从具有给定约束的三个组形成一个组的方法

给定三个数字(x、y 和 z)，分别表示第一组、第二组和第三组的人数。我们可以通过从第一组、第二组和第三组中选择人员来组成组，使得以下条件不无效。

• 必须从每个组中至少选择一个人。
• 从第一组中选出的人数必须至少比从第三组中选出的人数多一个。

任务是找到形成不同组的方法的数量。
例子：

Input: x = 3, y = 2, z = 1 
Output: 9 
Lets say x has people (a, b, c) 
Y has people (d, e) 
Z has people (f) 
Then the 9 ways are {a, b, d, f}, {a, b, e, f}, {a, c, d, f}, {a, c, e, f}, 
{b, c, d, f}, {b, c, e, f}, {a, b, c, d, f}, {a, b, c, e, f} and 
{a, b, c, d, e, f}
Input: x = 4, y = 2, z = 1 
Output: 27 
 

这个问题可以用组合学来解决。需要填补三个职位(就来自不同群体的人而言)。第一个必须填充一个比第二个位置大一个或多个的数字。第三个可以填充任意数字。我们知道如果我们需要用 N 个人来填补 k 个职位，那么这样做的方法的数量是${N \choose K}$ .因此，可以按照以下步骤来解决上述问题。

• 第二个位置可以填 i=1 到 i=y 的人。
• 第一个位置可以填充 j = i+1 到 j = x 人。
• 第三个位置可以由任意数量的 k = 1 到 k = z 人填补。
• 因此，常见的事情是用 k 个人填充第三个位置。因此，我们可以将那部分视为常见的。
• 运行两个循环(i 和 j)分别填充第二个位置和第一个位置。
• 填补职位的方式数是${y \choose i}$ * ${x \choose j}$ .
• 在计算了填补这两个位置的所有方法之后，我们可以简单地乘以总和${z \choose 1}$ + ${z \choose 2}$ + … ${z \choose z}$因为这是两者的共同部分。

${N \choose K}$可以使用动态规划预先计算以降低时间复杂度。此处讨论该方法。
下面是上述方法的实现。

C++

// C++ program to find the number of
// ways to form the group of peopls
#include 
using namespace std;
 
int C[1000][1000];
 
// Function to pre-compute the
// Combination using DP
void binomialCoeff(int n)
{
    int i, j;
 
    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for (i = 0; i <= n; i++) {
        for (j = 0; j <= i; j++) {
 
            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;
 
            // Calculate value using previously
            // stored values
            else
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
        }
    }
 
    // return C[n][k];
}
 
// Function to find the number of ways
int numberOfWays(int x, int y, int z)
{
    // Function to pre-compute
    binomialCoeff(max(x, max(y, z)));
 
    // Sum the Zci
    int sum = 0;
    for (int i = 1; i <= z; i++) {
        sum = (sum + C[z][i]);
    }
 
    // Iterate for second position
    int sum1 = 0;
    for (int i = 1; i <= y; i++) {
 
        // Iterate for first position
        for (int j = i + 1; j <= x; j++) {
            sum1 = (sum1 + (C[y][i] * C[x][j]));
        }
    }
 
    // Multiply the common Combination value
    sum1 = (sum * sum1);
 
    return sum1;
}
 
// Driver Code
int main()
{
    int x = 3;
    int y = 2;
    int z = 1;
 
    cout << numberOfWays(x, y, z);
 
    return 0;
}

Java

// Java program to find the number of
// ways to form the group of peopls
class GFG
{
     
static int C[][] = new int [1000][1000];
 
// Function to pre-compute the
// Combination using DP
static void binomialCoeff(int n)
{
    int i, j;
 
    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for (i = 0; i <= n; i++)
    {
        for (j = 0; j <= i; j++)
        {
 
            // Base Cases
            if (j == 0 || j == i)
                C[i][j] = 1;
 
            // Calculate value using previously
            // stored values
            else
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
        }
    }
 
    // return C[n][k];
}
 
// Function to find the number of ways
static int numberOfWays(int x, int y, int z)
{
    // Function to pre-compute
    binomialCoeff(Math.max(x, Math.max(y, z)));
 
    // Sum the Zci
    int sum = 0;
    for (int i = 1; i <= z; i++)
    {
        sum = (sum + C[z][i]);
    }
 
    // Iterate for second position
    int sum1 = 0;
    for (int i = 1; i <= y; i++)
    {
 
        // Iterate for first position
        for (int j = i + 1; j <= x; j++)
        {
            sum1 = (sum1 + (C[y][i] * C[x][j]));
        }
    }
 
    // Multiply the common Combination value
    sum1 = (sum * sum1);
 
    return sum1;
}
 
// Driver Code
public static void main(String args[])
{
    int x = 3;
    int y = 2;
    int z = 1;
 
    System.out.println(numberOfWays(x, y, z));
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 program to find the number of
# ways to form the group of peopls
C = [[0 for i in range(1000)]
        for i in range(1000)]
 
# Function to pre-compute the
# Combination using DP
def binomialCoeff(n):
    i, j = 0, 0
 
    # Calculate value of Binomial Coefficient
    # in bottom up manner
    for i in range(n + 1):
        for j in range(i + 1):
 
            # Base Cases
            if (j == 0 or j == i):
                C[i][j] = 1
 
            # Calculate value using previously
            # stored values
            else:
                C[i][j] = C[i - 1][j - 1] + \
                          C[i - 1][j]
 
    # return C[n][k]
 
# Function to find the number of ways
def numberOfWays(x, y, z):
     
    # Function to pre-compute
    binomialCoeff(max(x, max(y, z)))
 
    # Sum the Zci
    sum = 0
    for i in range(1, z + 1):
        sum = (sum + C[z][i])
 
    # Iterate for second position
    sum1 = 0
    for i in range(1, y + 1):
 
        # Iterate for first position
        for j in range(i + 1, x + 1):
            sum1 = (sum1 + (C[y][i] * C[x][j]))
 
    # Multiply the common Combination value
    sum1 = (sum * sum1)
 
    return sum1
 
# Driver Code
x = 3
y = 2
z = 1
 
print(numberOfWays(x, y, z))
 
# This code is contributed by Mohit Kumar

C#

// C# program to find the number of
// ways to form the group of peopls
using System;
     
class GFG
{
     
static int [,]C = new int [1000,1000];
 
// Function to pre-compute the
// Combination using DP
static void binomialCoeff(int n)
{
    int i, j;
 
    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for (i = 0; i <= n; i++)
    {
        for (j = 0; j <= i; j++)
        {
 
            // Base Cases
            if (j == 0 || j == i)
                C[i,j] = 1;
 
            // Calculate value using previously
            // stored values
            else
                C[i,j] = C[i - 1,j - 1] + C[i - 1,j];
        }
    }
 
    // return C[n,k];
}
 
// Function to find the number of ways
static int numberOfWays(int x, int y, int z)
{
    // Function to pre-compute
    binomialCoeff(Math.Max(x, Math.Max(y, z)));
 
    // Sum the Zci
    int sum = 0;
    for (int i = 1; i <= z; i++)
    {
        sum = (sum + C[z,i]);
    }
 
    // Iterate for second position
    int sum1 = 0;
    for (int i = 1; i <= y; i++)
    {
 
        // Iterate for first position
        for (int j = i + 1; j <= x; j++)
        {
            sum1 = (sum1 + (C[y,i] * C[x,j]));
        }
    }
 
    // Multiply the common Combination value
    sum1 = (sum * sum1);
 
    return sum1;
}
 
// Driver Code
public static void Main(String []args)
{
    int x = 3;
    int y = 2;
    int z = 1;
 
    Console.WriteLine(numberOfWays(x, y, z));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

// javascript program to find the number of
// ways to form the group of peopls    
var C = Array(1000).fill().map(()=>Array(1000).fill(0));
 
    // Function to pre-compute the
    // Combination using DP
    function binomialCoeff(n) {
        var i, j;
 
        // Calculate value of Binomial Coefficient
        // in bottom up manner
        for (i = 0; i <= n; i++) {
            for (j = 0; j <= i; j++) {
 
                // Base Cases
                if (j == 0 || j == i)
                    C[i][j] = 1;
 
                // Calculate value using previously
                // stored values
                else
                    C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
            }
        }
 
        // return C[n][k];
    }
 
    // Function to find the number of ways
    function numberOfWays(x , y , z) {
        // Function to pre-compute
        binomialCoeff(Math.max(x, Math.max(y, z)));
 
        // Sum the Zci
        var sum = 0;
        for (i = 1; i <= z; i++) {
            sum = (sum + C[z][i]);
        }
 
        // Iterate for second position
        var sum1 = 0;
        for (i = 1; i <= y; i++) {
 
            // Iterate for first position
            for (j = i + 1; j <= x; j++) {
                sum1 = (sum1 + (C[y][i] * C[x][j]));
            }
        }
 
        // Multiply the common Combination value
        sum1 = (sum * sum1);
 
        return sum1;
    }
 
    // Driver Code
     
        var x = 3;
        var y = 2;
        var z = 1;
 
        document.write(numberOfWays(x, y, z));
 
// This code contributed by aashish1995

输出：

9

时间复杂度： O(K * K )，其中 K 是 (x, y 和 z) 的最大值。

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。