hdu 1212 Big Number（大数取模）

 

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

 

//重最高位开始不断取余，简单来说模拟手算

//比如  985%6

//   9%6=3

//  38%6=2

//  25%6=1

//计算完毕

#include <iostream>

using namespace std;
char a[12000];
int b;
int mod(char *a,int b)
{
    int ans=0;
    for(int i=0;a[i]!='\0';i++)
    ans=(ans*10+a[i]-'0')%b;
    return ans;
}
int main()
{
    while(cin>>a>>b)
    {
        cout<<mod(a,b)<<endl;
    }
    return 0;
}