As we know, Big Number is always troublesome. But it‘s really important in our ACM. And today, your task is to write a program to calculate A mod B.

To make the problem easier, I promise that B will be smaller than 100000.

Is it too hard? No, I work it out in 10 minutes, and my program contains less than 25 lines.

The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000, and B will be smaller than 100000. Process to the end of file.

For each test case, you have to ouput the result of A mod B.

2 3
12 7
152455856554521 3250

2
5
1521

Ignatius.L

杭电ACM省赛集训队选拔赛之热身赛

Eddy   |   We have carefully selected several similar problems for you:  1215 1211 1214 1213 1695 

Java大数算法，仅仅需一步!请叫我大水神！

import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class Main
{

	public static void main(String[] args)
	{
		// TODO Auto-generated method stub
		Scanner input = new Scanner(System.in);
		while (input.hasNext())
		{
			BigInteger A = input.nextBigInteger();
			BigInteger B = input.nextBigInteger();
			BigInteger N = A.remainder(B);
			System.out.println(N);
		}
	}

}