Rightmost Digit（HDU1061）

最新推荐文章于 2019-07-18 19:55:49 发布

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本文探讨了一道编程题目，即给定一个正整数N，如何快速找到N^N结果的最右侧数字。通过对大量数据的观察与分析，发现了计算结果尾数的规律，并提供了一个高效的解决方案。

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23184 Accepted Submission(s): 8836

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.

//解题思路：N(1<=N<=1,000,000,000).范围比较大，所以这题应该是有规律的，所以先打表找规律；
#include <cstdio>
#include <cstring>

int main()
{
	int num[25] = {0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0};
	int i , j; 
	/*freopen("data.txt" ,"w" ,stdout);//打表找规律，发现每20个数的尾数都是相同的；用数组储存每一个数的尾数，然后直接输出
	for(i = 1; i <= 100 ; i++)       
	{
		int sum = 1;
		for(j = 1 ; j <= i; j++)
		{
			sum *= i;
			if(sum > 100)
			sum %= 100;
		}
		printf("%d " ,sum%10);
		if(i % 20 == 0)
		printf("\n");
	}*/
	int n ;
	while(scanf("%d" ,&n) != EOF)
	{
		int nums;
		while(n--)
		{
		   scanf("%d" ,&nums);
		   while(nums > 20)
		   {
		   	nums %= 20;
	       }
	       printf("%d\n" , num[nums]);
		}
	}
}