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.

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.

For each test case, you have to ouput the result of A mod B.

2 3
12 7
152455856554521 3250

2
5
1521

Ignatius.L

 

这是在网上搜到的 应该记住的结论：

A*B % C = (A%C * B%C)%C
(A+B)%C = (A%C + B%C)%C

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int main()
{
    char ls[1004];int n;
    while(scanf("%s%d",ls,&n)!=EOF)
    {
        int sum=0;
        for(int i=0;i<strlen(ls);i++)
        {
            sum=(sum*10+(ls[i]-'0')%n)%n;

        }
        cout<<sum<<endl;
    }
    return 0;
}