大数取余

Big Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4680    Accepted Submission(s): 3247

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
  
  

    
  
  

    
  
  

   #include<stdio.h>
#include<string.h>
int mod(char *n1,int n2)
{
  int temp,i;
  int k=strlen(n1);
      temp=0;
  for(i=0;i<k;i++)
  {
    temp=temp*10+(n1[i]-'0');
    temp%=n2;
  }    
  return temp;
}
int main()
{
    int n,;
    char str[10000];
    while(~scanf("%s %d*c",str,&n))
    {
       printf("%d\n",mod(str,n));                             
    }
return 0;    
}