HDU_2053

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.
 1 #include <cstdio>
 2 int main()
 3 {
 4     int n, ans;
 5     while(~scanf("%d",&n))
 6         {
 7             ans=1;
 8             for(int i=1;i<=n;i++)
 9                 {
10                     if(n%i==0)
11                         {
12                             ans=-1*ans;    
13                         }    
14                 }
15             printf(ans==1?"0\n":"1\n");
16         }
17     return 0;    
18 }

 

posted @ 2013-07-11 22:20  瓶哥  Views(390)  Comments(0Edit  收藏  举报