思路很简单系列不过带坑【水水水】

最新推荐文章于 2023-01-23 15:30:00 发布

原创最新推荐文章于 2023-01-23 15:30:00 发布 · 341 阅读

0 ·

CC 4.0 BY-SA版权

ACM 同时被 3 个专栏收录

8 篇文章

订阅专栏

水题

6 篇文章

订阅专栏

字符串

5 篇文章

订阅专栏

本文介绍了一种高效算法来确定正整数N的N次方的最右侧数字。通过分析不同个位数在幂运算时的规律，文章提供了一个C语言实现的代码示例，能够处理高达十亿级别的输入。

Rightmost Digit

Time Limit: 1 Sec Memory Limit: 32 MB
Submit: 21 Solved: 5
[ Submit][ Status][ Web Board]

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.

Source

#include <stdio.h>

int main()
{
	long long n,x;
	long long m;
	
	while (~scanf("%lld",&n))
	{
		while(n--)
		{
			scanf("%lld",&x);
			m = x % 10;
			if ( m == 1 || m == 5 || m == 6 || m == 0)
			{
				printf("%lld\n", m);
			
			}
				
			if ( m == 2)
			{
				if ( x % 4 == 0)
				{
					printf("6\n");
				
				}
				if ( x % 4 == 1)
				{
					printf("2\n");
				}
				if ( x % 4 == 2)
				{
					printf("4\n");
				}
				
				if (x % 4 == 3)
				{
					printf("8\n");
				}
	 	  }
	 	  if ( m == 3)
			{
				if ( x % 4 == 0)
				{
					printf("1\n");
				
				}
				if ( x % 4 == 1)
				{
					printf("3\n");
				}
				if ( x % 4 == 2)
				{
					printf("9\n");
				}
				
				if (x % 4 == 3)
				{
					printf("7\n");
				}
	 	  }
	 	  if ( m == 4)
			{
				if ( x % 2 == 0)
				{
					printf("6\n");
				
				}
				if ( x % 2 == 1)
				{
					printf("4\n");
				}

	 	  }
	 	  if ( m == 7)
			{
				if ( x % 4 == 0)
				{
					printf("1\n");
				
				}
				if ( x % 4 == 1)
				{
					printf("7\n");
				}
				if ( x % 4 == 2)
				{
					printf("9\n");
				}
				
				if (x % 4 == 3)
				{
					printf("3\n");
				}
	 	  }
	 	  if ( m == 8)
			{
				if ( x % 4 == 0)
				{
					printf("6\n");
				
				}
				if ( x % 4 == 1)
				{
					printf("8\n");
				}
				if ( x % 4 == 2)
				{
					printf("4\n");
				}
				
				if (x % 4 == 3)
				{
					printf("2\n");
				}
	 	  }
	 	  if ( m == 9)
			{
				if ( x % 2 == 0)
				{
					printf("1\n");
				
				}
				if ( x % 2 == 1)
				{
					printf("9\n");
				}
			
	 	  }
		}
		
		
	
	}

	
	return 0;

}

有坑啊如果数非常非常大 long long 都不支持了。。。

题目有没有说范围。。。。

所以要用数组了。。。。。