E - Rightmost Digit

E - Rightmost Digit

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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. 

题意分析：

这道题本身的意思就是输入T的样例，对于每个样例N，做N^N，然后求N^N的最后一位的数值。

解题思路：

这道题本质上非常简单，而且以前也做过。只不过看起来比较吓人，因为N^N的数量级太大。不过也很明显，肯定不能直接暴力。来说说我的解题思路，对于0-9每一个数与自己相乘都是有规律的，只需要判断N的最后一位数，然后分类输出结果就好了，下面是编码。

编码：

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int n;
    cin >> n;
    while (n--)
    {
        int m;
        cin >> m;
        int a = m % 10;
        if (a == 1 || a == 5 || a == 6 || a == 0)
            cout << a << endl;
        else if (a == 2)
        {
            int num[4] = {6, 2, 4, 8};
            cout << num[m % 4] << endl;
        }
        else if (a == 3)
        {
            int num[4] = {1, 3, 9, 7};
            cout << num[m % 4] << endl;
        }
        else if (a == 4)
        {
            int num[2] = {6, 4};
            cout << num[m % 2] << endl;
        }
        else if (a == 7)
        {
            int num[4] = {1, 7, 9, 3};
            cout << num[m % 4] << endl;
        }
        else if (a == 8)
        {
            int num[4] = {6, 8, 4, 2};
            cout << num[m % 4] << endl;
        }
        else if (a == 9)
        {
            int num[2] = {1, 9};
            cout << num[m % 2] << endl;
        }
    }
    return 0;
}