快速幂

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

<span style="color:#6633ff;">/********************************************************
    author    :    Grant Yuan
    time      :    2014.7.28
    algorithm :    快速幂
    
*********************************************************/

#include <iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#define LL long long

using namespace std;

LL mod_pow(LL x)
{
    LL res=1;
    LL n=x;
    while(n>0){
        if(n&1) res=res*x%10;
        x=x*x%10;
        n>>=1;
    }
    return res%10;
}
int main()
{
    int t;LL n,ans;
    scanf("%d",&t);
    while(t--){
        scanf("%lld",&n);
        ans=mod_pow(n);
        printf("%lld
",ans);
    }
    return 0;
}
</span>