在组合算术中,Narayana数N(n,k) ,n = 1、2、3…,1≤k≤n形成自然数的三角形阵列,称为Narayana三角形。它由下式给出:
Narayana数字N(n,k)可用于查找包含n对括号的表达式的数目,这些括号正确匹配并且包含k个不同的嵌套。
例如,N(4,2)= 6以及四对括号可以创建六个序列,每个序列包含子模式’()’的两倍:
()((())) (())(()) (()(())) ((()())) ((())()) ((()))()
例子:
Input : n = 6, k = 2
Output : 15
Input : n = 8, k = 5
Output : 490
下面是找到N(n,k)的实现:
C++
// CPP program to find Narayana number N(n, k)
#include
using namespace std;
// Return product of coefficient terms in formula
int productofCoefficiet(int n, int k)
{
int C[n + 1][k + 1];
// 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 C[n][k] * C[n][k - 1];
}
// Returns Narayana number N(n, k)
int findNN(int n, int k)
{
return (productofCoefficiet(n, k)) / n;
}
// Driven Program
int main()
{
int n = 8, k = 5;
cout << findNN(n, k) << endl;
return 0;
}
Java
// Java program to find
// Narayana number N(n, k)
class GFG
{
// Return product of coefficient
// terms in formula
static int productofCoefficiet(int n,
int k)
{
int C[][] = new int[n + 1][k + 1];
// 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 C[n][k] * C[n][k - 1];
}
// Returns Narayana number N(n, k)
static int findNN(int n, int k)
{
return (productofCoefficiet(n, k)) / n;
}
// Driver code
public static void main (String[] args)
{
int n = 8, k = 5;
System.out.println(findNN(n, k));
}
}
// This code is contributed by Anant Agarwal.
Python3
# Python3 program to find Narayana number N(n, k)
# Return product of coefficient terms in formula
def productofCoefficiet(n, k):
C = [[0 for x in range(k+1)] for y in range(n+1)]
# Calculate value of Binomial Coefficient
# in bottom up manner
for i in range(0, n+1):
for j in range(0, min(i+1,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 C[n][k] * C[n][k - 1]
# Returns Narayana number N(n, k)
def findNN(n, k):
return (productofCoefficiet(n, k)) / n
# Driven Program
n = 8
k = 5
print(int(findNN(n, k)))
# This code is contributed by Prasad Kshirsagar
C#
// C# program to find
// Narayana number N(n, k)
using System;
class GFG {
// Return product of coefficient
// terms in formula
static int productofCoefficiet(int n,
int k)
{
int[, ] C = new int[n + 1, k + 1];
// 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 C[n, k] * C[n, k - 1];
}
// Returns Narayana number N(n, k)
static int findNN(int n, int k)
{
return (productofCoefficiet(n, k)) / n;
}
// Driver code
public static void Main()
{
int n = 8, k = 5;
Console.WriteLine(findNN(n, k));
}
}
// This code is contributed by vt_m.
PHP
Javascript
输出:
490