程序以找到恰好设置了两位的第 N 个自然数

给定一个整数N ，任务是找到恰好设置了两位的第 N 个自然数。
例子：

Input: N = 4 
Output: 9 
Explanation: 
Numbers with exactly two bits set: 3, 5, 6, 9, 10, 12, … 
4th number in this is 9
Input: N = 15 
Output: 48 
 

天真的方法：

1. 循环遍历所有自然数，对于每个数字，通过计算数字中的设置位来检查它是否设置了两位。
2. 打印具有两个设置位的第 N 个数字。

有效的方法：

1. 通过找到 N 所属的分区来找到最左边的设置位（分区 ‘i’ 中有 ‘i’ 个数字）。
2. 要找到另一个设置位，我们必须首先找到 N 与前一个分区的最后一个数字的距离。根据它们的不同，我们设置相应的位。

1. 注意：要在数字K 中设置第 i 位 (i = 0, 1, 2…)：

k = k | (1<<(i))

1. 下面是上述方法的实现：

C++

// C++ Code to  find the Nth number
// with exactly two bits set
 
#include 
using namespace std;
 
// Function to find the Nth number
// with exactly two bits set
void findNthNum(long long int N)
{
 
    long long int bit_L = 1, last_num = 0;
 
    // Keep incrementing until
    // we reach the partition of 'N'
    // stored in bit_L
    while (bit_L * (bit_L + 1) / 2 < N) {
        last_num = last_num + bit_L;
        bit_L++;
    }
 
    // set the rightmost bit
    // based on bit_R
    int bit_R = N - last_num - 1;
 
    cout << (1 << bit_L) + (1 << bit_R)
         << endl;
}
 
// Driver code
int main()
{
    long long int N = 13;
 
    findNthNum(N);
 
    return 0;
}

Java

// Java Code to  find the Nth number
// with exactly two bits set
class GFG{
  
// Function to find the Nth number
// with exactly two bits set
static void findNthNum(int N)
{
  
    int bit_L = 1, last_num = 0;
  
    // Keep incrementing until
    // we reach the partition of 'N'
    // stored in bit_L
    while (bit_L * (bit_L + 1) / 2 < N) {
        last_num = last_num + bit_L;
        bit_L++;
    }
  
    // set the rightmost bit
    // based on bit_R
    int bit_R = N - last_num - 1;
  
    System.out.print((1 << bit_L) + (1 << bit_R)
         +"\n");
}
  
// Driver code
public static void main(String[] args)
{
    int N = 13;
  
    findNthNum(N);
}
}
 
// This code is contributed by Princi Singh

Python3

# Python Code to  find the Nth number
# with exactly two bits set
 
 
# Function to find the Nth number
# with exactly two bits set
def findNthNum(N):
 
    bit_L = 1;
    last_num = 0;
 
    # Keep incrementing until
    # we reach the partition of 'N'
    # stored in bit_L
    while (bit_L * (bit_L + 1) / 2 < N):
        last_num = last_num + bit_L;
        bit_L+=1;
     
 
    # set the rightmost bit
    # based on bit_R
    bit_R = N - last_num - 1;
 
    print((1 << bit_L) + (1 << bit_R));
 
 
# Driver code
if __name__ == '__main__':
    N = 13;
 
    findNthNum(N);
 
 
# This code contributed by PrinciRaj1992

C#

// C# Code to  find the Nth number
// with exactly two bits set
using System;
 
class GFG{
   
// Function to find the Nth number
// with exactly two bits set
static void findNthNum(int N)
{
   
    int bit_L = 1, last_num = 0;
   
    // Keep incrementing until
    // we reach the partition of 'N'
    // stored in bit_L
    while (bit_L * (bit_L + 1) / 2 < N) {
        last_num = last_num + bit_L;
        bit_L++;
    }
   
    // set the rightmost bit
    // based on bit_R
    int bit_R = N - last_num - 1;
   
    Console.Write((1 << bit_L) + (1 << bit_R)
         +"\n");
}
   
// Driver code
public static void Main(String[] args)
{
    int N = 13;
   
    findNthNum(N);
}
}
 
// This code is contributed by Princi Singh

Javascript

 
// JavaScript Code to find the Nth number
// with exactly two bits set
 
// Function to find the Nth number
// with exactly two bits set
function findNthNum(N)
{
 
    let bit_L = 1, last_num = 0;
 
    // Keep incrementing until
    // we reach the partition of 'N'
    // stored in bit_L
    while (bit_L * (bit_L + 1) / 2 < N) {
        last_num = last_num + bit_L;
        bit_L++;
    }
 
    // set the rightmost bit
    // based on bit_R
    let bit_R = N - last_num - 1;
 
    document.write((1 << bit_L) + (1 << bit_R)
        + "<br>");
}
 
// Driver code
let N = 13;
findNthNum(N);
 
// This code is contributed by Mayank Tyagi

输出：

36

1. 时间复杂度： O（数字的分区）

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