简单&&大数取模

Big Number

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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 3

12 7

152455856554521 3250

 

Sample Output

2

5

1521

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提炼：数字转化为字符串处理，每位取模。

#include<stdio.h>
#include<string.h>
int main(void){
    char number[1001];
    int b,ans;
    while(scanf("%s %d",number,&b)!=EOF){
        int len=strlen(number);
        ans=0;
        for(int i=0;i<len;i++){
            ans=(ans*10+(number[i]-'0'))%b;
        }
        printf("%d\n",ans);
    }
    return 0;
}

题目来源：

http://acm.hdu.edu.cn/showproblem.php?pid=1212