1090: integer power (multi instance test)

Title Description

seek A^B An integer represented by the last three digits of （1<=A,B<=1000）

Input

n A test case , Each instance gives two positive integers A,B

Output

Output A^B The last three （ There is no forerunner 0）

The sample input

2
2 3
12 6

Sample output

8
984

Code ：

// Find the power of an integer ,A^B
// After the output number 3 position 
// Input ：
//2
//2 3
//12 6
// Output ：
//8
//984



#include <stdio.h>
int main ()
{
    int n;
    scanf("%d",&n);
    int a,b;
    int i,j;
    int sum;
    for(i=0;i<n;i++)
    {
        scanf("%d %d",&a,&b);
        sum=1;
        for(j=0;j<b;j++)
        {
            sum*=a;
            while(sum>1000)
            {
                sum=sum%1000;
            }
        }
        printf("%0d\n",sum);
    }
    return 0;
}

Running results ：

  Submit ：

  But I still have one thing I don't understand , As shown in the following figure , It should be right to write logic like this

  But submit , will not pass

  If that big guy knows , Hope to inform