the output is the returned. The next thing to do is create another function for the variance. loop through each of the numbers again and subtract to each number the mean. square the value then add to the previous. so that means you have another variable that starts from zero and just adds the answer. after that divide it by the length of the array again and what you are going to get is the value for the variance.

for int i=0; i< array.count; i++
{
sum+= array[i];

}

average = sum/5

Your answer is here:
Solution # 1
http://www.web-tech-india.com/software/jsp_chat.php

Solution # 2
http://www.topshareware.com/java-source-codes-for-chat-application/downloads/1.htm

Solution # 3
http://www.codeproject.com/KB/java/java_applet_chat_with_gui.aspx

Good luck

Do you need them to be sorted after the merge and are they already sorted? If not, then what you can do is create a temporary third array which will set the half of third array equal to the first array, and then set the set half equal to the second array. If they need to be sorted, I can help too but you need to tell me

public class ArrayMerge
{
public static int arr1[] = new int [20];
public static int arr2[] = new int [20];

public static void main()
{
setScores(arr1);
setScores(arr2);
printScores(arr1);
printScores(arr2);
printScores(merger(arr1,arr2));
}

public static void setScores(int a[])
{
for(int x = 0; x<a.length; x++)
{
a[x]= (int)(Math.random() * 100)+1;
}
}

public static void printScores(int a[])
{
for(int x = 0; x<a.length; x++)
{
System.out.print(a[x] + " ");
}
System.out.println();
}

public static int[] merger(int a1[], int a2[])
{
int a3[] = new int [(a1.length + a2.length)];

for(int x = 0; x<a1.length; x++)
{
a3[x]=a1[x];
}

for(int x = 0; x<a2.length; x++)
{
a3[x+a2.length]=a2[x];
}
return a3;
}
}

program for deletion


#include<iostream.h>
void main()
{
int a[10],i,n,x;
cout<<"\n\t\tPROGRAM TO DELETE AN NUMBER FROM AN ARRAY";
cout<<"\nEnter the size of the array:";
cin>>n;
for(i=0;i<n;i++)
{
cout<<"\nEnter the number:";
cin>>a[i];
}
cout<<"\nEnter the position to be deleted:";
cin>>x;
for(i=x;i<n;i++)
a[i]=a[i+1];
for(i=0;i<n-1;i++)
cout<<a[i]<<" ";
}