HDU 1212

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
 
#include<cstdio>#include<cstring>using namespace std;int main()         // 同余定理 {                    //(a+b)%c=(a%c+b%c)%c;    char a[1005];         //(a*b)%c=(a%c*b%c)%c;    int b,i,t;    while(scanf("%s %d",a,&b)!=EOF)    {        t=0;        for(i=0;i<strlen(a);i++)        {            t=(t*10%b+(a[i]-'0')%b)%b;        }        printf("%d\n",t);    }    return 0;}
非常简单的，利用同余定理
题意 大数取模  a%b=？