HDU 1212 Big Number

　　　　　　　　　　Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
　　　　　　　　　　　　　　　Total Submission(s): 9158 Accepted Submission(s): 6214

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

1521

思路：大数求余：

　　　　　　　　( a + b ) % c = ( ( a % c) + ( b % c ) ) % c;

　　　　　　　　( a * b ) % c = ( ( a % c ) * ( b % c ) ) % c;

（来自百度文库 https://wenku.baidu.com/view/3784592beefdc8d376ee32ba.html ）

#include<iostream>
#include<string>
using namespace std;
int main() {
    int n, i, j;
    string s;
    while(cin>>s>>n) {
        j = 0;
        for(i = 0; i < s.length(); i++) {
            j = ( j*10 % n + (s[i]-'0')%n ) % n;
        }
        cout<<j<<endl;
    }
    return 0;
}