n+2 条边的凸多边形通过连接顶点分裂成三角形的方式数
原文:https://www . geeksforgeeks . org/路数-一个由 n2 条边组成的凸多边形-可以通过连接顶点分割成三角形/
给定一个 n+2 边的凸多边形。任务是计算用不相交的线段连接顶点形成三角形的方法的数量。 例:
输入 : n = 1 输出 : 1 已经是三角形了，所以只能用 1 种方式形成。 输入 : n = 2 输出 : 2 可以用任意一对相对的顶点切割成 2 个三角形。
上面的问题是一个加泰罗尼亚数字的应用。所以，任务就是只找到第 n 个加泰罗尼亚号。前几个加泰罗尼亚数字是 1 1 2 5 14 42 132 429 1430 4862，…(从第 0 个数字开始考虑) 下面是查找第 n 个加泰罗尼亚数字的程序:
C++
// C++ program to find the
// nth catalan number
#include <bits/stdc++.h>
using namespace std;
// Returns value of Binomial Coefficient C(n, k)
unsigned long int binomialCoeff(unsigned int n,
unsigned int k)
{
unsigned long int res = 1;
// Since C(n, k) = C(n, n-k)
if (k > n - k)
k = n - k;
// Calculate value of
// [n*(n-1)*---*(n-k+1)] / [k*(k-1)*---*1]
for (int i = 0; i < k; ++i) {
res *= (n - i);
res /= (i + 1);
}
return res;
}
// A Binomial coefficient based function
// to find nth catalan
// number in O(n) time
unsigned long int catalan(unsigned int n)
{
// Calculate value of 2nCn
unsigned long int c = binomialCoeff(2 * n, n);
// return 2nCn/(n+1)
return c / (n + 1);
}
// Driver code
int main()
{
int n = 3;
cout << catalan(n) << endl;
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find the
// nth catalan number
class GFG
{
// Returns value of Binomial
// Coefficient C(n, k)
static long binomialCoeff(int n,
int k)
{
long res = 1;
// Since C(n, k) = C(n, n-k)
if (k > n - k)
k = n - k;
// Calculate value of
// [n*(n-1)*---*(n-k+1)] /
// [k*(k-1)*---*1]
for (int i = 0; i < k; ++i)
{
res *= (n - i);
res /= (i + 1);
}
return res;
}
// A Binomial coefficient
// based function to find
// nth catalan number in
// O(n) time
static long catalan( int n)
{
// Calculate value of 2nCn
long c = binomialCoeff(2 * n, n);
// return 2nCn/(n+1)
return c / (n + 1);
}
// Driver code
public static void main(String[] args)
{
int n = 3;
System.out.println(catalan(n));
}
}
// This code is contributed
// by Arnab Kundu
Python 3
# Python3 program to find the
# nth catalan number
# Returns value of Binomial
# Coefficient C(n, k)
def binomialCoeff(n, k):
res = 1;
# Since C(n, k) = C(n, n-k)
if (k > n - k):
k = n - k;
# Calculate value of
# [n*(n-1)*---*(n-k+1)] /
# [k*(k-1)*---*1]
for i in range(k):
res *= (n - i);
res /= (i + 1);
return res;
# A Binomial coefficient based
# function to find nth catalan
# number in O(n) time
def catalan(n):
# Calculate value of 2nCn
c = binomialCoeff(2 * n, n);
# return 2nCn/(n+1)
return int(c / (n + 1));
# Driver code
n = 3;
print(catalan(n));
# This code is contributed
# by mits
C
// C# program to find the
// nth catalan number
using System;
class GFG
{
// Returns value of Binomial
// Coefficient C(n, k)
static long binomialCoeff(int n,
int k)
{
long res = 1;
// Since C(n, k) = C(n, n-k)
if (k > n - k)
k = n - k;
// Calculate value of
// [n*(n-1)*---*(n-k+1)] /
// [k*(k-1)*---*1]
for (int i = 0; i < k; ++i)
{
res *= (n - i);
res /= (i + 1);
}
return res;
}
// A Binomial coefficient
// based function to find
// nth catalan number in
// O(n) time
static long catalan( int n)
{
// Calculate value of 2nCn
long c = binomialCoeff(2 * n, n);
// return 2nCn/(n+1)
return c / (n + 1);
}
// Driver code
public static void Main()
{
int n = 3;
Console.WriteLine(catalan(n));
}
}
// This code is contributed
// by Subhadeep
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find the
// nth catalan number
// Returns value of Binomial
// Coefficient C(n, k)
function binomialCoeff($n, $k)
{
$res = 1;
// Since C(n, k) = C(n, n-k)
if ($k > $n - $k)
$k = $n - $k;
// Calculate value of
// [n*(n-1)*---*(n-k+1)] /
// [k*(k-1)*---*1]
for ($i = 0; $i < $k; ++$i)
{
$res *= ($n - $i);
$res /= ($i + 1);
}
return $res;
}
// A Binomial coefficient based
// function to find nth catalan
// number in O(n) time
function catalan($n)
{
// Calculate value of 2nCn
$c = binomialCoeff(2 * $n, $n);
// return 2nCn/(n+1)
return $c / ($n + 1);
}
// Driver code
$n = 3;
echo catalan($n);
// This code is contributed
// by chandan_jnu.
?>
java 描述语言
<script>
// javascript program to find the
// nth catalan number// Returns value of Binomial
// Coefficient C(n, k)
function binomialCoeff(n, k)
{
var res = 1;
// Since C(n, k) = C(n, n-k)
if (k > n - k)
k = n - k;
// Calculate value of
// [n*(n-1)*---*(n-k+1)] /
// [k*(k-1)*---*1]
for (i = 0; i < k; ++i)
{
res *= (n - i);
res /= (i + 1);
}
return res;
}
// A Binomial coefficient
// based function to find
// nth catalan number in
// O(n) time
function catalan(n)
{
// Calculate value of 2nCn
var c = binomialCoeff(2 * n, n);
// return 2nCn/(n+1)
return c / (n + 1);
}
// Driver code
var n = 3;
document.write(catalan(n));
// This code is contributed by Princi Singh
</script>
Output:
5
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