Big Number

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　　

代码：

 1 #include<cstdio>
 2 #include<cstring>
 3
 4 int main()
 5 {
 6     int a[1005];
 7     char s[1005];
 8     int m,i,j;
 9     while(scanf("%s%d",s,&m)!=EOF)
10     {
11         int len=strlen(s);
12         for(j=0,i=len-1;i>=0;i--)
13         a[j++]=s[i]-‘0‘;
14         int sum=0;
15         for(j-=1;j>=0;j--)
16         sum=(sum*10+a[j])%m;
17         printf("%d\n",sum);
18     }
19     return 0;
20 }