5th personal contest-I - Switch Game

最新推荐文章于 2020-01-24 12:49:16 发布

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本文探讨了一种特殊情况下灯的状态变化问题。初始时所有灯都关闭，在一系列操作中，每轮将编号为当前轮次倍数的灯状态反转。经过无限轮后，最终灯的状态如何确定？文章提供了一个简洁的C语言程序来计算任意编号灯的最终状态。

Description

There are many lamps in a line. All of them are off at first. A series of operations are carried out on these lamps. On the i-th operation, the lamps whose numbers are the multiple of i change the condition ( on to off and off to on ).

Input

Each test case contains only a number n ( 0< n<= 10^5) in a line.

Output

Output the condition of the n-th lamp after infinity operations ( 0 - off, 1 - on ).

Sample Input

1
5

Sample Output

1
0

Hint

Consider the second test case:

The initial condition	   : 0 0 0 0 0 …
After the first operation  : 1 1 1 1 1 …
After the second operation : 1 0 1 0 1 …
After the third operation  : 1 0 0 0 1 …
After the fourth operation : 1 0 0 1 1 …
After the fifth operation  : 1 0 0 1 0 …

The later operations cannot change the condition of the fifth lamp any more. So the answer

思路：

第1轮，所有编号是1的倍数的灯转换状态
第2轮，所有编号是2的倍数的灯转换状态
第i轮...
经历了无数n轮后，灯的状态已经不会再变化了（因为没有是n的倍数的灯编号了）

欢迎交流：

#include<stdio.h>
#include<string.h>

int main()
{
	int n;
	while (scanf("%d", &n) != EOF)
	{
		int count = 0;
		for (int i = 1; i <= n; i++)
		{
			if (n%i==0) count++;
		}
		if (count%2==0)
			 printf("0\n");
		else printf("1\n");

	}


	return 0;
}