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

題意：輸入兩個m,n，讓m對n取餘，其中m是個大數
題解：將m以字元陣列形式輸入，從第一位數開始逐一對n取餘

程式碼（C）

#include<stdio.h>
#include<string.h>
int main()
{
    int m,n,i,ans;
    char a[2000];//陣列大小不能小於題意中的1000位
    while(scanf("%s",a)!=EOF)
    {
        scanf("%d",&n);
        m=strlen(a);
        ans=0
 
;
        for(i=0;i<m;i++)
        {
            ans=(ans*10+(a[i]-'0'))%n;//ans剩下的在下一次運算中乘以10加上
        }
        printf("%d\n",ans);
    }
    return 0;
}