Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) 
Total Submission(s): 8311 Accepted Submission(s): 5726

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

核心代码

for (int i =0; i < len; ++i){
            ans = (ans*10 + big_num[i] - '0') % num;
        }

#include <stdio.h>
#include <string.h>

int main()
{
    int num;
    char big_num[1024] = {0};
    while(scanf("%s%d",&big_num ,&num) == 2)
    {
        int ans = 0;
        int len = strlen(big_num);
        for (int i =0; i < len; ++i){
            ans = (ans*10 + big_num[i] - '0') % num;
        }
        printf("%d\n",ans);
    }
    return 0;
}