使用递归计算 nCr 值的程序

给定两个数字N和r ，使用递归找到^N C _r的值

例子：

Input: N = 5, r = 2
Output: 10
Explanation: The value of 5C2 is 10

Input: N = 3, r = 1
Output: 3

方法：这个想法只是基于以下公式。

^NC_r = N! / (r! * (N-r)!)
Also, 
^NC_r-1 = N! / ( (r-1)! * (N – (r-1))! )

Hence,

• ^NC_r * r! * (N – r)! = ^NC_r-1  * (r-1)! * (N – (r-1))!
• ^NC_r * r * (N-r)! = ^NC_r-1  * (N-r+1)!                          [eliminating (r – 1)! from both side]
• ^NC_r * r = ⁿC_r-1  * (N-r+1)

所以，

^NC_r = ^NC_r-1 * (N-r+1) / r

下面是上述方法的实现：

C++

// C++ code to implement above approach
#include 
using namespace std;
 
// Function to calculate the value of nCr
// using recursion
int nCr(int N, int r)
{
    int res = 0;
    if (r == 0) {
        return 1;
    }
    else {
        res = nCr(N, r - 1)
              * (N - r + 1) / r;
    }
    return res;
}
 
// Driver code
int main()
{
    int N = 5, r = 3;
    cout << nCr(N, r);
    return 0;
}

Java

// Java code for the above approach
import java.io.*;
class GFG {
 
  // Function to calculate the value of nCr
  // using recursion
  static int nCr(int N, int r)
  {
    int res = 0;
    if (r == 0) {
      return 1;
    }
    else {
      res = nCr(N, r - 1) * (N - r + 1) / r;
    }
    return res;
  }
 
  public static void main(String[] args)
  {
    int N = 5, r = 3;
 
    System.out.println(nCr(N, r));
  }
}
 
// This code is contributed by Potta Lokesh

Python3

# Python code to implement above approach
 
# Function to calculate the value Of nCr
# using recursion
def nCr(N, r):
    res = 0
    if(r == 0):
        return 1
    else:
        res = nCr(N, r-1)
* (N-r + 1) / r
    return res
 
 
# Driver code
if __name__ == "__main__":
    N = 5
    r = 3
    print(int(nCr(N, r)))

C#

using System;
 
public class GFG{
 
  // Function to calculate the value of nCr
  // using recursion
  static int nCr(int N, int r)
  {
    int res = 0;
    if (r == 0) {
      return 1;
    }
    else {
      res = nCr(N, r - 1)
        * (N - r + 1) / r;
    }
    return res;
  }
 
  // Driver code
  static public void Main (){
 
    int N = 5, r = 3;
    Console.WriteLine(nCr(N, r));
  }
}
 
// This code is contributed by hrithikgarg03188.

Javascript

//Javascript code to implement above approach
 
// Function to calculate the value of nCr
// using recursion
function nCr(N, r)
{
    let res = 0;
    if (r == 0) {
        return 1;
    }
    else {
        res = nCr(N, r - 1)
              * (N - r + 1) / r;
    }
    return res;
}
 
// Driver code
 
let N = 5, r = 3;
document.write(nCr(N,r));
     
// This code is contributed by Taranpreet
// 

输出

10

时间复杂度：O(N!)
辅助空间：O(1)