Switch Game

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7593 Accepted Submission(s): 4490

转载自：  http://blog.sina.com.cn/s/blog_ac5ed4f301019qj2.html

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.

这题思路：

一开始有n盏灯，且全部为关闭状态，都记为 0  就是  The initial condition :      0 0 0 0 0 …

然后之后进行i操作就是对这些灯以是否能被i整除，进行改变状态，如将 0 改为 1 或 将 1 改为 0

正如提醒里的

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 … 

然后题目的输出是 第i次操作的第i个的状态，正如上面红色标记的

代码如下：

#include<stdio.h> 

int count(int n) 

{ 

   int i,j=0; 

   for(i=1;i<=n;i++) 

   { 

       if(n%i==0) 

          j++; 

   } 

   return j; 

} 

int main() 

{ 

    int n; 

    while(~scanf("%d",&n)) 

    { 

        printf("%d\n",count(n)%2); 

    } 

    return 0; 

}

这题还有规律：

我们把前10个写出来就知道了

1.  1 1 1 1 1 1 1 1 1 1 1 1 1

2.  1 0 1 0 1 0 1 0 1 0 1 0 1 

3.  1 0 0 0 1 1 1 0 0 0 1 1 1

4.  1 0 0 1 1 1 1 1 0 0 1 0 1

5.  1 0 0 1 0 1 1 1 0 1 1 0 1

6.  1 0 0 1 0 0 1 1 0 1 1 1 1

7.  1 0 0 1 0 0 0 1 0 1 1 1 1

8.  1 0 0 1 0 0 0 0 0 1 1 1 1

9.  1 0 0 1 0 0 0 0 1 1 1 1 1

10. 1 0 0 1 0 0 0 0 1 0 1 1 1

从这里我们就可以规律了，就是除了 n*n 是 1 ，其余的都是 0 

代码如下：

#include<stdio.h> 

#include<string.h> 

int main() 

{ 

    int n,a[100001],i; 

    memset(a,0,sizeof(a)); 

    for(i=1;i*i<=100000;i++) 

        a[i*i]=1; 

    while(~scanf("%d",&n)) 

    { 

        if(n<0||n>100000) 

        break; 

        printf("%d\n",a[n]);

    }

    return 0;

}