HDU1212：Big Number

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

 

大数对小数取余的模板题

 

#include <stdio.h>#include <string.h>int mod(char *n1,int n2){    int tmp=0;    int len=strlen(n1);    for(int i=0; i<len; i++)    {        tmp=tmp*10+n1[i]-'0';        tmp=tmp%n2;    }    return tmp;}int main(){    char s[100000];    int n;    while(~scanf("%s%d%*c",s,&n))    {        printf("%d\n",mod(s,n));    }    return 0;}