HDU_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.

最先想到的是一个一个算，但是由于数据范围太大，O(n^2)的时间复杂度对于n = 10亿时，10亿^10亿次乘法运算实在是不能忍受的。因此下面的程序超时(Time Limit Exceeded)。

#include<stdio.h>
int main(void)
{
    int cases, n, copy_n, result;
    scanf("%d", &cases);
    while(cases--)
    {
        scanf("%d", &n);
        copy_n = n;
        n = n%10;
        result = 1;
        while(copy_n--)
        {
            result = (result*n)%10;
        }
        printf("%d\n", result);
    }

    return 0;
}

下面，快速幂一来就AC了：

#include<stdio.h>
int my_power(int m, int n); // 求m的n次方的尾数
int main(void)
{
    int cases, n;
    scanf("%d", &cases);
    while(cases--)
    {
        scanf("%d", &n);
        printf("%d\n", my_power(n, n));
    }

    return 0;
}

int my_power(int m, int n)
{
    m = m%10;
    if(n == 1)
        return m;
    if(n%2 == 0)
        return ( my_power(m*m, n/2) ) % 10;
    else
        return ( my_power(m*m, n/2)*m ) % 10;
}

可以看到，快速幂的时间复杂度是O(logn)，n = 10亿时，大约32次递归调用就能出结果，效率极大的提高了。