HDU 1061 - Rightmost Digit(快速幂)

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本文介绍了一种通过快速幂取模的方法来高效求解正整数N的N次方最右侧数字的问题，并提供了一份简洁的C++实现代码。

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 51099 Accepted Submission(s): 19312

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.

题意：
求N*N的最右一位。

解题思路：
快速幂%10，或者打表。

AC代码：

#include<bits/stdc++.h>
using namespace std;
int mp[10][4] =
{
0,0,0,0,
1,1,1,1,
6,2,4,8,
1,3,9,7,
6,4,6,4,
5,5,5,5,
6,6,6,6,
1,7,9,3,
6,8,4,2,
1,9,1,9
};
int main()
{
    int n;
    scanf("%d",&n);
    while(n--)
    {
        int tmp;
        scanf("%d",&tmp);
        if(!tmp)
        {
            printf("1\n");
            continue;
        }
        printf("%d\n",mp[tmp%10][tmp%4]);
    }
    return 0;
}