HDU OJ1061 Rightmost Digit

HDU OJ1061 Rightmost Digit

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.

法一：找规律，依次列出

这里需要注意
1】x的范围，所以定义为长整形__int64d输入，%I64d输出
2】输出时，sum的下标

#include <stdio.h>

int main(){
     __int64 x;
    int n,y;
    int sum[10][4]={{0},{1},{6,2,4,8},{1,3,9,7},{6,4},{5},{6},{1,7,9,3},{6,8,4,2},{1,9}};

    scanf("%d",&n);

    while(n--){
        scanf("%I64d",&x);
        y=x%10;
        if(y==0||y==1||y==5||y==6)
            printf("%d\n",sum[y][0]);
        else if(y==4||y==9)
            printf("%d\n",sum[y][x%2]);
        else 
            printf("%d\n",sum[y][x%4]); 

    }   
    return 0;
} 

快速幂取余可参看
http://www.cnblogs.com/PegasusWang/archive/2013/03/13/2958150.html