检查门是否打开或关闭

给定n门n人。门的编号为1到n，人员编号为1到n。每个门只能有2个状态打开和关闭。最初，所有门均处于关闭状态。如果某人更改了所有门的当前状态，即找到状态为打开，然后更改为关闭状态，反之亦然，则找到所有门的最终状态，并向他授权。如果“ j”是“ i”的倍数，则标识为“ i”的人有权更改编号为“ j”的门的状态。
笔记：
–一个人必须更改一次已被授权的所有门的当前状态。
–可能存在这样的情况，即在一个人更改门的状态之前，另一个也被授权使用同一门的人会更改门的状态。
例子 ：

Input : 3
Output : open closed closed

说明：由于n = 3，因此有
3门{1、2、3}和
3位ID为{1、2、3}的人
ID = 1的人可以更改门1、2、3的状态
ID = 2的人可以更改2号门的状态
ID = 3的人可以更改3号门的状态
所有门的当前状态：关闭关闭关闭
考虑一系列事件，

1. id = 1的人更改门2的状态
   所有门的当前状态：关闭打开关闭
2. ID = 3的人更改3号门的状态
   所有门的当前状态：关闭打开打开
3. id = 1的人更改1号门，3号门的状态
   所有门的当前状态：打开打开关闭
4. id = 2的人更改2号门的状态
   所有门的当前状态：打开关闭关闭

另一个例子 ：

Input : 5
Output : open closed closed open closed

Note: Sequence of open/closed is displayed in
increasing door number 

方法：这是数学和逻辑方法。如果我们正确观察，则发现如果i的因子数为奇数，则编号为i的门的最终状态为打开，而如果i的因子数为偶数，则最终状态为关闭。它不取决于门的状态更改顺序。要查找数字除数的计数是偶数还是奇数，我们可以查看检查除数的计数是偶数还是奇数。

C++

// C++ implementation of
// doors open or closed
#include 
using namespace std;
 
// Function to check whether 'n'
// has even number of factors or not
bool hasEvenNumberOfFactors(int n)
{
    int root_n = sqrt(n);
 
    // if 'n' is a perfect square
    // it has odd number of factors
    if ((root_n*root_n) == n)
        return false;
 
    // else 'n' has even
    // number of factors
    return true;
}
 
// Function to find and print
// status of each door
void printStatusOfDoors(int n)
{
    for (int i=1; i<=n; i++)
    {
        // If even number of factors
        // final status is closed
        if (hasEvenNumberOfFactors(i))
            cout << "closed" << " ";
 
        // else odd number of factors
        // final status is open
        else
            cout << "open" << " ";
    }
}
 
// Driver program
int main()
{
    int n = 5;
    printStatusOfDoors(n);
    return 0;
}

Java

// java implementation of
// doors open or closed
import java.io.*;
 
class GFG {
     
    // Function to check whether 'n'
    // has even number of factors or not
    static boolean hasEvenNumberOfFactors(int n)
    {
        double root_n = Math.sqrt(n);
     
        // if 'n' is a perfect square
        // it has odd number of factors
        if ((root_n*root_n) == n)
            return false;
     
        // else 'n' has even
        // number of factors
        return true;
    }
     
    // Function to find and print
    // status of each door
    static void printStatusOfDoors(int n)
    {
        for (int i = 1 ; i <= n; i++)
        {
            // If even number of factors
            // final status is closed
            if (hasEvenNumberOfFactors(i))
                System .out.print( "closed" + " ");
     
            // else odd number of factors
            // final status is open
            else
                System.out.print( "open" + " ");
        }
    }
     
    // Driver program
    public static void main (String[] args) {
        int n = 5;
        printStatusOfDoors(n);
         
    }
}
 
// This article is contributed by vt_m

Python3

# Python 3 implementation of
# doors open or closed
import math
 
# Function to check whether
# 'n' has even number of
# factors or not
def hasEvenNumberOfFactors(n):
 
    root_n = math.sqrt(n)
 
    # if 'n' is a perfect square
    # it has odd number of factors
    if ((root_n * root_n) == n):
        return False
 
    # else 'n' has even
    # number of factors
    return True
 
# Function to find and print
# status of each door
def printStatusOfDoors(n):
 
    for i in range(1, n + 1):
     
        # If even number of factors
        # final status is closed
        if (hasEvenNumberOfFactors(i) == True):
            print("closed", end =" ")
 
        # else odd number of factors
        # final status is open
        else:
            print("open", end =" ")
     
# Driver program
n = 5
 
printStatusOfDoors(n)
 
# This code is contributed by Smitha Dinesh Semwal

C#

// C# implementation of
// doors open or closed
using System;
 
class GFG
{
 
// Function to check whether
// 'n' has even number of
// factors or not
static bool hasEvenNumberOfFactors(int n)
{
    double root_n = Math.Sqrt(n);
 
    // if 'n' is a perfect square
    // it has odd number of factors
    if ((root_n * root_n) == n)
        return false;
 
    // else 'n' has even
    // number of factors
    return true;
}
 
// Function to find and print
// status of each door
static void printStatusOfDoors(int n)
{
    for (int i = 1 ; i <= n; i++)
    {
        // If even number of factors
        // final status is closed
        if (hasEvenNumberOfFactors(i))
            Console.Write("closed" + " ");
 
        // else odd number of factors
        // final status is open
        else
            Console.Write("open" + " ");
    }
}
 
// Driver Code
static public void Main ()
{
    int n = 5;
    printStatusOfDoors(n);
}
}
 
// This Code is contributed by ajit

Javascript

 
// JavaScript program for the above approach
 
    // Function to check whether 'n'
    // has even number of factors or not
    function hasEvenNumberOfFactors(n)
    {
        let root_n = Math.sqrt(n);
     
        // if 'n' is a perfect square
        // it has odd number of factors
        if ((root_n*root_n) == n)
            return false;
     
        // else 'n' has even
        // number of factors
        return true;
    }
     
    // Function to find and print
    // status of each door
    function printStatusOfDoors(n)
    {
        for (let i = 1 ; i <= n; i++)
        {
            // If even number of factors
            // final status is closed
            if (hasEvenNumberOfFactors(i))
                document.write( "closed" + " ");
     
            // else odd number of factors
            // final status is open
            else
                document.write( "open" + " ");
        }
    }
     
// Driver Code
     
    let n = 5;
    printStatusOfDoors(n);
 
// This code is contributed by susmitakundugoaldanga.

输出 ：

open closed closed open closed