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:
- 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则路数将等于 0。
- 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
输出:
2
时间复杂度: O(N)
辅助空间: O(N)