HDU 1061 Rightmost Digit (快速幂取余)

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 38903    Accepted Submission(s): 14660

Problem Description

Given a positive integer N, you should output the most right digit of N^N.

 

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

 

Output

For each test case, you should output the rightmost digit of N^N.

 

Sample Input

   
   
   
   

    
    
    
    2
3
4
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    7
6


    
    
    
    
 
     
 
      Hint 
     
 In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7. In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6. 
    
    
    
    
     
   
   
   
   

 

注意：

1.快速幂取余

#include<stdio.h>
int PowerMod(int a,int b,int c)// a的b 次方 取余数为c
{
    int ans=1;
    a=a%c;
    while(b>0)
    {
        if(b%2==1)
            ans=(ans*a)%c;
        b=b/2;
        a=(a*a)%c;
    }
    return ans;
}
int main (void)
{
    int n,a;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d",&a);
        printf("%d\n",PowerMod(a,a,10));
    }
    return 0;
}