HDOJ 1061快速幂

问题描述：

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

 234 

 

Sample Output

 76 

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. 

分析：感觉这道题目的思路特别好，首先求N^N得最后一位，

#include <iostream>#include <stdio.h>using namespace std;int main(){   int t,a,n,f;   while(scanf("%d",&t)!=EOF)   {   while(t--)   {   scanf("%d",&n);   f=n%10;   if(n%4==0)   {   n=4;}else   n%=4;a=1;for(int i=0;i<n;i++){  a=a*f;}printf("%d\n",a%10);}  } return 0;}