戴克路径

考虑一个左上角索引为(0，0)的anxn网格。 Dyck路径是从左下角(即(n-1，0)到右上角，即(0，n-1))位于对角单元格(或从左下角到右上角的线上的单元格)上方的阶梯步道。
任务是计算从(n-1，0)到(0，n-1)的Dyck路径数。
例子 ：

Input : n = 1
Output : 1

Input : n = 2
Output : 2

Input : n = 3
Output : 5

Input : n = 4
Output : 14

从(n-1，0)到(0，n-1)的Dyck路径数可以由加泰罗尼亚语数C(n)给出。
$C_n=\frac{(2n)!}{(n+1)!n1}=\prod_{k=2}^{n}\frac{n+k}{k} \ for\ n\geq 0$

我们强烈建议您单击此处并进行实践，然后再继续解决方案。

以下是查找Dyck路径(或第n个加泰罗尼亚语编号)的计数的实现。

C++

// C++ program to count
// number of Dyck Paths
#include
using namespace std;
 
// Returns count Dyck
// paths in n x n grid
int countDyckPaths(unsigned int n)
{
    // Compute value of 2nCn
    int res = 1;
    for (int i = 0; i < n; ++i)
    {
        res *= (2 * n - i);
        res /= (i + 1);
    }
 
    // return 2nCn/(n+1)
    return res / (n+1);
}
 
// Driver Code
int main()
{
    int n = 4;
    cout << "Number of Dyck Paths is "
         << countDyckPaths(n);
    return 0;
}

Java

// Java program to count
// number of Dyck Paths
class GFG
{
    // Returns count Dyck
    // paths in n x n grid
    public static int countDyckPaths(int n)
    {
        // Compute value of 2nCn
        int res = 1;
        for (int i = 0; i < n; ++i)
        {
            res *= (2 * n - i);
            res /= (i + 1);
        }
 
        // return 2nCn/(n+1)
        return res / (n + 1);
    }
 
    // Driver code
    public static void main(String args[])
    {
        int n = 4;
        System.out.println("Number of Dyck Paths is " +
                                    countDyckPaths(n));
    }
}

Python3

# Python3 program to count
# number of Dyck Paths
 
# Returns count Dyck
# paths in n x n grid
def countDyckPaths(n):
     
    # Compute value of 2nCn
    res = 1
    for i in range(0, n):
        res *= (2 * n - i)
        res /= (i + 1)
 
    # return 2nCn/(n+1)
    return res / (n+1)
 
# Driver Code
n = 4
print("Number of Dyck Paths is ",
    str(int(countDyckPaths(n))))
 
# This code is contributed by
# Prasad Kshirsagar

C#

// C# program to count
// number of Dyck Paths
using System;
 
class GFG {
     
    // Returns count Dyck
    // paths in n x n grid
    static int countDyckPaths(int n)
    {
         
        // Compute value of 2nCn
        int res = 1;
        for (int i = 0; i < n; ++i)
        {
            res *= (2 * n - i);
            res /= (i + 1);
        }
 
        // return 2nCn/(n+1)
        return res / (n + 1);
    }
 
    // Driver code
    public static void Main()
    {
        int n = 4;
        Console.WriteLine("Number of "
                  + "Dyck Paths is " +
                   countDyckPaths(n));
    }
}
 
// This code is contributed by anuj_67.

Javascript

 
// JavaScript program to count
// number of Dyck Paths
 
    // Returns count Dyck
    // paths in n x n grid
    function countDyckPaths(n)
    {
     
        // Compute value of 2nCn
        let res = 1;
        for (let i = 0; i < n; ++i)
        {
            res *= (2 * n - i);
            res /= (i + 1);
        }
   
        // return 2nCn/(n+1)
        return res / (n + 1);
    }
 
// Driver Code
 
        let n = 4;
        document.write("Number of Dyck Paths is " +
                                    countDyckPaths(n));
     
    // This code is contributed by target_2.

输出 ：

Number of Dyck Paths is 14

锻炼 ：

1. 找到1和-1的序列数，使每个序列遵循以下约束：
   a)序列的长度为2n
   b)1和-1的个数相等，即n 1个，n -1个
   c)每个序列的前缀总和大于或等于0。例如，1，-1，1，-1和1，1，-1，-1是有效的，但是-1，-1，1， 1无效。
2. 从(m-1，0)到(0，n-1)的长度为m + n的路径数，仅限于东和北台阶。