Find the winner by incrementing co-ordinates till Euclidean distance <= D

A and B play a game and A starts first. Consider a Token placed at (0,0) in a 2D plane, a player can perform the below moves on the token:

• increase x coordinate of the token to x+K
• increase y coordinate of the token to y+K

After each move the euclidean distance of the token from origin should be smaller than equal to D. i.e. x² + y² <= D² . The player who is unable to make a move looses.

Examples:

Input: D=2, K=1
Output: B
Explanation: From the initial position (0,0), player A has the option to move to either (0,1) or (1,0) From there, player B can move to any of the following coordinates: (0,2), (2,0) or (1,1). Now A cannot make any move. Therefore, B wins.

Input: D=5, K=2
Output: A
Explanation: From the initial position (0,0), player A has the option to move to either (0,2) or (2,0). From there, player B can move to any of the following coordinates: (0,4), (2,2) or (4,0). Player A can only move to the coordinates: (4,2) or (2,4). Now B can not make any move. Therefore, A wins.

Approach: To solve the problem, follow the idea below:

The idea is to find the numbers of steps taken by both the players mathematically. Both players, A and B, move a distance of K. Since player A starts the game, the optimal strategy for player B is to maintain the trajectory on the line (y=x). This means that if player A moves K distance along the x-axis, player B will move K distance along the y-axis, and vice versa. This movement forms the hypotenuse of a triangle, with value √ 2 K.

The total distance to be covered is D, so the total number of hypotenuses formed will be D/√ 2 K. Since each hypotenuse is contributed by both players, the total number of steps becomes 2*D/√ 2 K = √ (2*(D²/K²)).

If the total number of steps is odd, player A wins. If it's even, player B wins.

Step-by-step algorithm:

• Calculate the square of the maximum distance D that can be moved from the origin, divide it by the square of the step size K, and multiply the result by 2.
• Take the square root of the result which is the maximum number of steps that can be taken by both players combined.
• If the maximum number of steps is odd, player A wins. If it is even, player B wins.

Below is the implementation of the above algorithm:

// C++ Code
#include <bits/stdc++.h>
using namespace std;

// Function to determine the winner of the game
string game(long long D, long long K)
{
    // Calculate the square of the maximum distance D,
    // divide it by the square of the step size K,
    // and multiply the result by 2.

    long long x = 2 * ((D * D) / (K * K));

    // Take the square root of the result
    // If the maximum number of steps is odd,
    // player A wins else player B wins.
    return (int(sqrt(x)) & 1 ? "A" : "B");
}

// Driver code
int main()
{
    // Test Case 1
    long long D = 2, K = 1;
    cout << game(D, K) << endl;

    // Test Case 2
    D = 5, K = 2;
    cout << game(D, K) << endl;
    return 0;
}

// Java code

import java.util.*;
public class GFG {

    // Function to determine the winner of the game
    public static String game(long D, long K)
    {
        // Calculate the square of the maximum distance D,
        // divide it by the square of the step size K,
        // and multiply the result by 2.

        long x = 2 * ((D * D) / (K * K));

        // Take the square root of the result
        // If the maximum number of steps is odd,
        // player A wins else player B wins.
        return ((int)Math.sqrt(x) & 1) == 1 ? "A" : "B";
    }

    // Driver code
    public static void main(String[] args)
    {
        // Test Case 1
        long D = 2, K = 1;
        System.out.println(game(D, K));

        // Test Case 2
        D = 5;
        K = 2;
        System.out.println(game(D, K));
    }
}

# Python code

import math

# Function to determine the winner of the game


def game(D, K):
    # Calculate the square of the maximum distance D,
    # divide it by the square of the step size K,
    # and multiply the result by 2.
    x = 2 * ((D * D) / (K * K))

    # Take the square root of the result
    # If the maximum number of steps is odd,
    # player A wins else player B wins.
    return 'A' if int(math.sqrt(x)) & 1 else 'B'


# Driver code
if __name__ == "__main__":
    # Test Case 1
    D = 2
    K = 1
    print(game(D, K))

    # Test Case 2
    D = 5
    K = 2
    print(game(D, K))

using System;

class Program
{
    // Function to determine the winner of the game
    static string Game(long D, long K)
    {
        // Calculate the square of the maximum distance D,
        // divide it by the square of the step size K,
        // and multiply the result by 2.
        long x = 2 * ((D * D) / (K * K));

        // Take the square root of the result
        // If the integer part of the square root is odd,
        // player A wins; else, player B wins.
        return ((long)Math.Sqrt(x) % 2 == 1) ? "A" : "B";
    }

    // Driver code
    static void Main()
    {
        // Test Case 1
        long D1 = 2, K1 = 1;
        Console.WriteLine(Game(D1, K1));

        // Test Case 2
        long D2 = 5, K2 = 2;
        Console.WriteLine(Game(D2, K2));
    }
}


// This code is contributed by akshitaguprzj3

// Function to determine the winner of the game
function game(D, K) {
    // Calculate the square of the maximum distance D,
    // divide it by the square of the step size K,
    // and multiply the result by 2.
    const x = 2 * Math.floor((D * D) / (K * K));

    // Take the square root of the result
    // If the maximum number of steps is odd,
    // player A wins else player B wins.
    return Math.sqrt(x) % 2 === 1 ? "A" : "B";
}

// Test Case 1
let D = 2, K = 1;
console.log(game(D, K));

// Test Case 2
D = 5, K = 2;
console.log(game(D, K));

B
A

Time Complexity: O(1)
Auxiliary Space: O(1)