Big Numberモデリング

Big Number
Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 19   Accepted Submission(s) : 14
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

   
   
   
   

 
  アキュムレータ型

      ，           ；
  ：
12345 9
    （12340%9+5%9）%9；
 12340 9
（12300%9+40%9）%9;
  ...
  （10000%9+2000%9）%9；
 10000%9=（1%9*10000）%9
2000%9=（2%9*1000）%9
 （1*10+2）%9*1000%9；
    ：for（i=0;i<len;i++）
           {
             sum=sum*10+s[i]-'0';
              sum=sum%9;
            }
#include<stdio.h>
#include<string.h>
int main()
{
 int n,sum,i,k;
 char s[1001];
 while(scanf("%s%d",s,&n)!=EOF)
 {
  k=strlen(s);
  sum=0;
  for(i=0;i<k;i++)
  {
   sum=sum*10+s[i]-'0';
   sum=sum%n;
  }
  printf("%d
",sum);
 }
 return 0;
}