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
分析：
题目大意为：有一列数，刚开始全为0，第一次将1的倍数的位置的数反转为1
第二次将2的倍数的位置的数反转为0
第三次将3的倍数的位置的数反转为1
……
……
注意点：
1 #include<iostream> 2 #include<cmath> 3 using namespace std; 4 int main() 5 { 6 double n; 7 while(cin>>n) 8 { 9 n = sqrt(n); 10 if(n == (int)n) 11 cout<<1<<endl; 12 else 13 cout<<0<<endl; 14 } 15 }