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：
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.
程序代码：
#include<bits/stdc++.h>
using namespace std;
int quickmul(int n)
{
int ans=1;
int temp=n%10;
while(n)
{
if(n&1)//等价于n%2==1
ans=(ans*temp)%10;
temp=(temp*temp)%10;
n>>=1;//等价于n=n/2
}
return ans;
}
int main()
{
int a,n,count;
cin>>n;
while(n--)
{
cin>>a;
cout<<quickmul(a)<<endl;
}
return 0;
}
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