【杭电1212】大数除法

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7331    Accepted Submission(s): 5062

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

 

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#include <cstdlib>
#include<stdio.h>
#include <string.h>
using namespace std;
char s[1005];
int main()
{
    int b ,i ,l ,n;
    while (~scanf("%s%d",&s,&b))
    {
          n = 0;
          l = strlen (s);
          for (i = 0 ;i < l ;i++)
           n = (n * 10 + (s[i] -'0')) % b; 
          printf("%d\n",n); 
    }
    
    return 0;
}