Rightmost Digit

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.题目很好理解：求N的N次方的个位数。有多种做法：

第一种直接暴力，数太大，GG，想都不用想肯定会超时，运行WA

#include<bits/stdc++.h>//错误做法 using namespace std;int main(){int n;int x;int sum=0;while(cin>>n){cin>>x;    sum=pow(x,x);    sum=sum%10;cout<<sum<<endl;}     return 0;} 

第二种找规律：

当   n = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 27 28 29 30 31 ...
rdigit = 1 4 7 6 5 6 3 6 9   0   1   6   3   6   5    6   7   4    9   0   1    4   7   6    5   6   3   6    9    0  ...

*#include <stdio.h>#include<iostream>using namespace std;int  rdigit[25] = {0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0};int main (){    int T , n ;     cin>>T;    while (T -- )    {          cin>>n;          printf ( "%d\n" , rdigit[n%20] ) ;    }    return 0 ;}

第三种快速幂：（）

#include <iostream>#include <cstdio>using namespace std;int mod_exp(int a, int b, int c)  //快速幂取余a^b%c{    int res, t;    res = 1 % c;     t = a % c;    while (b)    {        if (b & 1)        {            res = res * t % c;        }        t = t * t % c;        b >>= 1;    }    return res;}int main(){    int T;    cin >> T;    while (T--)    {        int n;        cin >> n;        cout << mod_exp(n, n, 10) << endl;    }    system("pause");    return 0;}