在 N 个球的游戏中寻找获胜者，在该游戏中，玩家可以一步移动 [A, B] 范围内的任何球

给定两个整数A和B ，并假设 Alice 和 Bob 正在玩一个游戏，从一个包含N 个球的袋子开始，其中，在一次移动中，玩家可以移除范围[A, B]之间的任意数量的球如果玩家不能移除任何球，那么玩家输了，如果爱丽丝和鲍勃交替进行游戏并且爱丽丝开始游戏，则任务是找到游戏的赢家。

例子：

Input: N = 2, A = 1, B = 1
Output: Bob
Explanation: 
One way in which the game can be played is:

1. Alice removes 1 ball, so the remaining balls are 1.
2. Now, Bob removes the last ball.
3. Alice cannot remove any balls, so she loses.

Input: N = 3, A = 1, B = 2
Output: Bob

朴素的方法：最简单的方法是找到每个状态的 grundy 数，并使用 Sprague – Grundy 定理找到获胜和失败的状态。

时间复杂度： O(N*N!)
辅助空间： O(N)

高效的方法：上述方法可以基于以下观察进行优化：

1. Firstly, it can be observed that (A + B) is always losing state, because whatever X (A ≤ X ≤ B) number of balls the current player chooses, the opponent can always empty the bag as there will be (A + B – X) number of balls left where A ≤ (A + B – X) ≤ B.
2. Also, from previous observations, one can observe that, for any multiple of (A + B) say, m*(A + B), the opponent can always reduce the current player to a state of (m – 1)*(A + B) and from (m – 1)*(A + B) to (m – 2)*(A + B) and so on.
3. Thus, extending the above observation, one can say that for any multiple of (A + B), the opponent can always reduce the current player to a state of exactly (A + B), which is indeed a losing state. Therefore, all multiples of (A + B) are a losing state.
4. So, the optimal choice for any player is to reduce the opponent to a multiple of (A + B), because after this, the player can always win, no matter what the opponent’s moves are.
5. So, now losing states are the states from which one can never reduce the opponent to a multiple of (A + B).
6. Therefore, any player with the state of the form: ((A + B)*m + y), where (0 ≤  y ≤ A-1) can never force the opponent to reduce to a multiple of (A + B), as any player can only pick at least A and at most B number of balls.

请按照以下步骤解决问题：

• 如果N%(A+B)小于A，则打印“ Bob ”。
• 否则，打印“ Alice ”。

下面是上述方法的实现：

C++

// C++ program of the above approach
#include 
using namespace std;
 
// Function to find the winner of the
// game
string NimGame(int N, int A, int B)
{
    // Stores sum of A and B
    int sum = A + B;
 
    // If N is of the form
    // m*(A+B)+y
    if (N % sum <= A - 1)
        return "Bob";
    // Otherwise,
    else
        return "Alice";
}
 
// Driver code
int main()
{
    // Input
    int N = 3, A = 1, B = 2;
    // Function call
    cout << NimGame(N, A, B) << endl;
 
    return 0;
}

Java

// Java Program for the above approach
import java.io.*;
 
class GFG
{
 
  // Function to find the winner of the
  // game
  public static String NimGame(int N, int A, int B)
  {
 
    // Stores sum of A and B
    int sum = A + B;
 
    // If N is of the form
    // m*(A+B)+y
    if (N % sum <= A - 1)
      return "Bob";
 
    // Otherwise,
    else
      return "Alice";
  }
 
  public static void main (String[] args)
  {
 
    // Input
    int N = 3, A = 1, B = 2;
 
    // Function call
    System.out.println(NimGame(N, A, B));
 
  }
}
 
  // This code is contributed by Potta Lokesh

Python3

# Python3 program of the above approach
 
# Function to find the winner of the game
def NimGame(N, A, B):
     
    # Stores sum of A and B
    sum = A + B
     
    # If N is of the form
    # m*(A+B)+y
    if (N % sum <= A - 1):
        return "Bob"
         
    # Otherwise,
    else:
        return "Alice"
 
# Driver code
 
# Input
N = 3
A = 1
B = 2
 
# Function call
print(NimGame(N, A, B))
 
# This code is contributed by amreshkumar3

C#

// C# program of the above approach
using System;
 
class GFG{
 
// Function to find the winner of the
// game
public static String NimGame(int N, int A, int B)
{
     
    // Stores sum of A and B
    int sum = A + B;
     
    // If N is of the form
    // m*(A+B)+y
    if (N % sum <= A - 1)
        return "Bob";
     
    // Otherwise,
    else
        return "Alice";
}
 
// Driver code
static void Main()
{
     
    // Input
    int N = 3, A = 1, B = 2;
     
    // Function call
    Console.Write(NimGame(N, A, B));
}
}
 
// This code is contributed by SoumikMondal

Javascript

        // JavaScript program for the above approach
 
        // Function to find the winner of the
        // game
        function NimGame(N, A, B) {
            // Stores sum of A and B
            let sum = A + B;
 
            // If N is of the form
            // m*(A+B)+y
            if (N % sum <= A - 1)
                return "Bob";
            // Otherwise,
            else
                return "Alice";
        }
 
        // Driver code
 
        // Input
        let N = 3, A = 1, B = 2;
        // Function call
        document.write(NimGame(N, A, B) + "<br>");
 
  // This code is contributed by Potta Lokesh
    

输出

Bob

时间复杂度： O(1)
辅助空间： O(1)