从给定数组中精确选择 K 个偶数的方法数

给定一个由 n 个整数组成的数组arr[]和一个整数K ，任务是找到从给定数组中精确选择K 个偶数的方法数。

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：这个想法是应用组合学的规则。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。以下是步骤：

1. 计算给定数组中偶数元素的总数(比如cnt )。
2. 检查K的值是否大于 cnt则路数将等于 0。
3. 否则，答案将是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

Javascript

 
// Javascript program for the above approach
var f = Array(12).fill(0);
 
// Function for calculating factorial
function fact()
{
    // Factorial of n defined as:
    // n! = n * (n - 1) * ... * 1
    f[0] = f[1] = 1;
 
    for (var 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
function solve(arr, n, k)
{
    fact();
 
    // Count even numbers
    var even = 0;
    for (var 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)
        document.write( 0 );
 
    else {
        // The number of ways will be nCk
        document.write( f[even] / (f[k] * f[even - k]));
    }
}
 
// Driver Code
 
// Given array arr[]
var arr = [ 1, 2, 3, 4 ];
var n = arr.length;
 
// Given count of even elements
var k = 1;
 
// Function Call
solve(arr, n, k);
 
 

输出：

2

时间复杂度： O(N)
辅助空间： O(N)

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