Switch Game

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 <stdio.h> 
 2  
 3 int main(){
 4     int n;
 5     int flag[100001];
 6     int i;
 7     int j;
 8     
 9     while(scanf("%d",&n)!=EOF){
10         for(i=0;i<n;i++)
11             flag[i]=0;
12             
13         for(i=1;i<=n;i++){
14             j=i;
15             while(j<=n){
16                 if(flag[j-1]==0)
17                     flag[j-1]=1;
18                     
19                 else
20                     flag[j-1]=0;
21                     
22                 j+=i;
23             }
24         }
25         
26         printf("%d\n",flag[n-1]);
27         
28         
29     }
30         
31     return 0;
32 }

 

 

 

posted @ 2014-10-29 00:32  zqxLonely  阅读(255)  评论(0编辑  收藏  举报