Counter Game - Hackerrank Problem Solution in Python

Counter Game

Hackerrank Problem Solution

Counter Game Hackerrank Problem Solution

PROBLEM STATEMENT

Louise and Richard have developed a numbers game. They pick a number and check to see if it is a power of . If it is, they divide it by . If not, they reduce it by the next lower number which is a power of . Whoever reduces the number to  wins the game. Louise always starts.

Given an initial value, determine who wins the game.

Example

It's Louise's turn first. She determines that  is not a power of . The next lower power of  is , so she subtracts that from  and passes  to Richard.  is a power of , so Richard divides it by  and passes  to Louise. Likewise,  is a power so she divides it by  and reaches . She wins the game.

Update If they initially set counter to , Richard wins. Louise cannot make a move so she loses.

Function Description

Complete the counterGame function in the editor below.

counterGame has the following parameter(s):

• int n: the initial game counter value

Returns

• string: either Richard or Louise

Input Format

The first line contains an integer , the number of testcases.
Each of the next  lines contains an integer , the initial value for each game.

Constraints

Sample Input 0

1
6

Sample Output 0

Richard

Explanation 0

•  is not a power of  so Louise reduces it by the largest power of  less than :.
•  is a power of  so Richard divides by  to get  and wins the game.

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SOLUTION IN PYTHON

#!/bin/python3 import math import os import random import re import sys def counterGame(n): n = bin(n)[2:] n = n.split('1') turns = len(n)+len(n[-1])-2 return 'Louise' if turns&1 else 'Richard' if __name__ == '__main__': fptr = open(os.environ['OUTPUT_PATH'], 'w') t = int(input().strip()) for t_itr in range(t): n = int(input().strip()) result = counterGame(n) fptr.write(result + '\n') fptr.close()

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