hdu 2053 Switch Game

Switch Game

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15580    Accepted Submission(s): 9521

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

15

 

Sample Output

10
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

典型的开灯问题，刚开始模拟了一下，然后直接超时了，在这里附上模拟的代码，用的bool型

#include <stdio.h>#include <string.h>int main(){int i,j,n;bool a[100005];while(~scanf("%d",&n)){memset(a,0,sizeof(a));for(i=1;i<=n;i++)for(j=i;j<=n;j+=i){if(a[j]==0)a[j]=1;elsea[j]=0;}printf("%d\n",a[n]);}return 0;}

后来看了看网上他们的博客，原来有规律，约数个数和他的平方根有关系，略微利用一下

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