在一个特殊的家庭中找到职业

考虑一个特殊的工程师和医生家族，遵循以下规则：

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. 医生的第一个孩子是医生，第二个孩子是工程师。
4. 历代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情况：

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

在上面的祖先树中给定一个人的等级和位置，找到此人的职业。 例如：

Input : level = 4, pos = 2Output : DoctorInput : level = 3, pos = 4Output : Engineer

方法1（递归） 这个想法基于这样一个事实：一个人的职业取决于以下两个方面。

1. 父母的职业。
2. 节点位置：如果节点的位置是奇数，则其职业与其父节点相同。否则，这个职业就不同于它的父母。

我们递归地找到父节点的职业，然后使用上面的第2点找到当前节点的职业。 下面是上述想法的实现。

C++

// C++ program to find profession of a person at // given level and position. #include<bits/stdc++.h> 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

// 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.

Python3

# 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.

PHP

<?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 ?>

Javascript

<script> // 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" ); </script>

输出：

Doctor

方法2（使用位运算符）

Level 1: ELevel 2: EDLevel 3: EDDELevel 4: EDDEDEEDLevel 5: EDDEDEEDDEEDEDDE 

级别输入是不必要的（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-1的二进制表示中的1计数。如果1的计数为偶数，则结果为工程师，否则为医生。 当然，位置限制是2^（1级）

输出：

Engineer

感谢Furkan Uslu提出的这种方法。 本文由 苛刻的贱民 .如果你喜欢GeekSforgek，并想贡献自己的力量，你也可以使用 写极客。组织 或者把你的文章寄去评论-team@geeksforgeeks.org.看到你的文章出现在Geeksforgeks主页上，并帮助其他极客。 如果您发现任何不正确的地方，或者您想分享有关上述主题的更多信息，请写下评论。