Hduoj2053 【水题】

Switch Game

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10929    Accepted Submission(s): 6661

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

LL

 

Source

校庆杯Warm Up 

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

//最简代码

#include<cstdio>
#include<cmath>
int main()
{
	int n,k;
	while(~scanf("%d",&n)){
		k=sqrt(n);
		if(k*k==n)printf("1\n");
		else printf("0\n");
	}
	return 0;
}