杭电1212--Big Number

Big Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5961 Accepted Submission(s): 4167

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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//··A``

 1 #include <stdio.h>
 2 #include <string.h>
 3 char str[1010] ;
 4 int main()
 5 {
 6     int i, n ;
 7     while(~scanf("%s %d", str, &n))
 8     {
 9         int mod = 0 ;
10         int len = strlen(str) ;
11         for(i = 0; i < len; i++)
12         mod = (mod * 10 + str[i] - '0') % n ;
13         printf("%d
", mod) ;
14     }
15     return 0 ;    
16 }