HDU 1061 Rightmost Digit 【快速幂 Or 规律（瞎搞）】

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

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.

题意：给一个数n，输出n的n次方的个位数。

思路：一开始暴力，很显然的TLE，然后套用快速幂的模板，过了。完后又去搜了一下发现有规律，一是周期为20，另一个规律是可以打表，这里的话只给出周期为20的那份代码，另一个规律以及快速幂的参照下篇文章HDU1097题）

AC代码：

#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>using namespace std;int main(){    int ans[20] = {0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9};    int n, t;    while(~scanf("%d",&t))    {        while(t--)        {            scanf("%d",&n);            cout << ans[n%20] << endl;        }    }    return 0;}