HDU 2053 Switch Game

最新推荐文章于 2019-11-15 15:00:22 发布

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Switch Game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10561 Accepted Submission(s): 6426

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

Source

校庆杯Warm Up

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Statistic | Submit | Discuss | Note

/*只要是倍数就进行一次开关操作，而每个倍数可以写成两个数相乘的形式比如6：
1*6
2*3
3*2
6*1
中间两次其实没做  所以最后一次就关掉了。

而对于完全平方数又多了个a*a的运算，比上面多一个，那么最后一步一定是开灯操作，所以只要是完全平方数那么就是亮着的。
*/
#include<iostream>
#include<stdio.h>
#include<math.h>
using namespace std;
int main()
{
    double n;
    while(scanf("%lf",&n)!=EOF)
    {
        n=sqrt(n);
        if(n==(int)n)
        printf("1\n");
        else
        printf("0\n");
    }
    return 0;
}