2053——Switch Game

最新推荐文章于 2021-08-20 13:12:31 发布

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最新推荐文章于 2021-08-20 13:12:31 发布

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分类专栏： ACM 文章标签：基础题

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探讨一排初始关闭的灯，在经过一系列操作后第n盏灯的状态。每次操作改变编号为当前操作次数倍数的灯的状态。

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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.

#include <stdio.h>
main()
{
     int n;
     while(scanf("%d",&n)!=EOF)
     {
          int i,j=0;
          for(i=1;i<=n;i++)
              if(n%i==0)
                  j++;
          printf("%d\n",j%2);
     }
}