python
这个代码，怎么样修改可以做到。1.在一局游戏中如果猜了四次再给出是奇数或者偶数的提示，这个猜了四次包括如果任一玩家放弃的次数！！！
就是一局游戏猜六次，如果前四次猜的过程中有一次放弃，只猜了三个数字，但是依旧会给出提示是奇数还是偶数
2.修改guess的初始定义和位置，让每个人猜之前可以看到之前猜过的
但是第一次猜之前什么也看不到
#introduce the game and rules briefly
print("Three games, each game guess a number between 1 and 12.")
print("The game will end if you guess the right number or you run out of turns, each game you can guess 3 times.")
#Have each player enter their name
player1 = input("Enter the name of the two players: ")
player2 = input("Enter the name of the other player: ")
win1 = 0
win2 = 0
#play 3 games total
for i in range (3):
#pick a random number between 1 and 12
import random
correct_num = random.randint(1,12)
print(correct_num)
guess = correct_num + 1
count = 1
guessed_number = []
while guess != correct_num and count <= 6:
#take turns to guess the number
if count % 2 == 0:
player = player1
else:
player = player2
guess = float(input(player+" guess a number between 1 and 12: "))
guessed_number.append(guess)
print("You have already guessed: ",guessed_number)
#if one player win,the game ends
if guess == correct_num:
print(player,"congratulations, you are right!")
count = count + 6
if player == player1:
win1 = win1 + 1
else:
win2 = win2 + 1
#tell them if the guess is bigger or smaller
else:
if guess > correct_num:
print(player,"your number is bigger.")
else:
print(player,"your number is smaller.")
#let one player give up part way through the game
if count < 7:
if player == player1:
Y = input(player2 + " Do you want to give up? ")
else:
Y = input(player1 + " Do you want to give up? ")
if Y == "n":
count = count + 1
elif Y == "y":
count = count + 2
#give the hint
if count >= 4 and guess != correct_num :
if correct_num % 2 == 0:
print("Hint: the correct number is even")
else:
print("Hint: the correct number is odd")
#if no winner, tell them they are losers
if count >= 6 and guess != correct_num:
print(player1,"and",player2,"You are all losers.")
count = count + 6
if i <= 1 :
print("Now the next game begin! ")
# add an extra congratulations to the player who wins the most games
if win1 > win2:
print(player1,"congratulations! you win the most games!")
elif win1 < win2:
print(player2,"congratulations! you win the most games!") 1. 要实现在一局游戏中如果猜了四次再给出是奇数或者偶数的提示，需要在猜数的while循环中加入判断。当猜数次数count大于等于4时，再加上判断猜数是否等于正确数，如果不等于，则加上判断正确数的奇偶性并输出提示。
代码如下：
if count >= 4 and guess != correct_num:
if correct_num % 2 == 0:
print("Hint: the correct number is even")
else:
print("Hint: the correct number is odd")
2. 要实现修改guess的初始定义和位置，让每个人猜之前可以看到之前猜过的，需要在while循环前先定义guess为空列表，然后在每次循环前输出之前猜过的数字。
代码如下：
guessed_number = []
while guess != correct_num and count <= 6:
if count % 2 == 0:
player = player1
else:
player = player2
print("You have already guessed: ",guessed_number)
guess = float(input(player+" guess a number between 1 and 12: "))
guessed_number.append(guess)
注意：第一次猜之前不能输出之前猜过的数字，因此需要在while循环后加上一个判断，如果猜数次数count等于1，则不输出之前猜过的数字。
完整代码如下： 修改方案：
1. 添加一个计数器来记录每个游戏中的猜测次数，当猜测次数为4时，再给出是奇数还是偶数的提示。在每次猜测时，判断计数器是否小于等于4，如果是则进行提示；否则不提示。
2. 将guess的定义和位置移动到while循环之前，并将其初始化为一个空列表。在每次猜测时，先输出之前猜测过的数字，再让玩家猜测。
代码实现如下：
# introduce the game and rules briefly
print("Three games, each game guess a number between 1 and 12.")
print("The game will end if you guess the right number or you run out of turns, each game you can guess 3 times.")
# Have each player enter their name
player1 = input("Enter the name of the two players: ")
player2 = input("Enter the name of the other player: ")
win1 = 0
win2 = 0
# play 3 games total
for i in range(3):
# pick a random number between 1 and 12
import random
correct_num = random.randint(1, 12)
print(correct_num)
guess = []
count = 1
guessed_number = []
while count <= 6:
# take turns to guess the number
if count % 2 == 0:
player = player1
else:
player = player2
# 输出之前猜测的数字
if len(guess) > 0:
print(player + "，之前猜测的数字为：" + str(guess))
guess_num = float(input(player + " guess a number between 1 and 12: "))
guessed_number.append(guess_num)
guess.append(guess_num)
# give the hint
if count == 4:
if correct_num % 2 == 0:
print("Hint: the correct number is even")
else:
print("Hint: the correct number is odd")
# let one player give up part way through the game
if count < 6:
if player == player1:
Y = input(player2 + " Do you want to give up? ")
else:
Y = input(player1 + " Do you want to give up? ")
if Y == "n":
count += 1
elif Y == "y":
count += 2
break
# if one player win, the game ends
if guess_num == correct_num:
print(player, "congratulations, you are right!")
if player == player1:
win1 += 1
else:
win2 += 1
break
else:
if guess_num > correct_num:
print(player, "your number is bigger.")
else:
print(player, "your number is smaller.")
count += 1
# if no winner, tell them they are losers
if count >= 6 and guess_num != correct_num:
print(player1, "and", player2, "You are all losers.")
if i <= 1:
print("Now the next game begins!")
# add an extra congratulations to the player who wins the most games
if win1 > win2:
print(player1, "congratulations! you win the most games!")
elif win1 < win2:
print(player2, "congratulations! you win the most games!")
球一个最佳答案谢谢啦！这对我非常重要！{:10_254:}{:10_254:}{:10_254:} isdkz 发表于 2023-10-12 04:08
1. 要实现在一局游戏中如果猜了四次再给出是奇数或者偶数的提示，需要在猜数的while循环中加入判断。当猜数 ...
但是如果count 大于等于4的情况下，后面的每一次猜测都会给出hint提示 根据你的需求，我对你的代码进行了一些修改。以下是修改后的代码：
# introduce the game and rules briefly
print(\"Three games, each game guess a number between 1 and 12.\")
print(\"The game will end if you guess the right number or you run out of turns, each game you can guess 6 times.\")
# Have each player enter their name
player1 = input(\"Enter the name of the two players: \")
player2 = input(\"Enter the name of the other player: \")
win1 = 0
win2 = 0
# play 3 games total
for i in range (3):
# pick a random number between 1 and 12
import random
correct_num = random.randint(1,12)
print(correct_num)
guess = correct_num + 1
count = 1
guessed_number = []
while count <= 6:
# take turns to guess the number
if count % 2 == 0:
player = player1
else:
player = player2
if count > 1:
print(\"You have already guessed: \",guessed_number)
guess = float(input(player+\" guess a number between 1 and 12: \"))
guessed_number.append(guess)
# if one player win,the game ends
if guess == correct_num:
print(player,\"congratulations, you are right!\")
count = count + 6
if player == player1:
win1 = win1 + 1
else:
win2 = win2 + 1
# tell them if the guess is bigger or smaller
else:
if guess > correct_num:
print(player,\"your number is bigger.\")
else:
print(player,\"your number is smaller.\")
# let one player give up part way through the game
if count < 7:
if player == player1:
Y = input(player2 + \" Do you want to give up? \")
else:
Y = input(player1 + \" Do you want to give up? \")
if Y == \"n\":
count = count + 1
elif Y == \"y\":
count = count + 2
guessed_number.append('give up')
# give the hint
if len(guessed_number) >= 4 and guess != correct_num :
if correct_num % 2 == 0:
print(\"Hint: the correct number is even\")
else:
print(\"Hint: the correct number is odd\")
# if no winner, tell them they are losers
if count >= 6 and guess != correct_num:
print(player1,\"and\",player2,\"You are all losers.\")
count = count + 6
if i <= 1 :
print(\"Now the next game begin! \")
# add an extra congratulations to the player who wins the根据你的需求，我对你的代码进行了一些修改。以下是修改后的代码：
`python
# introduce the game and rules briefly
print(\"Three games, each game guess a number between 1 and 12.\")
print(\"The game will end if you guess the right number or you run out of turns, each game you can guess 6 times.\")
# Have each player enter their name
player1 = input(\"Enter the name of the two players: \")
player2 = input(\"Enter the name of the other player: \")
win1 = 0
win2 = 0
# play 3 games total
for i in range (3):
# pick a random number between 1 and 12
import random
correct_num = random.randint(1,12)
print(correct_num)
guess = correct_num + 1
count = 1
guessed_number = []
while count <= 6:
# take turns to guess the number
if count % 2 == 0:
player = player1
else:
player = player2
if count > 1:
print(\"You have already guessed: \",guessed_number)
guess = float(input(player+\" guess a number between 1 and 12: \"))
guessed_number.append(guess)
# if one player win,the game ends
if guess == correct_num:
print(player,\"congratulations, you are right!\")
count = count + 6
if player == player1:
win1 = win1 + 1
else:
win2 = win2 + 1
# tell them if the guess is bigger or smaller
else:
if guess > correct_num:
print(player,\"your number is bigger.\")
else:
print(player,\"your number is smaller.\")
# let one player give up part way through the game
if count < 7:
if player == player1:
Y = input(player2 + \" Do you want to give up? \")
else:
Y = input(player1 + \" Do you want to give up? \")
if Y == \"n\":
count = count + 1
elif Y == \"y\":
count = count + 2
guessed_number.append('give up')
# give the hint
if len(guessed_number) >= 4 and guess != correct_num :
if correct_num % 2 == 0:
print(\"Hint: the correct number is even\")
else:
print(\"Hint: the correct number is odd\")
# if no winner, tell them they are losers
if count >= 6 and guess != correct_num:
print(player1,\"and\",player2,\"You are all losers.\")
count = count + 6
if i <= 1 :
print(\"Now the next game begin! \")
# add an extra congratulations to the player who wins the most games
if win1 > win2:
print(player1,\"congratulations! you win the most games!\")
elif win1 < win2:
print(player2,\"congratulations! you win the most games!\")
主要的修改点有：
1. 将每局游戏的猜测次数从3次改为6次。
2. 在每次猜测前，如果不是第一次猜测，都会显示已经猜过的数字。
3. 如果玩家选择放弃，会在已猜过的数字列表中添加'give up'，并且放弃也算作一次猜测。
4. 在猜测次数（包括放弃）达到4次后，如果还没有猜中，会给出数字是奇数还是偶数的提示。
请注意，本回答来自GPT-4
如果问题已经解决，请设置最佳答案 陶远航 发表于 2023-10-12 04:10
根据你的需求，我对你的代码进行了一些修改。以下是修改后的代码：
但是如果前四次猜测的时候，有人两次都放弃，列表里只有两个数字，但是也已经猜过四次了，也应该给出hint'提示
pan11 发表于 2023-10-12 04:10
但是如果count 大于等于4的情况下，后面的每一次猜测都会给出hint提示
那就换成count%4呗 dan但是如果中途有人放弃，那到第四次猜完的时候就不是count==4了 isdkz 发表于 2023-10-12 09:06
那就换成count%4呗
dan但是如果中途有人放弃，那到第四次猜完的时候就不是count==4了
isdkz 发表于 2023-10-12 09:06
那就换成count%4呗
就如果前面四次猜测，一共有两次放弃了，那count就是5，但是我仍然希望hint出现
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