计算N次幂的奇偶二项式系数

给定一个整数N ，任务是计算直到N^次方的偶数和奇数二项式系数的数量。

例子：

Input: N = 4
Output:
Odd: 2
Even: 3
Explanation:
The binomial coefficients are as follows:
⁴C₀ = 1, ⁴C₁ = 4 , ⁴C₂ = 6 , ⁴C₃ = 4 , ⁴C₄ = 1.
Therefore, it can be observed that there exists exactly 2 odd and 3 even Binomial Coefficients.

Input: N = 5
Output:
Odd: 4
Even: 2
Explanation:
The binomial coefficients are as follows:
⁵C₀ = 1, ⁵C₁ = 5, ⁵C₂ = 10, ⁵C₃ = 10, ⁵C₄ = 5, ⁵C₅ = 1.
Therefore, there are 4 odd and 2 even coefficients.

解决方法：解决此问题的想法是使用位操作。在给定的整数N中找到设置的位。奇二项式系数的计数等于N中置位的2 ^计数。同样，偶数二项式系数的计数等于(N + 1-2 ^ N中的置位计数) 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
#include 
using namespace std;
 
// Function to count set bits in
// binary representation of number N
int countSetBits(int N)
{
    int count = 0;
 
    // Count set bits in N
    while (N) {
 
        N = N & (N - 1);
        count++;
    }
 
    // Return the final count
    return count;
}
 
// Driver Code
int main()
{
    int N = 4;
 
    int bits = countSetBits(N);
 
    // Print odd Binomial coefficents
    cout << "Odd "
         << ": " << pow(2, bits) << "\n";
 
    // Print even Binomial coefficents
    cout << "Even "
         << ": " << N + 1 - pow(2, bits)
         << "\n";
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
      
// Function to count set bits in
// binary representation of number N
static int countSetBits(int N)
{
    int count = 0;
  
    // Count set bits in N
    while (N != 0)
    {
         
        N = N & (N - 1);
        count++;
    }
  
    // Return the final count
    return count;
}
  
// Driver code
public static void main(String[] args)
{
    int N = 4;
  
    int bits = countSetBits(N);
  
    // Print odd Binomial coefficents
    System.out.println("Odd " + ": " +
               (int)(Math.pow(2, bits)));
  
    // Print even Binomial coefficents
    System.out.println("Even " + ": " +
               (N + 1 - (int)(Math.pow(2, bits))));
}
}
 
// This code is contributed by susmitakundugoaldanga

Python3

# Python3 program for the above approach
 
# Function to count set bits in
# binary representation of number N
def countSetBits(N: int) -> int:
 
    count = 0
 
    # Count set bits in N
    while (N):
        N = N & (N - 1)
        count += 1
 
    # Return the final count
    return count
 
# Driver Code
if __name__ == "__main__":
 
    N = 4
 
    bits = countSetBits(N)
 
    # Print odd Binomial coefficents
    print("Odd : {}".format(pow(2, bits)))
 
    # Print even Binomial coefficents
    print("Even : {}".format(N + 1 - pow(2, bits)))
 
# This code is contributed by sanjeev2552

C#

// C# program for the above approach
using System;
  
class GFG{
  
// Function to count set bits in
// binary representation of number N
static int countSetBits(int N)
{
    int count = 0;
   
    // Count set bits in N
    while (N != 0)
    {
        N = N & (N - 1);
        count++;
    }
   
    // Return the final count
    return count;
}
  
// Driver Code
public static void Main()
{
    int N = 4;
    int bits = countSetBits(N);
   
    // Print odd Binomial coefficents
    Console.WriteLine("Odd " + ": " +
                     (int)(Math.Pow(2, bits)));
   
    // Print even Binomial coefficents
    Console.WriteLine("Even " + ": " +
                     (N + 1 - (int)(Math.Pow(2, bits))));
}
}
 
// This code is contributed by sanjoy_62

输出：

Odd : 2
Even : 3

时间复杂度： O(1)
辅助空间： O(1)