n 条水平平行线与 m 条垂直平行线相交时的平行四边形数量
原文:https://www . geesforgeks . org/平行四边形数量-当-n-水平-平行-直线-相交-m-垂直-平行/
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。
示例:
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
想法是使用 组合 ，其中状态，从给定的 n 个项目中选择 k 个项目的方式数由 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。所以，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，可能平行四边形的总数将是nC2xmC2。 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<bits/stdc++.h>
#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 语言(一种计算机语言，尤用于创建网站)
// 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.
Python 3
# 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.
java 描述语言
<script>
// 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.
</script>
输出:
100
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