检查从 1 到 N 的因子总数的总和是偶数还是奇数

给定一个整数N 。任务是检查从1到N的因子总数的总和是偶数还是奇数，即sum % 2 = 0 或 1 。

例子：

Input: N = 3
Output: Odd
Explanation: 1 has 1 factor, 2 has 2 factors (1, 2) and 3 has 2 factors (1, 3), 
Hence, total number of factors is 5 which is odd. 

Input: N = 4
Output: Even
Explanation: 1 has 1 factor, 2 has 2 factors (1, 2), 3 has 2 factors (1, 3) and 4 has 3 factors (1, 2, 4),  
Hence, a total of 8 factors which is even.

朴素的方法：解决问题的基本思路是首先找到从1到N的数字的所有因数，然后存储它们的和并检查和是偶数还是奇数。

下面是上述方法的实现。

C++

// C++ code for the above approach:
 
#include 
using namespace std;
 
// Function to find the factors
int factor(int n)
{
    int count = 0;
 
    // Counting the factors for
    // all the numbers.
    for (int i = 1; i <= n; i++) {
 
        // Counting the factors
        // for number i
        for (int j = 1; j < sqrt(n) + 1;
             j++) {
            if (i % j == 0) {
                if (i / j == j) {
                    count += 1;
                }
                else {
                    count += 2;
                }
                count %= 2;
            }
        }
    }
 
    // Return the count
    return count;
}
 
// Driver Code
int main()
{
    int N = 4;
 
    // Function call
    if (factor(N))
        cout << "Odd";
    else
        cout << "Even";
    return 0;
}

Java

// Java code for the above approach:
import java.io.*;
class GFG {
 
  // Function to find the factors
  static int factor(int n)
  {
    int count = 0;
 
    // Counting the factors for
    // all the numbers.
    for (int i = 1; i <= n; i++) {
 
      // Counting the factors
      // for number i
      for (int j = 1; j < Math.sqrt(n) + 1; j++) {
        if (i % j == 0) {
          if (i / j == j) {
            count += 1;
          }
          else {
            count += 2;
          }
          count %= 2;
        }
      }
    }
 
    // Return the count
    return count;
  }
 
  // Driver Code
  public static void main (String[] args) {
    int N = 4;
 
    // Function call
    if (factor(N) != 0)
      System.out.print("Odd");
    else
      System.out.print("Even");
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python code for the above approach:
 
# Function to find the factors
import math
def factor(n):
 
   count = 0
 
    # Counting the factors for
    # all the numbers.
   for i in range(1, n + 1):
 
        # Counting the factors
        # for number i
        for j in range(1,math.floor(math.sqrt(n)+1)):
            if (i % j == 0):
               if (i / j == j):
                  count += 1
               else:
                  count += 2
               count %= 2
 
    # Return the count
   return count
 
# Driver Code
N = 4
 
# Function call
if(factor(N)):
   print("Odd")
else:
   print("Even")
 
# This code is contributed by shinjanpatra

C#

// C# code for the above approach:
using System;
class GFG {
 
  // Function to find the factors
  static int factor(int n)
  {
    int count = 0;
 
    // Counting the factors for
    // all the numbers.
    for (int i = 1; i <= n; i++) {
 
      // Counting the factors
      // for number i
      for (int j = 1; j < Math.Sqrt(n) + 1; j++) {
        if (i % j == 0) {
          if (i / j == j) {
            count += 1;
          }
          else {
            count += 2;
          }
          count %= 2;
        }
      }
    }
 
    // Return the count
    return count;
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 4;
 
    // Function call
    if (factor(N) != 0)
      Console.Write("Odd");
    else
      Console.Write("Even");
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

 
// JavaScripy code for the above approach:
 
// Function to find the factors
function factor(n)
{
    let count = 0;
 
    // Counting the factors for
    // all the numbers.
    for (let i = 1; i <= n; i++) {
 
        // Counting the factors
        // for number i
        for (let j = 1; j < Math.sqrt(n) + 1; j++) {
            if (i % j == 0) {
                if (i / j == j) {
                    count += 1;
                }
                else {
                    count += 2;
                }
                count %= 2;
            }
        }
    }
 
    // Return the count
    return count;
}
 
// Driver Code
let N = 4;
 
// Function call
if (factor(N))
    document.write("Odd");
else
    document.write("Even");
     
 // This code is contributed by shinjanpatra
  
 

输出

Even

时间复杂度： O(N*sqrt(N))
辅助空间： O(1)

有效方法：有效解决问题的想法基于以下观察。

As, all numbers except perfect squares, have an even number of factors. So, the concern is only perfect squares. 
Find the number of perfect squares that lie between 1 to N. 
If the number of perfect squares is odd then total number of factors is odd, otherwise, even.

请按照以下步骤解决问题：

• 声明任何计数变量来存储一个数的因数，并将其初始化为 0。
• 使用从1 到 N的完美正方形总数更新计数，这与 sqrt(N) 相同。
• 检查count是偶数还是奇数，并相应地返回 0 或 1。

下面是上述方法的实现：

C++

// C++ code for the above approach:
 
#include 
using namespace std;
 
// Function to find the count of factors
int factor(int N)
{
    int count = 0;
 
    // Finding out the number of
    // perfect squares that lie
    // between 1 to N.
    count = (int)sqrt(N);
    count = count % 2;
    return count;
}
 
// Driver code
int main()
{
    int N = 4;
 
    // Function call
    if (factor(N))
        cout << "Odd";
    else
        cout << "Even";
    return 0;
}

Java

// JAVA code for the above approach:
import java.util.*;
class GFG
{
 
  // Function to find the count of factors
  public static int factor(int N)
  {
    int count = 0;
 
    // Finding out the number of
    // perfect squares that lie
    // between 1 to N.
    count = (int)Math.sqrt(N);
    count = count % 2;
    return count;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int N = 4;
 
    // Function call
    if (factor(N) != 0)
      System.out.print("Odd");
    else
      System.out.print("Even");
  }
}
 
// This code is contributed by Taranpreet

Python3

# Python code for the above approach:
 
# Function to find the count of factors
import math
 
def factor(N):
 
    count = 0
 
    # Finding out the number of
    # perfect squares that lie
    # between 1 to N.
    count = math.floor(math.sqrt(N))
    count = count % 2
    return count
 
# Driver code
 
N = 4
 
# Function call
if (factor(N)):
    print("Odd")
else:
    print("Even")
     
    # This code is contributed by shinjanpatra.

C＃

// C# code for the above approach:
using System;
class GFG {
 
    // Function to find the count of factors
    static int factor(int N)
    {
        int count = 0;
 
        // Finding out the number of
        // perfect squares that lie
        // between 1 to N.
        count = (int)Math.Sqrt(N);
        count = count % 2;
        return count;
    }
 
    // Driver code
    public static void Main()
    {
        int N = 4;
 
        // Function call
        if (factor(N) != 0)
            Console.Write("Odd");
        else
            Console.Write("Even");
    }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

 
// JavaScript code for the above approach:
 
// Function to find the count of factors
function factor(N)
{
 
    let count = 0
 
    // Finding out the number of
    // perfect squares that lie
    // between 1 to N.
    count = Math.floor(Math.sqrt(N))
    count = count % 2
    return count
}
 
// Driver code
let N = 4
 
// Function call
if (factor(N))
    document.write("Odd","</br>")
else
    document.write("Even","</br>")
 
// This code is contributed by shinjanpatra
 

输出

Even

时间复杂度： O(log N)
辅助空间： O(1)