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
hintThe 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 ",flag[n-1]); 27 28 29 } 30 31 return 0; 32 }