n条水平平行线与m条垂直平行线相交时的平行四边形个数

给定两个正整数n和m 。任务是计算当 n 条水平平行线与 m 条垂直平行线相交时可以形成任意大小的平行四边形的数量。

例子：

Input : n = 3, m = 2
Output : 3
2 parallelograms of size 1x1 and 1 parallelogram 
of size 2x1.

Input : n = 5, m = 5
Output : 100

这个想法是使用Combination ，它的状态，从给定的 n 个项目中选择 k 个项目的方法数量由ⁿ C _r给出。
要形成平行四边形，我们需要两条水平平行线和两条垂直平行线。因此，选择两条水平平行线的方式数为ⁿ C ₂ ，选择两条垂直平行线的方式数为^m C ₂ 。因此，可能的平行四边形的总数将是ⁿ C ₂ x ^m C ₂ 。
下面是这种方法的 C++ 实现：

C++

// CPP Program to find number of parallelogram when
// n horizontal parallel lines intersect m vertical
// parallel lines.
#include
#define MAX 10
using namespace std;
 
// Find value of Binomial Coefficient
int binomialCoeff(int C[][MAX], int n, int k)
{
    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for (int i = 0; i <= n; i++)
    {
        for (int j = 0; j <= min(i, k); 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 number of parallelogram when n horizontal
// parallel lines intersect m vertical parallel lines.
int countParallelogram(int n, int m)
{
    int  C[MAX][MAX] = { 0 };   
    binomialCoeff(C, max(n, m), 2);   
    return C[n][2] * C[m][2];
}
 
// Driver Program
int main()
{
    int n = 5, m = 5;   
    cout << countParallelogram(n, m) << endl;
    return 0;
}

Java

// Java Program to find number of parallelogram when
// n horizontal parallel lines intersect m vertical
// parallel lines.
class GFG
{
    static final int MAX = 10;
     
    // Find value of Binomial Coefficient
    static void binomialCoeff(int C[][], int n, int k)
    {
        // Calculate value of Binomial Coefficient
        // in bottom up manner
        for (int i = 0; i <= n; i++)
        {
            for (int j = 0; j <= Math.min(i, k); 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 number of parallelogram when n horizontal
    // parallel lines intersect m vertical parallel lines.
    static int countParallelogram(int n, int m)
    {
        int C[][]=new int[MAX][MAX];
         
        binomialCoeff(C, Math.max(n, m), 2);
         
        return C[n][2] * C[m][2];
    }
     
    // Driver code
    public static void main(String arg[])
    {
        int n = 5, m = 5;
        System.out.println(countParallelogram(n, m));
    }
}
 
// This code is contributed By Anant Agarwal.

Python3

# Python Program to find number of parallelogram when
# n horizontal parallel lines intersect m vertical
# parallel lines.
MAX = 10;
 
# Find value of Binomial Coefficient
def binomialCoeff(C, n, k):
     
    # Calculate value of Binomial Coefficient
    # in bottom up manner
    for i in range(n + 1):
        for j in range(0, min(i, k) + 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 number of parallelogram when n horizontal
# parallel lines intersect m vertical parallel lines.
def countParallelogram(n, m):
    C = [[0 for i in range(MAX)] for j in range(MAX)]
 
    binomialCoeff(C, max(n, m), 2);
 
    return C[n][2] * C[m][2];
 
# Driver code
if __name__ == '__main__':
    n = 5;
    m = 5;
    print(countParallelogram(n, m));
 
# This code is contributed by 29AjayKumar

C#

// C# Program to find number of parallelogram when
// n horizontal parallel lines intersect m vertical
// parallel lines.
using System;
 
class GFG
{
    static int MAX = 10;
     
    // Find value of Binomial Coefficient
    static void binomialCoeff(int [,]C, int n, int k)
    {
        // Calculate value of Binomial Coefficient
        // in bottom up manner
        for (int i = 0; i <= n; i++)
        {
            for (int j = 0; j <= Math.Min(i, k); 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 number of parallelogram when n horizontal
    // parallel lines intersect m vertical parallel lines.
    static int countParallelogram(int n, int m)
    {
        int [,]C = new int[MAX, MAX];
         
        binomialCoeff(C, Math.Max(n, m), 2);
         
        return C[n, 2] * C[m, 2];
    }
     
    // Driver code
    public static void Main()
    {
        int n = 5, m = 5;
        Console.WriteLine(countParallelogram(n, m));
    }
}
 
// This code is contributed By vt_m.

Javascript

 
// Javascript Program to find number of parallelogram when
// n horizontal parallel lines intersect m vertical
// parallel lines.
var MAX = 10;
 
// Find value of Binomial Coefficient
function binomialCoeff(C, n, k)
{
    // Calculate value of Binomial Coefficient
    // in bottom up manner
    for (var i = 0; i <= n; i++)
    {
        for (var j = 0; j <= Math.min(i, k); 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 number of parallelogram when n horizontal
// parallel lines intersect m vertical parallel lines.
function countParallelogram(n, m)
{
    var C = Array.from(Array(MAX), () => Array(MAX).fill(0));
    binomialCoeff(C, Math.max(n, m), 2);   
    return C[n][2] * C[m][2];
}
 
// Driver Program
var n = 5, m = 5;   
document.write( countParallelogram(n, m));
 
// This code is contributed by rdtank.

输出：

100

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