Big Number------HDOJ杭电1212（大数运算）

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

题意简单，不用讲的呐！

要是你看不懂我非常操心你的四级肿么办！！！！！！

直接上代码，能够作为模板！！！！！！

这代码有点抄袭别人的感觉；

开毛线玩笑，模板啊！！！！！！

模板不一样怎么叫模板！！！！！

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;

int main()
{
    int i,j,k;
    int t,n,m;
    char ch[1006];
    while(cin>>ch>>n)
    {
        m=0;
        for(i=0;i<strlen(ch);i++)
        {
            m*=10;
            m+=ch[i]-‘0‘;
            m%=n;
        }
        printf("%d\n",m);
    }
    return 0;
}

写代码能力有限，如有编程爱好者发现bug，还请指出，不胜感激！！！！！！

Big Number------HDOJ杭电1212（大数运算）