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