在一系列tictactoe(x和o)回放中统计玩家的胜负和平局,可以通过编写一个程序来实现。以下是一个可能的解决方案:
下面是一个示例代码,使用Python语言实现上述解决方案:
def check_game_over(board):
# 检查每一行
for row in board:
if row[0] == row[1] == row[2] != '':
return True
# 检查每一列
for col in range(3):
if board[0][col] == board[1][col] == board[2][col] != '':
return True
# 检查对角线
if board[0][0] == board[1][1] == board[2][2] != '':
return True
if board[0][2] == board[1][1] == board[2][0] != '':
return True
# 检查是否平局
if all(board[i][j] != '' for i in range(3) for j in range(3)):
return True
return False
def count_results(replays):
player1_wins = 0
player2_wins = 0
draws = 0
for replay in replays:
board = [[''] * 3 for _ in range(3)]
for move in replay:
row, col, player = move
board[row][col] = player
if check_game_over(board):
if player == 'x':
player1_wins += 1
else:
player2_wins += 1
else:
draws += 1
return player1_wins, player2_wins, draws
# 示例回放数据
replays = [
[(0, 0, 'x'), (0, 1, 'o'), (1, 1, 'x'), (1, 0, 'o'), (2, 2, 'x')],
[(0, 0, 'x'), (0, 1, 'o'), (1, 1, 'x'), (1, 0, 'o'), (2, 0, 'x'), (2, 2, 'o')],
[(0, 0, 'x'), (0, 1, 'o'), (1, 1, 'x'), (1, 0, 'o'), (2, 0, 'x'), (2, 1, 'o'), (0, 2, 'x')],
[(0, 0, 'x'), (0, 1, 'o'), (1, 1, 'x'), (1, 0, 'o'), (2, 0, 'x'), (2, 1, 'o'), (2, 2, 'x')],
]
player1_wins, player2_wins, draws = count_results(replays)
print("Player 1 wins:", player1_wins)
print("Player 2 wins:", player2_wins)
print("Draws:", draws)
这段代码将输出每个玩家的胜利次数、平局次数和失败次数。你可以根据实际情况进行调整和扩展。
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