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
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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.
题目大意：给出一个数a，求a的a次方的结果，输出结果的个位数字。
思路：跑一遍是不可能的，这辈子都不可能的。该题是快速幂的模板题，快速幂的时间复杂度是O(log2 N)。
快速幂模板：https://blog.csdn.net/yopilipala/article/details/68952650
原理：https://www.jianshu.com/p/1c3f88f63dec
代码如下：
#include<set>
#include<map>
#include<list>
#include<deque>
#include<cmath>
#include<queue>
#include<string>
#include<vector>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<ext/rope>
#include<iostream>
#include<algorithm>
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
#define per(i,a,b) for(int i=a;i<=b;++i)
#define max(a,b) a>b?a:b
#define min(a,b) a<b?a:b
#define swap(a,b) {int t=a;a=b;b=t}
using namespace std;
using namespace __gnu_cxx;
const long long MOD = 10;
long long q_pow(long long int a,long long int b)
{
long long int sum=1;
while(b)
{
if(b&1)
sum=(sum*a)%MOD;
a=(a*a)%MOD;
b>>=1;
}
return sum;
}
int main()
{
int t,i,a;
scanf("%d",&t);
while(t--)
{
scanf("%d",&a);
printf("%d\n",q_pow(a,a));
}
return 0;
}