hdoj1212Big 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：
   
   
   
   
   
   
   
   

    
    
    
    #include<stdio.h>
int main()
{
    char a[1005];
    int b,c;
    while(scanf("%s %d",a,&b)!=EOF)
    {
        long long sum=0;
        for(int i=0;a[i]!='\0';i++)
        {
            sum=(sum*10+a[i]-'0')%b;            
    }
        printf("%d\n",sum%b);
    }
    return 0;
}

   
   
   
   
   
   
   
   

    
    
    
    思路：每一步都去余，防止爆掉