hdoj1212Big Number（大数）

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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

2 312 7152455856554521 3250

Sample Output

251521
代码：
#include<stdio.h>int main(){    char a[1005];    int b,c;    while(scanf("%s %d",a,&b)!=EOF)    {        long long sum=0;        for(int i=0;a[i]!='\0';i++)        {            sum=(sum*10+a[i]-'0')%b;                }        printf("%d\n",sum%b);    }    return 0;}
思路：每一步都去余，防止爆掉

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