HDOJ 2053 Switch Game【规律题】

Switch Game

                                                                      Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                          Total Submission(s): 12887    Accepted Submission(s): 7839

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.
   
   
   
   
   
   
   
   


   
   
   
   

题目意思：

有一些灯排成一条直线。所有的灯在刚开始都是关闭的，在对灯进行一系列操作后:在第 i 次操作的时候，调整所有标号是 i 的倍数的灯的状态(原本打开的灯将它关闭，原本关闭的将它打开)。下附代码。

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

一开始看到这个题，打算用数组，可是10^5太大了，用的数组话可能会超时，此路不通；因此用计数器，具体原理是 只要第n个数变过，第m(m>n)次第n个数不再变，而且输出只有0和1，所以计数器对2取余即可。