程序以找到正好设置了两位的第N个自然数|套装2

给定整数N ，任务是找到恰好有两个设置位的^第N^个自然数。

例子：

Input: N = 4
Output: 9
Explanation: Numbers with exactly two set bits are 3, 5, 6, 9, 10, 12, …. 
The 4^th term in this series is 9.

Input: N = 15
Output: 48
Explanation: Numbers with exactly two set bits are 3, 5, 6, 9, 10, 12, 17, 18, 20, 24, 33, 34, 36, 40, 48….
The 15^th term in this series is 48.

天真的方法：有关此问题的最简单解决方案，请参阅本文的上一篇文章。

时间复杂度： O(N)
辅助空间： O(1)

高效方法：为了优化上述方法，使用二进制搜索来查找第N^个数字的想法。请按照以下步骤解决问题：

• 给定序列中的任何数字均采用(2 ^a + 2 ^b )形式，其中a> b 。
• 可以通过识别值a和b来识别系列中的第N^个项。
• 查找使得(a *(A + 1))/ 2≥N的值，并保持一个约束，一个必须为最小
• 因此，找到一个在使用二进制搜索上面的步骤限制值。
• 完成上述步骤后，使用a和N查找b的值，并将结果打印为(2 ^a + 2 ^b ) 。

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 Nth number
// with exactly two bits set
void findNthNum(long long int N)
{
    // Initialize variables
    long long int a, b, left;
    long long int right, mid;
    long long int t, last_num = 0;
 
    // Initialize the range in which
    // the value of 'a' is present
    left = 1, right = N;
 
    // Perform Binary Search
    while (left <= right) {
 
        // Find the mid value
        mid = left + (right - left) / 2;
 
        t = (mid * (mid + 1)) / 2;
 
        // Update the range using the
        // mid value t
        if (t < N) {
            left = mid + 1;
        }
        else if (t == N) {
            a = mid;
            break;
        }
        else {
            a = mid;
            right = mid - 1;
        }
    }
 
    // Find b value using a and N
    t = a - 1;
    b = N - (t * (t + 1)) / 2 - 1;
 
    // Print the value 2^a + 2^b
    cout << (1 << a) + (1 << b);
}
 
// Driver Code
int main()
{
    long long int N = 15;
 
    // Function Call
    findNthNum(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the Nth number
// with exactly two bits set
static void findNthNum(int N)
{
    // Initialize variables
    int a = 0, b, left;
    int right, mid;
    int t, last_num = 0;
 
    // Initialize the range in which
    // the value of 'a' is present
    left = 1;
    right = N;
 
    // Perform Binary Search
    while (left <= right) {
 
        // Find the mid value
        mid = left + (right - left) / 2;
 
        t = (mid * (mid + 1)) / 2;
 
        // Update the range using the
        // mid value t
        if (t < N) {
            left = mid + 1;
        }
        else if (t == N) {
            a = mid;
            break;
        }
        else {
            a = mid;
            right = mid - 1;
        }
    }
 
    // Find b value using a and N
    t = a - 1;
    b = N - (t * (t + 1)) / 2 - 1;
 
    // Print the value 2^a + 2^b
    System.out.print((1 << a) + (1 << b));
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 15;
 
    // Function Call
    findNthNum(N);
}
}
 
// This code contributed by shikhasingrajput

Python3

# Python3 program for the above approach
 
# Function to find the Nth number
# with exactly two bits set
def findNthNum(N):
     
    # Initialize variables
    last_num = 0
 
    # Initialize the range in which
    # the value of 'a' is present
    left = 1
    right = N
 
    # Perform Binary Search
    while (left <= right):
         
        # Find the mid value
        mid = left + (right - left) // 2
 
        t = (mid * (mid + 1)) // 2
 
        # Update the range using the
        # mid value t
        if (t < N):
            left = mid + 1
 
        elif (t == N):
            a = mid
            break
 
        else:
            a = mid
            right = mid - 1
 
    # Find b value using a and N
    t = a - 1
    b = N - (t * (t + 1)) // 2 - 1
 
    # Print the value 2^a + 2^b
    print((1 << a) + (1 << b))
 
# Driver Code
if __name__ == "__main__":
 
    N = 15
 
    # Function Call
    findNthNum(N)
 
# This code is contributed by chitranayal

C#

// C# program for the above approach
using System;
   
class GFG{
   
// Function to find the Nth number
// with exactly two bits set
static void findNthNum(int N)
{
   
    // Initialize variables
    int a = 0, b, left;
    int right, mid;
    int t;
    //int last_num = 0;
  
    // Initialize the range in which
    // the value of 'a' is present
    left = 1; right = N;
  
    // Perform Binary Search
    while (left <= right)
    {
       
        // Find the mid value
        mid = left + (right - left) / 2;
  
        t = (mid * (mid + 1)) / 2;
  
        // Update the range using the
        // mid value t
        if (t < N)
        {
            left = mid + 1;
        }
        else if (t == N)
        {
            a = mid;
            break;
        }
        else
        {
            a = mid;
            right = mid - 1;
        }
    }
  
    // Find b value using a and N
    t = a - 1;
    b = N - (t * (t + 1)) / 2 - 1;
  
    // Print the value 2^a + 2^b
    Console.Write((1 << a) + (1 << b));
}
   
// Driver Code
public static void Main()
{
    int N = 15;
  
    // Function Call
    findNthNum(N);
}
}
 
// This code is contributed by sanjoy_62

输出：

48

时间复杂度： O(log N)
辅助空间： O(1)