网格中从一个点到另一个点的方式数量

给定水平和垂直道路的NxN网格。任务是找出人可以使用最短路径从点A到达点B的方式。
注意： A和B点是固定的，即A在左上角，B在右下角，如下图所示。

在上图中，以红色和浅绿色显示的路径是从点A到达点B的两条可能路径。
例子：

Input: N = 3
Output: Ways = 20

Input: N = 4
Output: Ways = 70

公式：
令网格为N x N，可以将多种方式写为。

以上公式如何运作?
让我们考虑如上所示的5×5网格的示例。为了在5×5网格中从点A到达点B，我们必须采取5个水平步长和5个垂直步长。每条路径将是10个步骤的排列，其中5个步骤是一种相同，其他5个步骤是第二种相同。所以
方式数= 10！ /(5！* 5！)，即252种方式。

C++

// C++ implementation of above approach
#include 
using namespace std;
 
// function that will
// calculate the factorial
long factorial(int N)
{
    int result = 1;
    while (N > 0) {
        result = result * N;
        N--;
    }
    return result;
}
 
long countWays(int N)
{
    long total = factorial(N + N);
    long total1 = factorial(N);
    return (total / total1) / total1;
}
 
// Driver code
int main()
{
    int N = 5;
    cout << "Ways = " << countWays(N);
    return 0;
}

Java

// Java implementation of above approach
class GfG {
 
    // function that will
    // calculate the factorial
    static long factorial(int N)
    {
        int result = 1;
        while (N > 0) {
            result = result * N;
            N--;
        }
        return result;
    }
 
    static long countWays(int N)
    {
        long total = factorial(N + N);
        long total1 = factorial(N);
        return (total / total1) / total1;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int N = 5;
        System.out.println("Ways = " + countWays(N));
    }
}

Python3

# Python3 implementation of above approach
 
# function that will calculate the factorial
def factorial(N) :
     
    result = 1;
     
    while (N > 0) :
         
        result = result * N;
        N -= 1;
     
    return result;
 
def countWays(N) :
 
    total = factorial(N + N);
    total1 = factorial(N);
     
    return (total // total1) // total1;
 
# Driver code
if __name__ == "__main__" :
 
    N = 5;
     
    print("Ways =", countWays(N));
 
# This code is contributed by Ryuga

C#

// C# implementation of above approach
using System;
class GfG
{
 
    // function that will
    // calculate the factorial
    static long factorial(int N)
    {
        int result = 1;
        while (N > 0)
        {
            result = result * N;
            N--;
        }
        return result;
    }
 
    static long countWays(int N)
    {
        long total = factorial(N + N);
        long total1 = factorial(N);
        return (total / total1) / total1;
    }
 
    // Driver code
    public static void Main(String []args)
    {
        int N = 5;
        Console.WriteLine("Ways = " + countWays(N));
    }
}
 
// This code is contributed by Arnab Kundu

PHP

 0)
    {
        $result = $result * $N;
        $N--;
    }
    return $result;
}
 
function countWays($N)
{
    $total = factorial($N + $N);
    $total1 = factorial($N);
    return ($total / $total1) / $total1;
}
 
// Driver code
$N = 5;
echo "Ways = ", countWays($N);
     
// This code is contributed by ajit
?>

Javascript

 
// Javascript implementation of above approach
 
// function that will
// calculate the factorial
function factorial(N)
{
    var result = 1;
    while (N > 0) {
        result = result * N;
        N--;
    }
    return result;
}
 
function countWays(N)
{
    var total = factorial(N + N);
    var total1 = factorial(N);
    return (total / total1) / total1;
}
 
// Driver code
var N = 5;
document.write( "Ways = " + countWays(N));
 
// This code is contributed by rutvik_56.

输出：

Ways = 252