HDU_oj_2053 Switch Game

Posted 2020-10-17 T丶jl

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 }