给定数字n，请找到用关联运算将n个元素相乘的方式的数量。
例子 ：
Input : 2
Output : 2
For a and b there are two ways to multiply them.
1. (a * b)
2. (b * a)
Input : 3
Output : 12说明(示例2)：
For a, b and c there are 12 ways to multiply them.
1. ((a * b) * c) 2. (a * (b * c))
3. ((a * c) * b) 4. (a * (c * b))
5. ((b * a) * c) 6. (b * (a * c))
7. ((b * c) * a) 8. (b * (c * a))
9. ((c * a) * b) 10. (c * (a * b))
11. ((c * b) * a) 12. (c * (b * a))方法：首先，我们尝试找出递归关系。从上面的示例中，我们可以看到h(1)= 1，h(2)= 2，h(3)= 12。现在，对于n个元素，将有n – 1个乘法和n – 1个括号。并且，(a1，a2，…，an)可以通过以下两种方式之一从(a1，a2，…，a(n – 1))获得：
- 取一个乘法(a1，a2，…，a(n – 1))(具有n – 2个乘法和n – 2个括号)，然后将第n个元素“ an”插入任一n之一的任一侧。 2个乘法。因此，对于n个数字的每个方案，以这种方式给出n个数字的2 * 2 *(n-2)= 4 *(n-2)个方案。
- 采取(a1，a2，..，a(n-1))的乘法方案，并在左边或右边乘以(’an’)。因此，对于n – 1个数字的每个方案，以这种方式给出了n个数字的两个方案。
因此，将以上两个相加后，我们得到h(n)=(4 * n – 8 + 2)* h(n – 1)，h(n)=(4 * n – 6)* h(n – 1) 。具有相同初始值的该递归关系由伪加泰罗尼亚数满足。因此，h(n)=(2 * n – 2)！ /(n – 1)！
C++
// C++ code to find number of ways to multiply n
// elements with an associative operation
# include
using namespace std;
// Function to find the required factorial
int fact(int n)
{
if (n == 0 || n == 1)
return 1 ;
int ans = 1;
for (int i = 1 ; i <= n; i++)
ans = ans * i ;
return ans ;
}
// Function to find nCr
int nCr(int n, int r)
{
int Nr = n , Dr = 1 , ans = 1;
for (int i = 1 ; i <= r ; i++ ) {
ans = ( ans * Nr ) / ( Dr ) ;
Nr-- ;
Dr++ ;
}
return ans ;
}
// function to find the number of ways
int solve ( int n )
{
int N = 2*n - 2 ;
int R = n - 1 ;
return nCr (N, R) * fact(n - 1) ;
}
// Driver code
int main()
{
int n = 6 ;
cout << solve (n) ;
return 0 ;
} Java
// Java code to find number of
// ways to multiply n elements
// with an associative operation
import java.io.*;
class GFG
{
// Function to find the
// required factorial
static int fact(int n)
{
if (n == 0 || n == 1)
return 1 ;
int ans = 1;
for (int i = 1 ; i <= n; i++)
ans = ans * i ;
return ans ;
}
// Function to find nCr
static int nCr(int n, int r)
{
int Nr = n , Dr = 1 , ans = 1;
for (int i = 1 ; i <= r ; i++ )
{
ans = ( ans * Nr ) / ( Dr ) ;
Nr-- ;
Dr++ ;
}
return ans ;
}
// function to find
// the number of ways
static int solve ( int n )
{
int N = 2 * n - 2 ;
int R = n - 1 ;
return nCr (N, R) * fact(n - 1) ;
}
// Driver Code
public static void main (String[] args)
{
int n = 6 ;
System.out.println( solve (n)) ;
}
}
// This code is contributed by anuj_67.Python3
# Python3 code to find number
# of ways to multiply n
# elements with an
# associative operation
# Function to find the
# required factorial
def fact(n):
if (n == 0 or n == 1):
return 1;
ans = 1;
for i in range(1, n + 1):
ans = ans * i;
return ans;
# Function to find nCr
def nCr(n, r):
Nr = n ; Dr = 1 ; ans = 1;
for i in range(1, r + 1):
ans = int((ans * Nr) / (Dr));
Nr = Nr - 1;
Dr = Dr + 1;
return ans;
# function to find
# the number of ways
def solve ( n ):
N = 2* n - 2;
R = n - 1 ;
return (nCr (N, R) *
fact(n - 1));
# Driver code
n = 6 ;
print(solve (n) );
# This code is contributed
# by mitsC#
// C# code to find number of
// ways to multiply n elements
// with an associative operation
using System;
class GFG {
// Function to find the
// required factorial
static int fact(int n)
{
if (n == 0 || n == 1)
return 1 ;
int ans = 1;
for (int i = 1 ; i <= n; i++)
ans = ans * i ;
return ans ;
}
// Function to find nCr
static int nCr(int n, int r)
{
int Nr = n , Dr = 1 , ans = 1;
for (int i = 1 ; i <= r ; i++ )
{
ans = ( ans * Nr ) / ( Dr ) ;
Nr-- ;
Dr++ ;
}
return ans ;
}
// function to find
// the number of ways
static int solve ( int n )
{
int N = 2 * n - 2 ;
int R = n - 1 ;
return nCr (N, R) * fact(n - 1) ;
}
// Driver Code
public static void Main ()
{
int n = 6 ;
Console.WriteLine( solve (n)) ;
}
}
// This code is contributed by anuj_67.PHP
Javascript
输出 ：
30240