Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 4594    Accepted Submission(s): 3175

Problem Description

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.

Input

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.

Output

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

Sample Input

2 3
12 7
152455856554521 3250

Sample Output

2
5
1521

用到同余定理：（a+b）%c=(a%c+b%c)%c=（a+b%c）%c;    附：（a*b）%c=(a%c*b%c)%c;

#include <stdio.h>
#define maxn 1002

char str[maxn];

int main()
{
    int m, ans, i;
    while(scanf("%s%d", str, &m) != EOF){
        ans = 0;
        for(i = 0; str[i]; ++i){
            ans = (ans * 10 + (str[i] - '0') % m) %m;
        }
        printf("%d\n", ans);
    }
    return 0;
}