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.
这题是水题，找出n的约数的个数count，若为偶数则是0，若为奇数则为1；代码如下：
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int n,i,count;
while(scanf("%d",&n)!=EOF)
{
count=0;
for(i=1;i<=sqrt(n);i++)//求n的约数的个数
{
if(n%i==0)
{
count+=2;
if(i==n/i)
count--;
}
}
if(count%2==0) printf("0\n");
else printf("1\n");
}
return 0;
}
