在特殊家庭找职业
原文:https://www . geesforgeks . org/find-profession-in-a-假想-特殊情况/
考虑一个特殊的工程师和医生家族，其规则如下:
- 每个人都有两个孩子。
- 工程师的第一个孩子是工程师，第二个孩子是医生。
- 医生的第一个孩子是医生，第二个孩子是工程师。
- 所有一代的医生和工程师都是从工程师开始的。
我们可以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
给定一个人在上面祖先树中的级别和位置，找到这个人的职业。 例:
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。
- 父母的职业。
- 节点的位置:如果一个节点的位置是奇数，那么它的职业就是它的父节点。其他职业与其父母不同。
我们递归找到父节点的职业，然后用上面的点 2 找到当前节点的职业。 以下是上述思路的实现。
C++
``` // 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; } ```
Java language (a computer language, Especially for creating websites)
``` // 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. ```
Python 3
```
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
```
C
``` // 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. ```
Server-side programming language (abbreviation of professional hypertext preprocessor)
``` <?php // PHP 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 ($pos % 2) ? 'd' : 'e';
// If parent is an engineer, then // current node will be an engineer // if at odd position and doctor // if even position. return ($pos % 2) ? 'e' : 'd'; }
// Driver code $level = 4; $pos = 2; if((findProffesion($level, $pos) == 'e') == true) echo "Engineer"; else echo "Doctor" ;
// This code is contributed by ajit ?> ```
java description language
```
// 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");
```
输出:
Doctor
方法 2(使用按位运算符)
Level 1: E
Level 2: ED
Level 3: EDDE
Level 4: EDDEDEED
Level 5: EDDEDEEDDEEDEDDE
级别输入不是必需的(如果我们忽略最大位置限制)，因为第一个元素是相同的。 结果基于位置减 1 的二进制表示中 1 的计数。如果 1 的计数是偶数，那么结果是工程师，否则是医生。 当然位置限制是 2^(Level-1)
C++
// C++ program to find profession of a person at
// given level and position.
#include<bits/stdc++.h>
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;
}
Java 语言(一种计算机语言，尤用于创建网站)
// 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);
}
}
Python 3
# Python3 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. """
def countSetBits(n):
count = 0
while n > 0:
n &= (n-1)
count+=1
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.
def findProffesion(level, pos):
# Count set bits in 'pos-1'
c = countSetBits(pos-1)
# If set bit count is odd, then doctor, else engineer
if c%2 == 0:
return 'e'
else:
return 'd'
level, pos = 3, 4
if findProffesion(level, pos) == 'e':
print("Engineer")
else:
print("Doctor")
# This code is contributed by divyeshrabadiya07.
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
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP 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)
{
$count = 0;
while ($n)
{
$n &= ($n - 1) ;
$count++;
}
return $count;
}
// Returns 'e' if profession of
// node at given level and position
// is engineer. Else doctor. The
// function assumes that given
// position and level have valid values.
function findProffesion($level, $pos)
{
// Count set bits in 'pos-1'
$c = countSetBits($pos - 1);
// If set bit count is odd,
// then doctor, else engineer
return ($c % 2) ? 'd' : 'e';
}
// Driver Code
$level = 3;
$pos = 4;
if((findProffesion($level,
$pos) == 'e') == true)
echo "Engineer \n";
else
echo "Doctor \n";
// This code is contributed by aj_36
?>
java 描述语言
<script>
// 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);
</script>
输出:
Engineer
感谢富尔坎·乌斯卢提出这个方法。 本文由严酷帕里克供稿。如果你喜欢 GeeksforGeeks 并想投稿，你也可以使用write.geeksforgeeks.org写一篇文章或者把你的文章邮寄到 review-team@geeksforgeeks.org。看到你的文章出现在极客博客主页上，帮助其他极客。 如果发现有不正确的地方，或者想分享更多关于上述话题的信息，请写评论。
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