hdu 2053 Switch Game

最新推荐文章于 2021-03-28 01:15:14 发布

原创最新推荐文章于 2021-03-28 01:15:14 发布 · 913 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#hdu #模拟

HDU 专栏收录该内容

43 篇文章

订阅专栏

本文探讨了一系列操作对线性排列灯泡状态的最终影响，通过分析特定操作序列，揭示了灯泡状态的周期性和规律性，特别是对于第五个灯泡状态的详细解释。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Switch Game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8850 Accepted Submission(s): 5286

Problem 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
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 is 0.

题意：就是初始时有很多灯的状态都是关闭的，

然后第n次改变处于n的倍数的位置的灯的状态，

即：由0到1，由1到0；

思路：以下是我的代码，直接循环，当j大于n时，

即再也无法改变其状态，所以可以循环到n结束。

但是还可以找规律：只有处于n*n状态的灯才会是亮的。

#include <iostream>
#include <cstring>
#include <cstdio>
#include<cmath>
#include <algorithm>
using namespace std;


int main()
{
 int n,i,j,a[100010];
 while(~scanf("%d",&n))
 {
    memset(a,0,sizeof(a));
     int j=1;
   while(j<=n)
   {
     for(i=j;i<=n+2;i+=j)
        a[i]=!a[i];
      j++;
   }
  //for(i=1;i<=n;i++)
    //cout<<a[i]<<' ';
    printf("%d\n",a[n]);
   }

    return 0;
}