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📜  计算斯特林数,该数代表在n个不同的圆周围排列r个对象的方式的数量

📅  最后修改于: 2021-06-25 11:31:34             🧑  作者: Mango

S(r,n)表示我们可以在长度为n的不可区分的圆周围排列r个对象的方式的数量,并且每个圆n周围都必须至少有一个对象。

例子:

Input: r = 9, n = 2
Output: 109584

Input: r = 6, n = 3
Output: 225

特殊情况是:

  • S(r,0)= 0,微不足道。
  • S(r,1)表示等于(r – 1)的循环置换!
  • S(r,n)其中r = n等于1。
  • S(r,r -1)= rC2

S(r,n)= S(r – 1,n – 1)+(r – 1)* S(r – 1,n)的斯特林数的重要标识

方法:为简单起见,用1、2,…,r表示r个不同的对象。考虑对象“ 1”。在对象的任何排列中,

  1. “ 1”是圆圈中的唯一对象或
  2. “ 1”与其他人围成一圈。

在情况1中,有s(r – 1,n – 1)种方式构成这种安排。在情况2中,首先,以s(r_1,n)的方式将r_1个对象2、3,…,r放入n个圆中;然后可以将“ 1”放置在r-1个不同对象之一中,位于相应r-1个不同对象的“直接权利”附近。根据乘法原理,在情况2中有(r-1)s(r_1,n)种形式的排列方式。现在,根据s(r,n)的定义和加法原理来确定恒等式。
使用初始值S(0,0)= 1,当r> 1且s(r,1)=(r_1)时,s(r,0)= 0!对于r> 1,并应用我们证明的身份,我们可以通过递归方式轻松计算出斯特林数。

在代码中,我们使用了三个函数来生成斯特林数,分别是nCr(n,r),这是一个计算我们所谓的函数(n –选择– r),我们可以使用r的方式数来自n个对象的对象,而没有排序的重要性。毫不奇怪,阶乘(int n)用于计算数字n的阶乘。函数Stirling number(r,n)使用上面讨论的四种基本情况递归工作,然后使用我们证明的恒等式递归工作。

下面是上述方法的实现:

C++
// C++ program to implement above approach
#include 
using namespace std;
  
// Calculating factorial of an integer n.
long long factorial(int n)
{
    // Our base cases of factorial 0! = 1! = 1
    if (n == 0)
        return 1;
  
    // n can't be less than 0.
    if (n < 0)
        return -1;
    long long res = 1;
    for (int i = 2; i < n + 1; ++i)
        res *= i;
    return res;
}
  
// Function to compute the number of combination
// of r objects out of n objects.
int nCr(int n, int r)
{
    // r cant be more than n so we'd like the
    // program to crash if the user entered
    // wrong input.
    if (r > n)
        return -1;
  
    if (n == r)
        return 1;
  
    if (r == 0)
        return 1;
  
    // nCr(n, r) = nCr(n - 1, r - 1) + nCr(n - 1, r)
    return nCr(n - 1, r - 1) + nCr(n - 1, r);
}
  
// Function to calculate the Stirling numbers.
// The base cases which were discussed above are handled
// to stop the recursive calls.
long long stirlingNumber(int r, int n)
{
  
    // n can't be more than
    // r, s(r, 0) = 0.
    if (n > r)
        return -1;
  
    if (n == 0)
        return 0;
  
    if (r == n)
        return 1;
  
    if (n == 1)
        return factorial(r - 1);
  
    if (r - n == 1)
        return nCr(r, 2);
    else
        return stirlingNumber(r - 1, n - 1)
               + (r - 1) * stirlingNumber(r - 1, n);
}
  
// Driver program
int main()
{
    // Calculating the stirling number s(9, 2)
    int r = 9, n = 2;
  
    long long val = stirlingNumber(r, n);
    if (val == -1)
        cout << " No stirling number";
    else
        cout << "The Stirling Number s(" << r
             << ", " << n << ") is : "  << val;
    return 0;
}


Java
// Java program to implement
// above approach
import java.io.*;
  
class GFG 
{
  
// Calculating factorial of 
// an integer n.
static long factorial(int n)
{
    // Our base cases of factorial 
    // 0! = 1! = 1
    if (n == 0)
        return 1;
  
    // n can't be less than 0.
    if (n < 0)
        return -1;
    long res = 1;
    for (int i = 2; i < n + 1; ++i)
        res *= i;
    return res;
}
  
// Function to compute the number 
// of combination of r objects 
// out of n objects.
static int nCr(int n, int r)
{
    // r cant be more than n so 
    // we'd like the program to 
    // crash if the user entered
    // wrong input.
    if (r > n)
        return -1;
  
    if (n == r)
        return 1;
  
    if (r == 0)
        return 1;
  
    return nCr(n - 1, r - 1) + 
           nCr(n - 1, r);
}
  
// Function to calculate the Stirling 
// numbers. The base cases which were 
// discussed above are handled to stop
// the recursive calls.
static long stirlingNumber(int r, int n)
{
  
    // n can't be more than
    // r, s(r, 0) = 0.
    if (n > r)
        return -1;
  
    if (n == 0)
        return 0;
  
    if (r == n)
        return 1;
  
    if (n == 1)
        return factorial(r - 1);
  
    if (r - n == 1)
        return nCr(r, 2);
    else
        return stirlingNumber(r - 1, n - 1) + 
                                    (r - 1) * 
               stirlingNumber(r - 1, n);
}
  
// Driver Code
public static void main (String[] args) 
{
    // Calculating the stirling number s(9, 2)
    int r = 9, n = 2;
      
    long val = stirlingNumber(r, n);
    if (val == -1)
        System.out.println(" No stirling number");
    else
        System.out.println( "The Stirling Number s(" + 
                      r + ", " + n + ") is : " + val);
}
}
  
// This Code is Contributed by anuj_67


Python 3
# Python 3 program to implement above approach
  
# Function to compute the number of combination
# of r objects out of n objects.
# nCr(n, n) = 1, nCr(n, 0) = 1, and these are
# the base cases.
  
def nCr(n, r):
    if(n == r):
        return 1
    if(r == 0):
        return 1
    # nCr(n, r) = nCr(n - 1, r - 1) + nCr(n - 1, r)
    return nCr(n - 1, r - 1) + nCr(n - 1, r)
      
# This function is used to calculate the 
# factorial of a number n. 
def factorial(n):
    res = 1
      
    # 1 ! = 0 ! = 1
    if(n <= 1):
        return res
    for i in range(1, n + 1):
        res *= i
    return res
      
# Main function to calculate the Stirling numbers.
# the base cases which were discussed above are
# handled to stop the recursive call, n can't be
# more than r, s(r, 0) = 0.
# s(r, r) = 1. s(r, 1) = (r - 1)!.
# s(r, r - 1) = nCr(r, 2)
# else as we proved, s(r, n) = s(r - 1, n - 1) 
# + (r - 1) * s(r - 1, n) 
  
def stirlingNumber(r, n):
    if(r == n):
        return 1
    if(n == 0):
        return 0
    if(n == r -1):
        return nCr(r, 2)
    if(r - n == 1):
        return factorial(r - 1)
    return (stirlingNumber(r - 1, n - 1) 
        + (r - 1) * stirlingNumber(r - 1, n))
          
r, n = 9, 2
  
print(stirlingNumber(r, n))


C#
// C# program to implement
// above approach
using System;
  
class GFG 
{
  
// Calculating factorial of 
// an integer n.
static long factorial(int n)
{
    // Our base cases of factorial 
    // 0! = 1! = 1
    if (n == 0)
        return 1;
  
    // n can't be less than 0.
    if (n < 0)
        return -1;
    long res = 1;
    for (int i = 2; i < n + 1; ++i)
        res *= i;
    return res;
}
  
// Function to compute the number 
// of combination of r objects 
// out of n objects.
static int nCr(int n, int r)
{
    // r cant be more than n so 
    // we'd like the program to 
    // crash if the user entered
    // wrong input.
    if (r > n)
        return -1;
  
    if (n == r)
        return 1;
  
    if (r == 0)
        return 1;
  
    return nCr(n - 1, r - 1) + 
        nCr(n - 1, r);
}
  
// Function to calculate the Stirling 
// numbers. The base cases which were 
// discussed above are handled to stop
// the recursive calls.
static long stirlingNumber(int r, int n)
{
  
    // n can't be more than
    // r, s(r, 0) = 0.
    if (n > r)
        return -1;
  
    if (n == 0)
        return 0;
  
    if (r == n)
        return 1;
  
    if (n == 1)
        return factorial(r - 1);
  
    if (r - n == 1)
        return nCr(r, 2);
    else
        return stirlingNumber(r - 1, n - 1) + 
                                    (r - 1) * 
            stirlingNumber(r - 1, n);
}
  
// Driver Code
public static void Main () 
{
    // Calculating the stirling 
    // number s(9, 2)
    int r = 9, n = 2;
      
    long val = stirlingNumber(r, n);
    if (val == -1)
        Console.WriteLine(" No stirling number");
    else
        Console.WriteLine( "The Stirling Number s(" + 
                     r + ", " + n + ") is : " + val);
}
}
  
// This code is contributed by inder_verma..


PHP
 $n)
        return -1;
  
    if ($n == $r)
        return 1;
  
    if ($r == 0)
        return 1;
  
    // nCr($n, $r) = nCr($n - 1, $r - 1) + nCr($n - 1, $r)
    return nCr($n - 1, $r - 1) + nCr($n - 1, $r);
}
  
// Function to calculate the Stirling numbers.
// The base cases which were discussed above are handled
// to stop the recursive calls.
function stirlingNumber($r, $n)
{
  
    // n can't be more than
    // r, s(r, 0) = 0.
    if ($n > $r)
        return -1;
  
    if ($n == 0)
        return 0;
  
    if ($r == $n)
        return 1;
  
    if ($n == 1)
        return factorial($r - 1);
  
    if ($r - $n == 1)
        return nCr($r, 2);
    else
        return stirlingNumber($r - 1, $n - 1)
               + ($r - 1) * stirlingNumber($r - 1, $n);
}
  
     // Calculating the stirling number s(9, 2)
    $r = 9;
    $n = 2;
  
    $val = stirlingNumber($r, $n);
    if ($val == -1)
        echo " No stirling number";
    else
        echo  "The Stirling Number s(", $r
             ,", " , $n , ") is : " , $val;
               
// This code is contributed by ANKITRAI1
?>


输出:
The Stirling Number s(9, 2) is : 109584


注意:
可以使用动态编程来优化上述解决方案。例如,请参阅响铃编号(对集合进行分区的方式编号)。

请参阅第一种斯特林编号,以了解有关斯特林编号的更多信息。