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 ).
Each test case contains only a number n ( 0< n<= 10^5) in a line.
Output the condition of the n-th lamp after infinity operations ( 0 - off, 1 - on ).
1
0
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 }
