从给定数组中精确选择K个偶数的方法数量

给定n个整数和整数K的数组arr [] ，任务是找到从给定数组中精确选择K个偶数的方法。

例子：

Input: arr[] = {1, 2, 3, 4} k = 1 
Output: 2 
Explanation:
The number of ways in which we can select one even number is 2.

Input: arr[] = {61, 65, 99, 26, 57, 68, 23, 2, 32, 30}  k = 2
Output:10
Explanation:
The number of ways in which we can select 2 even number is 10.

方法：想法是应用组合规则。为了从给定的n个对象中选择r个对象，选择方式的总数由ⁿ C _r给出。步骤如下：

1. 计算给定数组(例如cnt )中偶数元素的总数。
2. 检查K的值是否大于cnt，那么路数将等于0。
3. 否则，答案将为ⁿ C _k 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
long long f[12];
  
// Function for calculating factorial
void fact()
{
    // Factorial of n defined as:
    // n! = n * (n - 1) * ... * 1
    f[0] = f[1] = 1;
  
    for (int i = 2; i <= 10; i++)
        f[i] = i * 1LL * f[i - 1];
}
  
// Function to find the number of ways to
// select exactly K even numbers
// from the given array
void solve(int arr[], int n, int k)
{
    fact();
  
    // Count even numbers
    int even = 0;
    for (int i = 0; i < n; i++) {
  
        // Check if the current
        // number is even
        if (arr[i] % 2 == 0)
            even++;
    }
  
    // Check if the even numbers to be
    // choosen is greater than n. Then,
    // there is no way to pick it.
    if (k > even)
        cout << 0 << endl;
  
    else {
        // The number of ways will be nCk
        cout << f[even] / (f[k] * f[even - k]);
    }
}
  
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof arr / sizeof arr[0];
  
    // Given count of even elements
    int k = 1;
  
    // Function Call
    solve(arr, n, k);
    return 0;
}

Java

// Java program for the above approach
class GFG{
      
static int []f = new int[12];
  
// Function for calculating factorial
static void fact()
{
      
    // Factorial of n defined as:
    // n! = n * (n - 1) * ... * 1
    f[0] = f[1] = 1;
  
    for(int i = 2; i <= 10; i++)
        f[i] = i * 1 * f[i - 1];
}
  
// Function to find the number of ways to
// select exactly K even numbers
// from the given array
static void solve(int arr[], int n, int k)
{
    fact();
  
    // Count even numbers
    int even = 0;
    for(int i = 0; i < n; i++)
    {
          
        // Check if the current
        // number is even
        if (arr[i] % 2 == 0)
            even++;
    }
  
    // Check if the even numbers to be
    // choosen is greater than n. Then,
    // there is no way to pick it.
    if (k > even)
        System.out.print(0 + "\n");
  
    else 
    {
        // The number of ways will be nCk
        System.out.print(f[even] / 
                        (f[k] * f[even - k]));
    }
}
  
// Driver Code
public static void main(String[] args)
{
      
    // Given array arr[]
    int arr[] = { 1, 2, 3, 4 };
    int n = arr.length;
  
    // Given count of even elements
    int k = 1;
  
    // Function call
    solve(arr, n, k);
}
}
  
// This code is contributed by Rajput-Ji

Python3

# Python3 program for the above approach
f = [0] * 12
  
# Function for calculating factorial
def fact():
      
    # Factorial of n defined as:
    # n! = n * (n - 1) * ... * 1
    f[0] = f[1] = 1
  
    for i in range(2, 11):
        f[i] = i * 1 * f[i - 1]
  
# Function to find the number of ways to
# select exactly K even numbers
# from the given array
def solve(arr, n, k):
      
    fact()
  
    # Count even numbers
    even = 0
    for i in range(n):
  
        # Check if the current
        # number is even
        if (arr[i] % 2 == 0):
            even += 1
      
    # Check if the even numbers to be
    # choosen is greater than n. Then,
    # there is no way to pick it.
    if (k > even):
        print(0)
    else:
          
        # The number of ways will be nCk
        print(f[even] // (f[k] * f[even - k]))
      
# Driver Code
  
# Given array arr[]
arr = [ 1, 2, 3, 4 ]
  
n = len(arr)
  
# Given count of even elements
k = 1
  
# Function call
solve(arr, n, k)
  
# This code is contributed by code_hunt

C#

// C# program for the above approach
using System;
class GFG{
      
static int []f = new int[12];
  
// Function for calculating factorial
static void fact()
{
      
    // Factorial of n defined as:
    // n! = n * (n - 1) * ... * 1
    f[0] = f[1] = 1;
  
    for(int i = 2; i <= 10; i++)
        f[i] = i * 1 * f[i - 1];
}
  
// Function to find the number of ways to
// select exactly K even numbers
// from the given array
static void solve(int []arr, int n, int k)
{
    fact();
  
    // Count even numbers
    int even = 0;
    for(int i = 0; i < n; i++)
    {
          
        // Check if the current
        // number is even
        if (arr[i] % 2 == 0)
            even++;
    }
  
    // Check if the even numbers to be
    // choosen is greater than n. Then,
    // there is no way to pick it.
    if (k > even)
        Console.Write(0 + "\n");
  
    else 
    {
        // The number of ways will be nCk
        Console.Write(f[even] / 
                        (f[k] * f[even - k]));
    }
}
  
// Driver Code
public static void Main(String[] args)
{
      
    // Given array []arr
    int []arr = { 1, 2, 3, 4 };
    int n = arr.Length;
  
    // Given count of even elements
    int k = 1;
  
    // Function call
    solve(arr, n, k);
}
}
  
// This code is contributed by sapnasingh4991

输出：

2

时间复杂度： O(N)
辅助空间： O(N)