HDOJ 1061 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.

大概题意就是输出一个数的本身次方的个位数，例如3的3次方的个位数是7.
这个题有两种解法：
第一种是经典的快速幂算法：
快速幂我这里就不多介绍了，关于快速幂算法我单独出l一个博客做详细介绍的了感兴趣的朋友可以去看看。

快速幂介绍

#include <stdio.h>

// 公式:(a * b) % p = (a % p * b % p) % p
//快速求幂

long long npower4(long long base, long long power);//快速求幂优化版pro

int main(void)
{

    int n;
    scanf("%d", &n);
    while (n--)
    {
        int num;
        scanf("%d", &num);
        printf("%lld\n",npower4(num, num));
    }
    return 0;
}

long long npower4(long long base, long long power)
{
    long long result = 1;
    while (power > 0)
    {
        if (power & 1)
            result = result * base % 10;
        power >>= 1;
        base = base * base % 10;
    }

    return result;
}

第二种解法是找规律：

不难看出个位数出现是固定组合，每二十个数一次循环，根据这个我们可以写代码了。

#include <stdio.h>

const int arr[21] = { 0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0 };        
//保存个位数出现组合方便调用

int main(void)
{

    int n;
    scanf("%d", &n);
    while (n--)
    {
        int num;
        scanf("%d", &num);
        printf("%d\n",arr[num % 20]);
    }
    return 0;
}