BlueJ Program On Kaprekar Numbers

Given the two positive integers p and q, where p kaprekar numbers

 are there in the range between 'p' and 'q'(both inclusive )and output them.About 'kaprekar' number:

A posotive whole number 'n' that has 'd' number of digits is squared and split into

2 pieces, a right hand piece that has 'd' digits and a left hand piece that has remaining 'd' or 'd-1' digits. If sum of the pieces is equal to the number then it's a kaprekar number.

SAMPLE DATA:

INPUT:

p=1

Q=1000

OUTPUT:

THE KAPREKAR NUMBERS ARE:

1,9,45,55,99,297,999

FREQUENCY OF KAPREKAR NUMBERS IS:8

import java.io.*;

class Bank

{

int i,p,q,c=0;

int num;

BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

public void take() throws IOException

{

System.out.println("Enter the Lower Range:"); p=Integer.parseInt(br.readLine()); System.out.println("Enter the Upper Range:"); q=Integer.parseInt(br.readLine()); if(p >=q)

{

System.out.println("Wrong Entry...");

return;

}

System.out.println("THE KAPREKAR NUMBERS ARE:");

for(i=p;i<=q;i++)

{

show(i);

}

System.out.println("FREQUENCY OF KAPREKAR NUMBERS IS:"+c);

}

public void show(int x)

{

int digit,rev=0;int no;num=x*x;digit=0;no=x;

while(no >0)

{

digit++;

no=no/10;

}

no=num;

while(digit >0)

{

rev=rev*10+no%10;

no=no/10;

digit--;

} 

//  'rev' holds the right part in reverse order and 'no' holds the left part rev=reverse(rev);

if((rev+no)==x)

{

System.out.print(" "+x);

c++;

}

}

private int reverse(int n)

{

int r=0;

while(n >0)

{

r=r*10+n%10;

n=n/10;

}

return r;

}

}

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