Switch Game
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
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.
Hint
hint
AC Code
#include<stdio.h>
#include<string.h>
int main(void)
{
int n,arr[100001],i;
memset(arr,0,sizeof(arr));
for(i = 1; i * i <= 100000; i ++)
arr[i * i] = 1;
while(scanf("%d",&n) != EOF)
{
if(n<0||n>100000)
break;
printf("%d\n",arr[n]);
}
return 0;
}
Thinking
N has a start lamp, and all the off state, are referred to as 0 is The initial condition: 0 0 0 0 0 ...
After then i operating these lamps is whether i is divisible, change state, as it will be 0 to 1 or 1 to 0
As a reminder in the
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 …
Then the subject is the i-th output of the i-th state of operation, the red mark as above
The first ten to write the laws can be found
- 1 1 1 1 1 1 1 1 1 1 1 1 1
- 1 0 1 0 1 0 1 0 1 0 1 0 1
- 1 0 0 0 1 1 1 0 0 0 1 1 1
- 1 0 0 1 1 1 1 1 0 0 1 0 1
- 1 0 0 1 0 1 1 1 0 1 1 0 1
- 1 0 0 1 0 0 1 1 0 1 1 1 1
- 1 0 0 1 0 0 0 1 0 1 1 1 1
- 1 0 0 1 0 0 0 0 0 1 1 1 1
- 1 0 0 1 0 0 0 0 1 1 1 1 1
- 1 0 0 1 0 0 0 0 1 0 1 1 1
In addition to n * n is 1, and the rest are all 0
