hdu2053

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Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14215    Accepted Submission(s): 8681

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.

水题

给出代码：

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>


using namespace std;


int n;


int main(){
while(scanf("%d",&n)!=EOF){
int ans = 1;
double a = n/2;
for(int i = 1;i <= a;i++){
if(n % i == 0)ans++;
}
if(n == 1)printf("%d\n",1);
else if(ans % 2)printf("%d\n",1);
else printf("%d\n",0);
}
return 0;
}