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

意解：大整数取模运算，类似于整数除法（手算），比如512 % 10，我们手算是51 % 10 = 1，之后1 * 10 + 2 = 12，在12 % 10  =  2，就结束了；
AC代码：
      

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>

using namespace std;
typedef long long ll;

int main()
{
    char s[1100];
    int mod,len;
    while(cin>>s>>mod)
    {
        ll sum = 0;
        len = strlen(s);
        for(int i = 0; i < len; i++)
        {
            sum = (sum * 10 + s[i] - '0') % mod;
        }
        cout<<sum<<endl;
    }
    return 0;
}