笔分配问题

给定一个整数N表示一支笔的盒子数，并且两个玩家P1和P2玩游戏，按照以下规则在他们之间分配N支笔：

• P1抢先用2支^X笔。 (最初， X = 0)
• P2采用^3×笔。
• 每次移动后， X的值增加1。
• P1和P2交替移动。
• 如果当前玩家所需要的笔数多于框中剩余的笔数，则他们退出。
• 当两个玩家都退出游戏或者当盒子变空时，游戏将结束。

游戏结束后，将打印以下详细信息的任务：

1. 盒子中剩余的笔数。
2. P1收集的笔数。
3. P2收集的笔数。

例子：

Input: N = 22
Output: 
Number of pens remaining in the box: 14
Number of pens collected by P1 : 5
Number of pens collected by P2 : 3
Explanation:

• Move 1: X = 0, P1 takes 1 pen from the box. Therefore, N = 22 – 1 = 21.
• Move 2: X = 1, P2 takes 3 pens from the box. Therefore, N = 21 – 3 = 18.
• Move 3: X = 2, P1 takes 4 pens from the box. Therefore, N = 18 – 4 = 14.
• Move 4: X = 3, P2 quits as 27 > 14.
• Move 5: X = 4, P1 quits as 16 > 14.
• Game Over! Both players have quit.

Input: N = 1
Output: 
Number of pens remaining in the box : 0
Number of pens collected by P1 : 1
Number of pens collected by P2 : 0

方法：想法是使用递归。请按照以下步骤解决问题：

1.定义一个递归函数：

Game_Move(N, P1, P2, X, Move, QuitP1, QuitP2) 
where, 
N : Total number of Pens 
P1 : Score of P1 
P2 : Score of P2 
X : Initialized to zero 
Move = 0 : P1’s turn 
Move = 1 : P2’s turn 
QuitP1 : Has P1 quit 
QuitP2 : Has P2 quit 

2.最后，在游戏结束后打印最终值

下面是上述方法的实现：

C++

// C++ implementation of the
// above approach
 
#include 
using namespace std;
 
// N = Total number of Pens
// P1 : Score of P1
// P2 : Score of P2
// X : Initialized to zero
// Move = 0 : P1's turn
// Move = 1 : P2's turn
// QuitP1 : Has P1 quit
// QuitP2 : Has P2 quit
 
// Recursive function to play Game
void solve(int& N, int& P1, int& P2, int& X, bool Move,
           bool QuitP1, bool QuitP2)
{
    if (N == 0 or (QuitP1 and QuitP2)) {
 
        // Box is empty, Game Over! or
        // Both have quit, Game Over!
        cout << "Number of pens remaining"
             << " in the box: " << N << endl;
        cout << "Number of pens collected"
             << " by P1: " << P1 << endl;
        cout << "Number of pens collected"
             << " by P2: " << P2 << endl;
        return;
    }
 
    if (Move == 0 and QuitP1 == false) {
 
        // P1 moves
        int req_P1 = pow(2, X);
 
        if (req_P1 <= N) {
            P1 += req_P1;
            N -= req_P1;
        }
        else {
            QuitP1 = true;
        }
    }
 
    else if (Move == 1 and QuitP2 == false) {
 
        // P2 moves
        int req_P2 = pow(3, X);
 
        if (req_P2 <= N) {
            P2 += req_P2;
            N -= req_P2;
        }
        else {
            QuitP2 = true;
        }
    }
 
    // Increment X
    X++;
 
    // Switch moves between P1 and P2
    Move = ((Move == 1) ? 0 : 1);
 
    solve(N, P1, P2, X, Move, QuitP1, QuitP2);
}
 
// Function to find the number of
// pens remaining in the box and
// calculate score for each player
void PenGame(int N)
{
    // Score of P1
    int P1 = 0;
 
    // Score of P2
    int P2 = 0;
 
    // Initialized to zero
    int X = 0;
 
    // Move = 0, P1's turn
    // Move = 1, P2's turn
    bool Move = 0;
 
    // Has P1 quit
    bool QuitP1 = 0;
 
    // Has P2 quit
    bool QuitP2 = 0;
 
    // Recursively continue the game
    solve(N, P1, P2, X, Move,
          QuitP1, QuitP2);
}
 
// Driver Code
int main()
{
    int N = 22;
    PenGame(N);
 
    return 0;
}

Java

// Java implementation of the
// above approach
import java.util.*;
import java.lang.*;
public class GFG
{
 
  // N = Total number of Pens
  // P1 : Score of P1
  // P2 : Score of P2
  // X : Initialized to zero
  // Move = 0 : P1's turn
  // Move = 1 : P2's turn
  // QuitP1 : Has P1 quit
  // QuitP2 : Has P2 quit
 
  // Recursive function to play Game
  static void solve(int N, int P1, int P2, int X,
                    int Move, boolean QuitP1, boolean QuitP2)
  {
    if (N == 0 || (QuitP1 && QuitP2))
    {
 
      // Box is empty, Game Over! or
      // Both have quit, Game Over!
      System.out.println("Number of pens remaining"
                         + " in the box: " + N);
      System.out.println("Number of pens collected"
                         + " by P1: " + P1);
      System.out.println("Number of pens collected"
                         + " by P2: " + P2);
      return;
    }
    if (Move == 0 && QuitP1 == false)
    {
 
      // P1 moves
      int req_P1 = (int)(Math.pow(2, X));
      if (req_P1 <= N)
      {
        P1 += req_P1;
        N -= req_P1;
      }
      else
      {
        QuitP1 = true;
      }
    }
    else if (Move == 1 && QuitP2 == false)
    {
 
      // P2 moves
      int req_P2 = (int)(Math.pow(3, X));
      if (req_P2 <= N)
      {
        P2 += req_P2;
        N -= req_P2;
      }
      else
      {
        QuitP2 = true;
      }
    }
 
    // Increment X
    X++;
 
    // Switch moves between P1 and P2
    Move = ((Move == 1) ? 0 : 1);
    solve(N, P1, P2, X, Move, QuitP1, QuitP2);
  }
 
  // Function to find the number of
  // pens remaining in the box and
  // calculate score for each player
  static void PenGame(int N)
  {
 
    // Score of P1
    int P1 = 0;
 
    // Score of P2
    int P2 = 0;
 
    // Initialized to zero
    int X = 0;
 
    // Move = 0, P1's turn
    // Move = 1, P2's turn
    int Move = 0;
 
    // Has P1 quit
    boolean QuitP1 = false;
 
    // Has P2 quit
    boolean QuitP2 = false;
 
    // Recursively continue the game
    solve(N, P1, P2, X, Move, QuitP1, QuitP2);
  }
 
  // Driver Code
  public static void main (String[] args)
  {
    int N = 22;
    PenGame(N);
  }
}
 
// This code is contributed by jana_sayantan.

Python3

# Python3 implementation of the
# above approach
 
# N = Total number of Pens
# P1 : Score of P1
# P2 : Score of P2
# X : Initialized to zero
# Move = 0 : P1's turn
# Move = 1 : P2's turn
# QuitP1 : Has P1 quit
# QuitP2 : Has P2 quit
 
# Recursive function to play Game
def solve(N, P1, P2, X, Move,
          QuitP1, QuitP2):
 
    if (N == 0 or (QuitP1 and QuitP2)):
 
        # Box is empty, Game Over! or
        # Both have quit, Game Over!
        print("Number of pens remaining in the box: ", N)
        print("Number of pens collected by P1: ", P1)
        print("Number of pens collected by P2: ", P2)
        return
    if (Move == 0 and QuitP1 == False):
 
        # P1 moves
        req_P1 = int(pow(2, X))
 
        if (req_P1 <= N):
            P1 += req_P1
            N -= req_P1
        else:
            QuitP1 = True
    elif (Move == 1 and QuitP2 == False):
 
        # P2 moves
        req_P2 = int(pow(3, X))
        if (req_P2 <= N):
            P2 += req_P2
            N -= req_P2
        else:
            QuitP2 = True
 
    # Increment X
    X += 1
 
    # Switch moves between P1 and P2
    if(Move == 1):
        Move = 0
    else:
        Move = 1
    solve(N, P1, P2, X, Move, QuitP1, QuitP2)
 
# Function to find the number of
# pens remaining in the box and
# calculate score for each player
def PenGame(N):
 
    # Score of P1
    P1 = 0
 
    # Score of P2
    P2 = 0
 
    # Initialized to zero
    X = 0
 
    # Move = 0, P1's turn
    # Move = 1, P2's turn
    Move = False
 
    # Has P1 quit
    QuitP1 = False
 
    # Has P2 quit
    QuitP2 = False
 
    # Recursively continue the game
    solve(N, P1, P2, X, Move,
          QuitP1, QuitP2)
 
# Driver Code
N = 22
PenGame(N)
 
# This code is contributed by Dharanendra L V.

C#

// C# implementation of the
// above approach
using System;
class GFG {
 
    // N = Total number of Pens
    // P1 : Score of P1
    // P2 : Score of P2
    // X : Initialized to zero
    // Move = 0 : P1's turn
    // Move = 1 : P2's turn
    // QuitP1 : Has P1 quit
    // QuitP2 : Has P2 quit
 
    // Recursive function to play Game
    static void solve(int N, int P1, int P2, int X,
                      int Move, bool QuitP1, bool QuitP2)
    {
        if (N == 0 || (QuitP1 && QuitP2)) {
 
            // Box is empty, Game Over! or
            // Both have quit, Game Over!
            Console.WriteLine("Number of pens remaining"
                              + " in the box: " + N);
            Console.WriteLine("Number of pens collected"
                              + " by P1: " + P1);
            Console.WriteLine("Number of pens collected"
                              + " by P2: " + P2);
            return;
        }
 
        if (Move == 0 && QuitP1 == false) {
 
            // P1 moves
            int req_P1 = (int)(Math.Pow(2, X));
 
            if (req_P1 <= N) {
                P1 += req_P1;
                N -= req_P1;
            }
            else {
                QuitP1 = true;
            }
        }
 
        else if (Move == 1 && QuitP2 == false)
        {
 
            // P2 moves
            int req_P2 = (int)(Math.Pow(3, X));
 
            if (req_P2 <= N)
            {
                P2 += req_P2;
                N -= req_P2;
            }
            else
            {
                QuitP2 = true;
            }
        }
 
        // Increment X
        X++;
 
        // Switch moves between P1 and P2
        Move = ((Move == 1) ? 0 : 1);
        solve(N, P1, P2, X, Move, QuitP1, QuitP2);
    }
 
    // Function to find the number of
    // pens remaining in the box and
    // calculate score for each player
    static void PenGame(int N)
    {
        // Score of P1
        int P1 = 0;
 
        // Score of P2
        int P2 = 0;
 
        // Initialized to zero
        int X = 0;
 
        // Move = 0, P1's turn
        // Move = 1, P2's turn
        int Move = 0;
 
        // Has P1 quit
        bool QuitP1 = false;
 
        // Has P2 quit
        bool QuitP2 = false;
 
        // Recursively continue the game
        solve(N, P1, P2, X, Move, QuitP1, QuitP2);
    }
 
    // Driver Code
    public static void Main()
    {
        int N = 22;
        PenGame(N);
    }
}
 
// This code is contributed by chitranayal.

Javascript

// javascript program of the above approach
 
  // N = Total number of Pens
  // P1 : Score of P1
  // P2 : Score of P2
  // X : Initialized to zero
  // Move = 0 : P1's turn
  // Move = 1 : P2's turn
  // QuitP1 : Has P1 quit
  // QuitP2 : Has P2 quit
 
  // Recursive function to play Game
  function solve(N, P1, P2, X,
                    Move, QuitP1, QuitP2)
  {
    if (N == 0 || (QuitP1 && QuitP2))
    {
 
      // Box is empty, Game Over! or
      // Both have quit, Game Over!
      document.write("Number of pens remaining"
                         + " in the box: " + N + "<br/>");
      document.write("Number of pens collected"
                         + " by P1: " + P1+ "<br/>");
      document.write("Number of pens collected"
                         + " by P2: " + P2+ "<br/>");
      return;
    }
    if (Move == 0 && QuitP1 == false)
    {
 
      // P1 moves
      let req_P1 = (Math.pow(2, X));
      if (req_P1 <= N)
      {
        P1 += req_P1;
        N -= req_P1;
      }
      else
      {
        QuitP1 = true;
      }
    }
    else if (Move == 1 && QuitP2 == false)
    {
 
      // P2 moves
      let req_P2 = (Math.pow(3, X));
      if (req_P2 <= N)
      {
        P2 += req_P2;
        N -= req_P2;
      }
      else
      {
        QuitP2 = true;
      }
    }
 
    // Increment X
    X++;
 
    // Switch moves between P1 and P2
    Move = ((Move == 1) ? 0 : 1);
    solve(N, P1, P2, X, Move, QuitP1, QuitP2);
  }
 
  // Function to find the number of
  // pens remaining in the box and
  // calculate score for each player
  function PenGame(N)
  {
 
    // Score of P1
    let P1 = 0;
 
    // Score of P2
    let P2 = 0;
 
    // Initialized to zero
    let X = 0;
 
    // Move = 0, P1's turn
    // Move = 1, P2's turn
    let Move = 0;
 
    // Has P1 quit
    let QuitP1 = false;
 
    // Has P2 quit
    let QuitP2 = false;
 
    // Recursively continue the game
    solve(N, P1, P2, X, Move, QuitP1, QuitP2);
  }
 
    // Driver Code
     
    let N = 22;
    PenGame(N);
 

输出：

Number of pens remaining in the box: 14
Number of pens collected by P1: 5
Number of pens collected by P2: 3

时间复杂度： O(Log(N))
辅助空间： O(1)