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 偶数的方法数。
示例:
输入: arr[] = {1，2，3，4} k = 1 输出: 2 解释: 我们可以选择一个偶数的方式数是 2。
输入: arr[] = {61，65，99，26，57，68，23，2，32，30} k = 2 输出: 10 说明: 我们可以选择 2 偶数的方式数是 10。
方法:思路是应用组合学的规则。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:
- 计算给定数组中偶数元素的总数(比如 cnt )。
- 检查 K 的值是否大于 cnt ，则路数等于 0。
- 否则，答案将是nCkT5。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
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
// chosen 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 语言(一种计算机语言，尤用于创建网站)
// 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
// chosen 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
Python 3
# 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
# chosen 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
// chosen 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
java 描述语言
<script>
// 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
// chosen 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);
</script>
Output:
2
时间复杂度:O(N) T5】辅助空间: O(N)
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