hdu2053 Switch Game（C语言）

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概述

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.

Author

C语言AC代码

#include<stdio.h>
#include<math.h>
int main()
{
    int n;
    double t;
    while(scanf("%d",&n)!=EOF)
    {
        t=sqrt(n*1.0);
        if(t==(int)t)
            printf("1n");
        else
            printf("0n");
    }
    return 0;
}

思路：1 模拟所有情况，超时

　　　　 2 n的的状态由n的约数个数决定。奇数时，开启，偶数时，关闭。

3 约数的个数为奇数时，n为平方数（平方数的两个约数相同），由此，判断n是否为平方数即可。