铅笔游戏代码无法输出获胜结果及测试报错排查

问题概述 #

我开发的铅笔游戏无法打印最后一条判定玩家获胜的语句，同时测试#16失败，提示「当玩家输入要拿取的铅笔数量时，应输出恰好2行非空内容」。以下是相关信息：

开发需求 #

• 需校验输入合法性，限制玩家每次最多拿3支铅笔；输入错误时输出对应提示并重新请求输入。
• 游戏规则：拿最后一支铅笔的玩家落败，结束时需打印「Winner-name won!」。
• 输入错误场景及提示：
  • 初始铅笔数非数字字符串：The number of pencils should be numeric
  • 初始铅笔数为0：The number of pencils should be positive
  • 输入负数：按场景1处理
  • 所选玩家非John/Jack：Choose between John and Jack
  • 拿取铅笔数非1/2/3：Possible values: '1', '2' or '3'
  • 拿取铅笔数超过剩余数量：Too many pencils were taken

测试用例 #

提供6组测试示例，涵盖各类错误输入处理及获胜判定场景。

我的代码 #

print('How many pencils would you like to use:')
pen_num: int = 0
first_player = 'John'
second_player = 'Jack'

while True:
    pen_num = input()
    if not pen_num.isnumeric():
        print('The number of pencils should be numeric')
        continue
    if int(pen_num) == 0:
        print('The number of pencils should be positive')
        continue
    if int(pen_num) == 0:
        print('The number of pencils should be numeric')
        continue
    pen_num = int(pen_num)
    break
print('Who will be the first (John, Jack):')
while True:
    first_player = input()
    if not str(first_player) in ['John', 'Jack']:
        print('Choose between John and Jack')
        continue
    if first_player == 'Jack':
        second_player = 'John'
    break

cur_turn: int = 1
cur_player = first_player
while pen_num > 0:
    print('|' * pen_num)
    if cur_turn % 2 == 1:
        cur_player = first_player
    else:
        cur_player = second_player
    print(cur_player + "'s turn:")
    cut_num = 0
    while True:
        cut_num = input()
        if not cut_num.isnumeric():
            print("Possible values: '1', '2' or '3'")
            continue
        cut_num = int(cut_num)
        if not 1 <= cut_num <= 3:
            print("Possible values: '1', '2' or '3'")
            continue

        if cut_num > pen_num:
            print("Too many pencils were taken")
            continue
        break

    cur_turn += 1
    pen_num -= cut_num
    if pen_num == 0:
        print(cur_player + " won!")
    else:
        break

测试报错 #

测试#16失败，提示「当玩家输入要拿取的铅笔数量时，应输出恰好2行非空内容」，并附上测试过程的输出记录。

问题修复分析 #

1. 核心问题点 #

• 游戏循环逻辑错误：外层while pen_num > 0循环中，每次处理完一轮拿取后就执行else: break，导致游戏仅进行一轮就终止，既无法完成多轮游戏，也无法触发正确的获胜判定。
• 获胜判定逻辑错误：按照规则，拿最后一支铅笔的玩家落败，获胜者应为另一位玩家，但当前代码错误地将拿最后一支铅笔的玩家判定为胜者。

修复后的代码 #

print('How many pencils would you like to use:')
pen_num: int = 0
first_player = 'John'
second_player = 'Jack'

# 初始铅笔数校验逻辑优化
while True:
    pen_num_input = input()
    if not pen_num_input.isnumeric():
        print('The number of pencils should be numeric')
        continue
    pen_num = int(pen_num_input)
    if pen_num <= 0:
        print('The number of pencils should be positive')
        continue
    break

# 玩家选择逻辑优化
print('Who will be the first (John, Jack):')
while True:
    selected_player = input()
    if selected_player not in ['John', 'Jack']:
        print('Choose between John and Jack')
        continue
    first_player = selected_player
    second_player = 'Jack' if first_player == 'John' else 'John'
    break

cur_turn: int = 1
while pen_num > 0:
    print('|' * pen_num)
    cur_player = first_player if cur_turn % 2 == 1 else second_player
    print(f"{cur_player}'s turn:")
    
    # 拿取数量校验
    while True:
        cut_num_input = input()
        if not cut_num_input.isnumeric():
            print("Possible values: '1', '2' or '3'")
            continue
        cut_num = int(cut_num_input)
        if not 1 <= cut_num <= 3:
            print("Possible values: '1', '2' or '3'")
            continue
        if cut_num > pen_num:
            print("Too many pencils were taken")
            continue
        break
    
    pen_num -= cut_num
    cur_turn += 1
    
    # 正确判定获胜者：拿最后一支的玩家落败，胜者为另一位玩家
    if pen_num == 0:
        winner = second_player if cur_player == first_player else first_player
        print(f"{winner} won!")

修复要点 #

• 移除外层循环中错误的else: break，让游戏可以持续进行直到铅笔耗尽
• 修正获胜判定逻辑，将拿最后一支铅笔的玩家判定为落败者，另一位玩家为胜者
• 优化初始铅笔数和玩家选择的校验逻辑，消除重复判断和变量混淆

内容的提问来源于stack exchange，提问作者Mustafa Anandwala