HDU - 2053 Switch Game

最新推荐文章于 2021-03-20 23:16:41 发布

luer9

最新推荐文章于 2021-03-20 23:16:41 发布

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分类专栏：模拟题思绪题

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本文探讨了一道关于灯状态变化的算法题。初始时所有灯均为关闭状态，通过一系列操作改变灯的状态，每次操作会翻转特定编号灯的状态。文章详细解析了如何确定任意一盏灯经过无限次操作后的最终状态，并提供了实现这一逻辑的C语言代码。

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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 ).

InputEach test case contains only a number n ( 0< n<= 10^5) in a line.
OutputOutput the condition of the n-th lamp after infinity operations ( 0 - off, 1 - on ).Sample Input

1
5

Sample Output

1
0


        
  

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 is 0.

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 …
说下题目大意吧,
先把这些灯标上号,1 2 3 4 5 6 7 8 ……无穷
首先全是关的,也就是全是0
第一次操作 ,标号是1的倍数,全都变成相反的状态,也就是全变成1..
第二次操作 ,标号是2的倍数,全都变成相反的状态,你可以看下,2 4 6……变成了0..
第三次操作 ,标号是3的倍数,全都变成相反的状态,你可以看下,3 6 9……
他问你 N 号台灯最后 变成了 什么状态,
例如 1号灯,最后变成了1,不管多少次操作都是1..
例如 5号灯 最后变成了0,不管多少次操作都是0..
当操作次数大于N的时候 N的状态就不会改变了,因为N不会是M（M>N）的倍数.
代码：
#include <stdio.h>
int judge(int n)
{
	int change=0;
	for(int i=1;i<=n;i++)
	{
		if(n%i==0)change++;
	}
	return change%2;
}
int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	printf("%d\n",judge(n));
	return 0;
}