将给定的球分成具有相同数量不同颜色的两半的概率

给定一个大小为N的数组arr[] ，代表N 种不同颜色的球的数量，任务是找到将所有球分布到两个盒子中的概率，这样两个盒子都包含相同数量的不同颜色球。

例子：

Input: arr[] = {1, 1}, N = 2
Output: 1.00000
Explanation: Two possible ways to distribute them are as follows: 
Put the ball of 0^th color type in the 1st box and the ball of 1^st color type into the 2nd box. 
Put the color of 1^st type in the 1^st box and the color of 0^th type in the 2nd box. 
Therefore, the probability is equal to (2 / 2) = 1.00000

Input: arr[] = {2, 1, 1}, N = 3
Output: 0.66667

方法：这个想法是使用组合学和回溯来解决问题。可以根据以下观察解决给定的问题：

• Suppose X is the total number of ways to distribute K balls into two equal halves.
• It can be observed that X is the same as choosing K/2 balls from the K balls, which is equal to ^KC_(K/2).
• The total number of ways of distributing the balls such that both the boxes contain the same number of balls and same number of total distinct colors, say Y, can be calculated using Backtracking, by choosing j balls from the arr[i] and putting it in a box for every 0 ≤ i < N and j ≤ arr[i], and place the remaining balls in the other box.
• Therefore, the required probability is Y/X.

请按照以下步骤解决问题：

• 计算一个变量数组arr[]中所有元素的总和，比如K。
• 如果K是奇数，则打印0 。
• 计算选择K/2 个球的方式数并将其存储在变量中，例如X 。
• 定义一个递归函数，比如validPermutations(pos, usedBalls, box1, box2) ，并执行以下步骤：
  • 定义基本情况：如果usedBalls等于K/2 ，则在box1 = box2 时返回1 。否则，返回0 。
  • 如果pos ≥ N ，则返回0 。
  • 现在，初始化一个变量，例如res ，以存储分配剩余球的方式数。
  • 迭代范围[0, arr[pos]]：
    • 将box1和box2分配给变量，分别说newbox1和newbox2 。
    • 如果j > 0和newbox2 ，如果j < arr[pos] ，则将newbox1加一。
    • 现在，更新资源作为解析度= RES + ^{ARR _[POS]ÇĴ*} validPermutations(位置+ 1，usedBalls + J，newbox1，newbox2)。
  • 返回res的值。
• 调用函数validPermutations(0, 0, 0, 0)并将其存储在一个变量中，比如Y。
• 最后，将获得的结果打印为Y/X 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Stores the count of
// distinct colors in box1
static int box1 = 0;
 
// Stores the count of
// distinct colors in box2
static int box2 = 0;
 
static int fact[11];
 
// Function to calculate NcR
long comb(int n, int r)
{
    long res = fact[n] / fact[r];
    res /= fact[n - r];
    return res;
}
 
// Function to calculate factorial of N
void factorial(int N)
{
     
    // Base Case
    fact[0] = 1;
 
    // Iterate over the range [1, N]
    for(int i = 1; i <= N; i++)
        fact[i] = fact[i - 1] * i;
}
 
// Function to calculate total number
// of possible distributions which
// satisfies the given conditions
long validPermutations(int n, int balls[],
                       int usedBalls,
                       int i, int M)
{
     
    // If used balls is equal to K/2
    if (usedBalls == n)
    {
         
        // If box1 is equal to box2
        return box1 == box2 ? 1 : 0;
    }
 
    // Base condition
    if (i >= M)
        return 0;
 
    // Stores the number of ways of
    // distributing remaining balls without
    // including the current balls in box1
    long res = validPermutations(n, balls,
                                 usedBalls,
                                 i + 1, M);
 
    // Increment box1 by one
    box1++;
 
    // Iterate over the range [1, balls[i]]
    for(int j = 1; j <= balls[i]; j++)
    {
         
        // If all the balls goes to box1,
        // then decrease box2 by one
        if (j == balls[i])
 
            box2--;
 
        // Total number of ways of
        // selecting j balls
        long combinations = comb(balls[i], j);
 
        // Increment res by total number of valid
        // ways of distributing the remaining balls
        res += combinations * validPermutations(n, balls,
                                                usedBalls + j,
                                                i + 1, M);
    }
 
    // Decrement box1 by one
    box1--;
 
    // Increment box2 by 1
    box2++;
 
    return res;
}
 
// Function to calculate the required probability
double getProbability(int balls[], int M)
{
     
    // Calculate factorial from [1, 10]
    factorial(10);
 
    // Assign all distinct balls to second box
    box2 = M;
 
    // Total number of balls
    int K = 0;
 
    // Calculate total number of balls
    for(int i = 0; i < M; i++)
        K += balls[i];
 
    // If K is an odd number
    if (K % 2 == 1)
        return 0;
 
    // Total ways of distributing the balls
    // in two equal halves
    long all = comb(K, K / 2);
 
    // Required number of ways
    long validPermutation = validPermutations(K / 2, balls,
                                              0, 0, M);
 
    // Return the required probability
    return (double)validPermutation / all;
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 1, 1 };
    int N = 4;
    int M = sizeof(arr) / sizeof(arr[0]);
 
    // Print the result
    cout << (getProbability(arr, M));
     
    return 0;
}
 
// This code is contributed by ukasp

Java

// Java program for the above approach
 
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Stores the count of
    // distinct colors in box1
    static int box1 = 0;
 
    // Stores the count of
    // distinct colors in box2
    static int box2 = 0;
 
    static int[] fact = new int[11];
 
    // Function to calculate the required probability
    public static double getProbability(int[] balls)
    {
 
        // Calculate factorial from [1, 10]
        factorial(10);
 
        // Assign all distinct balls to second box
        box2 = balls.length;
 
        // Total number of balls
        int K = 0;
 
        // Calculate total number of balls
        for (int i = 0; i < balls.length; i++)
            K += balls[i];
 
        // If K is an odd number
        if (K % 2 == 1)
            return 0;
 
        // Total ways of distributing the balls
        // in two equal halves
        long all = comb(K, K / 2);
 
        // Required number of ways
        long validPermutations = validPermutations(K / 2, balls, 0, 0);
 
        // Return the required probability
        return (double)validPermutations / all;
    }
 
    // Function to calculate total number
    // of possible distributions which
    // satisfies the given conditions
    static long validPermutations(int n, int[] balls,
                          int usedBalls, int i)
    {
 
        // If used balls is equal to K/2
        if (usedBalls == n) {
 
            // If box1 is equal to box2
            return box1 == box2 ? 1 : 0;
        }
 
        // Base condition
        if (i >= balls.length)
            return 0;
 
        // Stores the number of ways of
        // distributing remaining balls without
        // including the current balls in box1
        long res = validPermutations(n, balls, usedBalls, i + 1);
 
        // Increment box1 by one
        box1++;
 
        // Iterate over the range [1, balls[i]]
        for (int j = 1; j <= balls[i]; j++) {
 
            // If all the balls goes to box1,
            // then decrease box2 by one
            if (j == balls[i])
                box2--;
 
            // Total number of ways of
            // selecting j balls
            long combinations = comb(balls[i], j);
 
            // Increment res by total number of valid
            // ways of distributing the remaining balls
            res += combinations * validPermutations(n, balls,
                                            usedBalls + j, i + 1);
        }
 
        // Decrement box1 by one
        box1--;
 
        // Increment box2 by 1
        box2++;
 
        return res;
    }
 
    // Function to calculate factorial of N
    static void factorial(int N)
    {
 
        // Base Case
        fact[0] = 1;
 
        // Iterate over the range [1, N]
        for (int i = 1; i <= N; i++)
            fact[i] = fact[i - 1] * i;
    }
 
    // Function to calculate NcR
    static long comb(int n, int r)
    {
 
        long res = fact[n] / fact[r];
        res /= fact[n - r];
        return res;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 2, 1, 1 };
        int N = 4;
 
        // Print the result
        System.out.println(getProbability(arr));
    }
}

C#

// C# program for the above approach
using System;
public class GFG
{
 
    // Stores the count of
    // distinct colors in box1
    static int box1 = 0;
 
    // Stores the count of
    // distinct colors in box2
    static int box2 = 0;
    static int[] fact = new int[11];
 
    // Function to calculate the required probability
    public static double getProbability(int[] balls)
    {
 
        // Calculate factorial from [1, 10]
        factorial(10);
 
        // Assign all distinct balls to second box
        box2 = balls.Length;
 
        // Total number of balls
        int K = 0;
 
        // Calculate total number of balls
        for (int i = 0; i < balls.Length; i++)
            K += balls[i];
 
        // If K is an odd number
        if (K % 2 == 1)
            return 0;
 
        // Total ways of distributing the balls
        // in two equal halves
        long all = comb(K, K / 2);
 
        // Required number of ways
        long validPermutationss = validPermutations((K / 2), balls, 0, 0);
 
        // Return the required probability
        return (double)validPermutationss / all;
    }
 
    // Function to calculate total number
    // of possible distributions which
    // satisfies the given conditions
    static long validPermutations(int n, int[] balls,
                          int usedBalls, int i)
    {
 
        // If used balls is equal to K/2
        if (usedBalls == n)
        {
 
            // If box1 is equal to box2
            return box1 == box2 ? 1 : 0;
        }
 
        // Base condition
        if (i >= balls.Length)
            return 0;
 
        // Stores the number of ways of
        // distributing remaining balls without
        // including the current balls in box1
        long res = validPermutations(n, balls, usedBalls, i + 1);
 
        // Increment box1 by one
        box1++;
 
        // Iterate over the range [1, balls[i]]
        for (int j = 1; j <= balls[i]; j++)
        {
 
            // If all the balls goes to box1,
            // then decrease box2 by one
            if (j == balls[i])
                box2--;
 
            // Total number of ways of
            // selecting j balls
            long combinations = comb(balls[i], j);
 
            // Increment res by total number of valid
            // ways of distributing the remaining balls
            res += combinations * validPermutations(n, balls,
                                            usedBalls + j, i + 1);
        }
 
        // Decrement box1 by one
        box1--;
 
        // Increment box2 by 1
        box2++;
 
        return res;
    }
 
    // Function to calculate factorial of N
    static void factorial(int N)
    {
 
        // Base Case
        fact[0] = 1;
 
        // Iterate over the range [1, N]
        for (int i = 1; i <= N; i++)
            fact[i] = fact[i - 1] * i;
    }
 
    // Function to calculate NcR
    static long comb(int n, int r)
    {
 
        long res = fact[n] / fact[r];
        res /= fact[n - r];
        return res;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int[] arr = { 2, 1, 1 };
        int N = 4;
 
        // Print the result
        Console.WriteLine(getProbability(arr));
    }
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript program for the above approach
 
// Stores the count of
// distinct colors in box1
var box1 = 0;
 
// Stores the count of
// distinct colors in box2
var box2 = 0;
 
var fact = Array(11);
 
// Function to calculate NcR
function comb(n, r)
{
    var res = fact[n] / fact[r];
    res /= fact[n - r];
    return res;
}
 
// Function to calculate factorial of N
function factorial(N)
{
     
    // Base Case
    fact[0] = 1;
 
    // Iterate over the range [1, N]
    for(var i = 1; i <= N; i++)
        fact[i] = fact[i - 1] * i;
}
 
// Function to calculate total number
// of possible distributions which
// satisfies the given conditions
function validPermutations(n, balls,usedBalls, i, M)
{
     
    // If used balls is equal to K/2
    if (usedBalls == n)
    {
         
        // If box1 is equal to box2
        return box1 == box2 ? 1 : 0;
    }
 
    // Base condition
    if (i >= M)
        return 0;
 
    // Stores the number of ways of
    // distributing remaining balls without
    // including the current balls in box1
    var res = validPermutations(n, balls,
                                 usedBalls,
                                 i + 1, M);
 
    // Increment box1 by one
    box1++;
 
    // Iterate over the range [1, balls[i]]
    for(var j = 1; j <= balls[i]; j++)
    {
         
        // If all the balls goes to box1,
        // then decrease box2 by one
        if (j == balls[i])
 
            box2--;
 
        // Total number of ways of
        // selecting j balls
        var combinations = comb(balls[i], j);
 
        // Increment res by total number of valid
        // ways of distributing the remaining balls
        res += combinations * validPermutations(n, balls,
                                                usedBalls + j,
                                                i + 1, M);
    }
 
    // Decrement box1 by one
    box1--;
 
    // Increment box2 by 1
    box2++;
 
    return res;
}
 
// Function to calculate the required probability
function getProbability(balls, M)
{
     
    // Calculate factorial from [1, 10]
    factorial(10);
 
    // Assign all distinct balls to second box
    box2 = M;
 
    // Total number of balls
    var K = 0;
 
    // Calculate total number of balls
    for(var i = 0; i < M; i++)
        K += balls[i];
 
    // If K is an odd number
    if (K % 2 == 1)
        return 0;
 
    // Total ways of distributing the balls
    // in two equal halves
    var all = comb(K, K / 2);
 
    // Required number of ways
    var validPermutation = validPermutations(K / 2, balls,
                                              0, 0, M);
 
    // Return the required probability
    return validPermutation / all;
}
 
// Driver Code
var arr = [2, 1, 1];
var N = 4;
var M = arr.length;
// Print the result
document.write(getProbability(arr, M));
 

输出：

0.6666666666666666

时间复杂度： O(N!)
辅助空间： O(1)