杭电OJ——1212 Big Number

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 3
12 7
152455856554521 3250
 
 
Sample Output
2
5
1521
 
 
Author
Ignatius.L
 
 
Source
杭电ACM省赛集训队选拔赛之热身赛
 
 
Recommend
Eddy
 
     在做题之前，先了解这样一些结论：
A*B % C = (A%C * B%C)%C
(A+B)%C = (A%C + B%C)%C
如 532 mod 7 =（500%7+30%7+2%7)%7;
当然还有a*b mod c=(a mod c+b mod c)mod c;
如35 mod 3=((5%3)*(7%3))%3
   有了这一些结论，题目就好做了！
   代码如下：
[cpp]  
#include<iostream>  
#include<string.h>  
using namespace std;  
const int MAX=100010;  
  
  
int main()  
{  
    char str[MAX];  
    int s,len,sum;  
    while(scanf("%s%d",str,&s)!=EOF)  
    {  
        len=strlen(str);  
        sum=0;  
        for(int i=0;i<len;i++)  
        sum=(sum*10+(str[i]-'0')%s)%s;  
        cout<<sum<<endl;  
    }  
    return 0;  
}

补充：软件开发 , C++ ,