给定范围内的二项式系数 (nCr) 之和

给定三个值N 、 L和R ，任务是计算从L到R的所有r值的二项式系数的总和( ⁿ C _r ) 。

例子：

Input: N = 5, L = 0, R = 3
Output: 26
Explanation: Sum of ⁵C₀ + ⁵C₁ + ⁵C₂ + ⁵C₃ = 1 + 5 + 10 + 10 = 26.

Input: N = 3, L = 3, R = 3
Output: 1

方法：通过使用公式n!直接计算ⁿ C _r来解决这个问题！ / (r!(n−r)!)并递归计算从L到R的每个r值的阶乘。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the factorial
// of a given number
long long factorial(long long num)
{
    if (num == 0 || num == 1)
        return 1;
    else
        return num * factorial(num - 1);
}
 
// Function to calculate the sum
// of binomial coefficients(nCr) for
// all values of r from L to R
long long sumOfnCr(int n, int R, int L)
{
    long long r;
    long long res = 0;
 
    for (r = L; r <= R; r++)
        res += (factorial(n)
                / (factorial(r)
                   * factorial(n - r)));
 
    return res;
}
 
// Driver Code
int main()
{
    int N = 5, L = 0, R = 3;
    cout << sumOfnCr(N, R, L);
    return 0;
}

Java

// JAVA program for the above approach
import java.io.*;
 
class GFG {
 
    // Function to find the factorial
    // of a given number
    static long factorial(long num)
    {
        if (num == 0 || num == 1)
            return 1;
        else
            return num * factorial(num - 1);
    }
 
    // Function to calculate the sum
    // of binomial coefficients(nCr) for
    // all values of r from L to R
    static long sumOfnCr(int n, int R, int L)
    {
        long r;
        long res = 0;
 
        for (r = L; r <= R; r++)
            res += (factorial(n)
                    / (factorial(r) * factorial(n - r)));
 
        return res;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 5, L = 0, R = 3;
        long ans = sumOfnCr(N, R, L);
        System.out.println(ans);
    }
}
 
// This code is contributed by Taranpreet

Python3

# Python code for the above approach
 
# Function to find the factorial
# of a given number
def factorial(num):
    if (num == 0 or num == 1):
        return 1;
    else:
        return num * factorial(num - 1);
 
# Function to calculate the sum
# of binomial coefficients(nCr) for
# all values of r from L to R
def sumOfnCr(n, R, L):
    res = 0;
 
    for r in range(L, R + 1):
        res += (factorial(n) / (factorial(r) * factorial(n - r)));
 
    return res;
 
# Driver Code
N = 5
L = 0
R = 3;
print((int)(sumOfnCr(N, R, L)))
 
# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
class GFG {
 
  // Function to find the factorial
  // of a given number
  static long factorial(long num)
  {
    if (num == 0 || num == 1)
      return 1;
    else
      return num * factorial(num - 1);
  }
 
  // Function to calculate the sum
  // of binomial coefficients(nCr) for
  // all values of r from L to R
  static long sumOfnCr(int n, int R, int L)
  {
    long r;
    long res = 0;
 
    for (r = L; r <= R; r++)
      res += (factorial(n)
              / (factorial(r) * factorial(n - r)));
 
    return res;
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 5, L = 0, R = 3;
    Console.Write(sumOfnCr(N, R, L));
  }
}
 
// This code is contributed by ukasp.

Javascript

        // JavaScript code for the above approach
 
        // Function to find the factorial
        // of a given number
        function factorial(num) {
            if (num == 0 || num == 1)
                return 1;
            else
                return num * factorial(num - 1);
        }
 
        // Function to calculate the sum
        // of binomial coefficients(nCr) for
        // all values of r from L to R
        function sumOfnCr(n, R, L) {
            let r;
            let res = 0;
 
            for (r = L; r <= R; r++)
                res += (factorial(n)
                    / (factorial(r)
                        * factorial(n - r)));
 
            return res;
        }
 
        // Driver Code
        let N = 5, L = 0, R = 3;
        document.write(sumOfnCr(N, R, L));
 
         // This code is contributed by Potta Lokesh
    

输出

26

时间复杂度： O(N * (R – L))
辅助空间： 在）