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 longusing 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\n",ans);    }    return 0;}</span>