【HDU】1061 - Rightmost Digit（快速幂）

点击打开题目

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 47404 Accepted Submission(s): 17900

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.

Author

Ignatius.L

以前用观察找规律然后打表的方法做的，其实用快速幂也可以的，时间上也亏不了多少，代码相对就短很多了。

打表的代码：点击打开链接

代码如下：

#include <cstdio>
int cal(int n,int m)
{
	int ans = 1;
	n %= 10;
	while (m)
	{
		if (m & 1)
			ans = (ans * n) % 10;
		n = (n * n) % 10;
		m >>= 1;
	}
	return ans % 10;
}
int main()
{
	int n;
	int u;
	scanf ("%d",&u);
	while (u--)
	{
		scanf ("%d",&n);
		printf ("%d\n",cal(n,n));
	}
	return 0;
}