在一个特殊的家庭中找到职业

考虑具有以下规则的特殊的工程师和医生家族：

1. 每个人都有两个孩子。
2. 工程师的第一个孩子是工程师，第二个孩子是医生。
3. Doctor的第一个孩子是Doctor，第二个孩子是工程师。
4. 所有世代的医生和工程师都始于工程师。

我们可以使用下图来表示这种情况：

E
           /        
          E          D
        /          /  
       E     D     D    E
      /    /    /    / 
     E   D D   E  D  E  E  D

给定一个人在祖先树中的级别和位置，找到该人的职业。
例子 ：

Input : level = 4, pos = 2
Output : Doctor

Input : level = 3, pos = 4
Output : Engineer

方法1(递归)
这个想法是基于一个人的职业取决于以下两个事实。

1. 父母的职业。
2. 节点的位置：如果节点的位置是奇数，则其职业与其父代相同。其他职业不同于其父职业。

我们递归地找到父行业，然后使用上面的第2点来找到当前节点的行业。
以下是上述想法的实现。

C#

using System;
 
// c# program to find profession of a person at 
// given level and position. 
public class GFG
{
/* Function to get no of set bits in binary 
representation of passed binary no. */
public static int countSetBits(int n)
{
    int count = 0;
    while (n != 0)
    {
    n &= (n - 1);
    count++;
    }
    return count;
}
 
// Returns 'e' if profession of node at given level 
// and position is engineer. Else doctor. The function 
// assumes that given position and level have valid values. 
public static char findProffesion(int level, int pos)
{
    // Count set bits in 'pos-1' 
    int c = countSetBits(pos - 1);
 
    // If set bit count is odd, then doctor, else engineer 
    return (c % 2 != 0)? 'd' : 'e';
}
 
// Driver code 
public static void Main(string[] args)
{
    int level = 3, pos = 4;
    string prof = (findProffesion(level, pos) == 'e')? "Engineer" : "Doctor";
    Console.Write(prof);
}
}
 
  // This code is contributed by Shrikant13

Javascript

 
    // Javascript program to find
    // profession of a person at
    // given level and position.
     
    /* Function to get no of set bits in binary
    representation of passed binary no. */
    function countSetBits(n)
    {
        let count = 0;
        while (n != 0)
        {
          n &= (n - 1);
          count++;
        }
        return count;
    }
 
    // Returns 'e' if profession of node at given level
    // and position is engineer. Else doctor.
    // The function assumes that given position and
    // level have valid values.
    function findProffesion(level, pos)
    {
        // Count set bits in 'pos-1'
        let c = countSetBits(pos - 1);
 
        // If set bit count is odd, then doctor,
        // else engineer
        return (c % 2 != 0)? 'd' : 'e';
    }
     
    let level = 3, pos = 4;
    let prof = (findProffesion(level, pos) == 'e')?
                "Engineer" : "Doctor";
    document.write(prof);
     

输出 ：

// C++ program to find profession of a person at
// given level and position.
#include
using namespace std;
 
// Returns 'e' if profession of node at given level
// and position is engineer. Else doctor. The function
// assumes that given position and level have valid values.
char findProffesion(int level, int pos)
{
    // Base case
    if (level == 1)
        return 'e';
 
    // Recursively find parent's profession. If parent
    // is a Doctor, this node will be a Doctor if it is
    // at odd position and an engineer if at even position
    if (findProffesion(level-1, (pos+1)/2) == 'd')
        return (pos%2)? 'd' : 'e';
 
    // If parent is an engineer, then current node will be
    // an engineer if at add position and doctor if even
    // position.
    return (pos%2)?  'e' : 'd';
}
 
// Driver code
int main(void)
{
    int level = 4, pos = 2;
    (findProffesion(level, pos) == 'e')? cout << "Engineer"
                                       : cout << "Doctor" ;
    return 0;
}

方法2(使用按位运算符)

// Java program to find
// profession of a person
// at given level and position
import java.io.*;
 
class GFG
{
 
// Returns 'e' if profession
// of node at given level
// and position is engineer.
// Else doctor. The function
// assumes that given position
// and level have valid values.
static char findProffesion(int level,
                           int pos)
{
    // Base case
    if (level == 1)
        return 'e';
 
    // Recursively find parent's
    // profession. If parent
    // is a Doctor, this node
    // will be a Doctor if it
    // is at odd position and an
    // engineer if at even position
    if (findProffesion(level - 1,
                      (pos + 1) / 2) == 'd')
        return (pos % 2 > 0) ?
                         'd' : 'e';
 
    // If parent is an engineer,
    // then current node will be
    // an engineer if at add
    // position and doctor if even
    // position.
    return (pos % 2 > 0) ?
                     'e' : 'd';
}
 
// Driver code
public static void main (String[] args)
{
    int level = 4, pos = 2;
    if(findProffesion(level,
                      pos) == 'e')
    System.out.println("Engineer");
    else
    System.out.println("Doctor");
}
}
 
// This code is contributed
// by anuj_67.

级别输入不是必需的(如果忽略最大位置限制)，因为第一个元素是相同的。
结果基于位置减一的二进制表示形式中的1计数。如果计数为1则结果为工程师，否则为Doctor。
当然，头寸限制为2 ^(Level-1)

C++

# python 3 program to find profession of a person at
# given level and position.
 
# Returns 'e' if profession of node at given level
# and position is engineer. Else doctor. The function
# assumes that given position and level have valid values.
def findProffesion(level, pos):
    # Base case
    if (level == 1):
        return 'e'
 
    # Recursively find parent's profession. If parent
    # is a Doctor, this node will be a Doctor if it is
    # at odd position and an engineer if at even position
    if (findProffesion(level-1, (pos+1)//2) == 'd'):
        if (pos%2):
            return 'd'
        else:
            return 'e'
 
    # If parent is an engineer, then current node will be
    # an engineer if at add position and doctor if even
    # position.
    if(pos%2):
        return 'e'
    else:
        return 'd'
 
# Driver code
if __name__ == '__main__':
    level = 3
    pos = 4
    if(findProffesion(level, pos) == 'e'):
        print("Engineer")
    else:
        print("Doctor")
         
# This code is contributed by
# Surendra_Gangwar

Java

// C# program to find
// profession of a person
// at given level and position
using System;
 
class GFG
{
 
// Returns 'e' if profession
// of node at given level
// and position is engineer.
// Else doctor. The function
// assumes that given position
// and level have valid values.
static char findProffesion(int level,
                           int pos)
{
    // Base case
    if (level == 1)
        return 'e';
 
    // Recursively find parent's
    // profession. If parent
    // is a Doctor, this node
    // will be a Doctor if it
    // is at odd position and an
    // engineer if at even position
    if (findProffesion(level - 1,
                      (pos + 1) / 2) == 'd')
        return (pos % 2 > 0) ?
                         'd' : 'e';
 
    // If parent is an engineer,
    // then current node will be
    // an engineer if at add
    // position and doctor if even
    // position.
    return (pos % 2 > 0) ?
                     'e' : 'd';
}
 
// Driver code
public static void Main ()
{
    int level = 4, pos = 2;
    if(findProffesion(level,
                    pos) == 'e')
    Console.WriteLine("Engineer");
    else
    Console.WriteLine("Doctor");
}
}
 
// This code is contributed
// by anuj_67.

的PHP

 
// JavaScript program to find
// profession of a person
// at given level and position
 
// Returns 'e' if profession
// of node at given level
// and position is engineer.
// Else doctor. The function
// assumes that given position
// and level have valid values.
function findProffesion(level,
                           pos)
{
    // Base case
    if (level == 1)
        return 'e';
  
    // Recursively find parent's
    // profession. If parent
    // is a Doctor, this node
    // will be a Doctor if it
    // is at odd position and an
    // engineer if at even position
    if (findProffesion(level - 1,
                      (pos + 1) / 2) == 'd')
        return (pos % 2 > 0) ?
                         'd' : 'e';
  
    // If parent is an engineer,
    // then current node will be
    // an engineer if at add
    // position and doctor if even
    // position.
    return (pos % 2 > 0) ?
                     'e' : 'd';
}
 
// Driver Code
 
    let level = 4, pos = 2;
    if(findProffesion(level,
                      pos) == 'e')
    document.write("Engineer");
    else
    document.write("Doctor");
                         

Java脚本

// C++ program to find profession of a person at
// given level and position.
#include
using namespace std;
 
/* Function to get no of set bits in binary
   representation of passed binary no. */
int countSetBits(int n)
{
    int count = 0;
    while (n)
    {
      n &= (n-1) ;
      count++;
    }
    return count;
}
 
// Returns 'e' if profession of node at given level
// and position is engineer. Else doctor. The function
// assumes that given position and level have valid values.
char findProffesion(int level, int pos)
{
    // Count set bits in 'pos-1'
    int c = countSetBits(pos-1);
 
    // If set bit count is odd, then doctor, else engineer
    return (c%2)?  'd' : 'e';
}
 
// Driver code
int main(void)
{
    int level = 3, pos = 4;
    (findProffesion(level, pos) == 'e')? cout << "Engineer"
                                       : cout << "Doctor" ;
    return 0;
}

输出 ：

// Java program to find profession of a person at
// given level and position.
class GFG{
/* Function to get no of set bits in binary
representation of passed binary no. */
static int countSetBits(int n)
{
    int count = 0;
    while (n!=0)
    {
    n &= (n-1) ;
    count++;
    }
    return count;
}
 
// Returns 'e' if profession of node at given level
// and position is engineer. Else doctor. The function
// assumes that given position and level have valid values.
static char findProffesion(int level, int pos)
{
    // Count set bits in 'pos-1'
    int c = countSetBits(pos-1);
 
    // If set bit count is odd, then doctor, else engineer
    return (c%2 !=0)? 'd' : 'e';
}
 
// Driver code
public static void main(String [] args)
{
    int level = 3, pos = 4;
    String prof = (findProffesion(level, pos) == 'e')? "Engineer"
                                    : "Doctor"  ;
    System.out.print(prof);
}
}