将 N 个元素平均分成至少 2 个组的方法数

给定一个整数N表示元素的数量，任务是找到将这些元素平均分成组的方法数，使得每个组至少有 2 个元素。

例子：

Input: N = 2 
Output: 1
Explanation: There can be only one group.

Input: N = 10
Output: 3
Explanation: There are 3 ways to divide elements:
One group having all the 10 elements.
Two groups where each group has 5 elements.
Five groups where each group has 2 elements.

方法：上述问题可以使用下面给出的蛮力方法来解决。在循环的每次迭代中，我 表示组数。如果N 完全是 可被i整除，因此，元素可以在组之间平均分配。请按照以下步骤解决问题：

• 声明可变方式并将其初始化为0。
• 使用变量i遍历范围[1, N/2]并执行以下任务：
  • 检查N是否可以被完全整除 我。
  • 如果是，则将方式增加1 。
• 执行上述步骤后，打印ways的值作为答案。

下面是上述方法的实现：

C++

// C++ program for the given approach
#include 
using namespace std;
 
// Function to find the number of ways
int numberofWays(int N)
{
    // Variable to store the number of ways
    int ways = 0;
    int i;
 
    // Loop to find total number of ways
    for (i = 1; i <= N / 2; i++) {
        if (N % i == 0)
            ways++;
    }
 
    // Returning the number of ways
    return ways;
}
 
// Driver Code
int main()
{
 
    // Declaring and initialising N
    int N = 10;
 
    // Function call
    int ans = numberofWays(N);
 
    // Displaying the answer on screen
    cout << ans;
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // Function to find the number of ways
  static int numberofWays(int N)
  {
    // Variable to store the number of ways
    int ways = 0;
    int i;
 
    // Loop to find total number of ways
    for (i = 1; i <= N / 2; i++) {
      if (N % i == 0)
        ways++;
    }
 
    // Returning the number of ways
    return ways;
  }
 
  public static void main (String[] args)
  {
 
    // Declaring and initialising N
    int N = 10;
 
    // Function call
    int ans = numberofWays(N);
 
    // Displaying the answer on screen
    System.out.print(ans);
  }
}
 
// This code is contributed by hrithikgarg03188

Python3

# Python code for the above approach
 
# Function to find the number of ways
def numberofWays(N):
 
    # Variable to store the number of ways
    ways = 0;
    i = None
 
    # Loop to find total number of ways
    for i in range(1, (N // 2) + 1):
        if (N % i == 0):
            ways += 1
 
    # Returning the number of ways
    return ways;
 
# Driver Code
 
# Declaring and initialising N
N = 10;
 
# Function call
ans = numberofWays(N);
 
# Displaying the answer on screen
print(ans);
 
# This code is contributed by Saurabh Jaiswal

C#

// C# program for the above approach
using System;
 
class GFG {
 
  // Function to find the number of ways
  static int numberofWays(int N)
  {
    // Variable to store the number of ways
    int ways = 0;
    int i;
 
    // Loop to find total number of ways
    for (i = 1; i <= N / 2; i++) {
      if (N % i == 0)
        ways++;
    }
 
    // Returning the number of ways
    return ways;
  }
 
  public static void Main(string[] args)
  {
 
    // Declaring and initialising N
    int N = 10;
 
    // Function call
    int ans = numberofWays(N);
 
    // Displaying the answer on screen
    Console.WriteLine(ans);
  }
}
 
// This code is contributed by ukasp.

Javascript

        // JavaScript code for the above approach
 
        // Function to find the number of ways
        function numberofWays(N)
        {
         
            // Variable to store the number of ways
            let ways = 0;
            let i;
 
            // Loop to find total number of ways
            for (i = 1; i <= Math.floor(N / 2); i++) {
                if (N % i == 0)
                    ways++;
            }
 
            // Returning the number of ways
            return ways;
        }
 
        // Driver Code
 
        // Declaring and initialising N
        let N = 10;
 
        // Function call
        let ans = numberofWays(N);
 
        // Displaying the answer on screen
        document.write(ans);
 
  // This code is contributed by Potta Lokesh
    

输出

3

时间复杂度： O(N)
辅助空间： O(1)