0

豆类游戏是两个玩家互相对战并且必须从包含 16 个豆子的一堆中拾取豆子的游戏。您每回合可以拿 1、2 或 3 个豆子。选择最后一个豆子的玩家输了。

 if you start 
 total = 16 you chose 3 making the total 13
 the opponent chooses 1,2,3 making the total 12,11,10
 to get your next answer you force the opponent to land on 9
 so in turn you take the total 12,11,10 -9 to get your output of 1,2 or 3

 Now you want them to land on 5 to ensure you victory.
 the iteration begins here, where the opponent chooses 1,2,3 making the total 8,7,6
 so in turn you take the total 8,7,6 -5 to get your output of 1,2,or 3.

 By doing this you now have control of the numbers 4,3,and 2.
 If they chose 1 total=4 you chose 3 you win.
 If they chose 2 total=3 you chose 2 you win.
 If they chose 3 total=2 you chose 1 you win.

这是测试它的代码。

Beans 测试驱动程序

from myPlayer_28 import player as otherPlayer  # load opponent's player code
from simple2  import player as me           # load your player code

def play(player0, player1):
    """Plays a single game of 'Heap of Beans'.  Winning player's # (0 or 1) is  returned"""
    beans = 16  # Starting bean heap
    while True:
        take = player0(beans) # find out how many beans player0 takes
        if 1 <= take <= 3 and take <= beans:  # legal # of beans taken?
            beans -= take                     # adjust count of remaining beans
        else:  # user made bad move
            print("Bad move:", beans, "bean(s) and", take, "taken.")
            beans = 0  # force lose for bad move
        if beans == 0: # does player0 lose?
            return 1   # player 1 wins
        take = player1(beans) # find out how many beans player1 takes
        if 1 <= take <= 3 and take <= beans:  # legal # of beans taken?
            beans -= take                     # adjust count of remaining beans
        else:  # user made bad move
            print("Bad move:", beans, "bean(s) and", take, "taken.")
            beans = 0  # force lose for bad move
        if beans == 0: # does player1 lose?
            return 0   # player 0 wins    

def main():
    """Play 100 games of 'Heap of Beans' with each player starting 50 times."""
    wins = [0,0] # win accumulators for player0 and player1
    for i in range(100):
        if i%2:
            winner = play(me,otherPlayer)
        else:  
            winner = play(otherPlayer,me)
            winner = (winner+1)%2
        wins[winner] += 1
    print("My wins:", wins[0], "  Other wins:", wins[1])

main()

这很简单 2 我正在运行的程序。

import random

def player(beans):
    if beans > 4:
        return random.randrange(1,4)
    if beans == 1:
        return 1
return beans - 1

新代码我已经显着改变了我的代码。

def player(beans):
    if beans==16:
        return 3
    if beans==15:
        return 2
    if beans==14:
        return 1
    if beans==13:
        return 1

    if beans <=12 and beans >=10:
        return beans-9
    if beans==9:
        return 1

    if beans >=6 and beans<9:
        return beans -5
    if beans ==5:
        return 1

    else:
        if beans ==4:
            return 3
        elif beans ==3:
            return 2
        elif beans==2:
            return 1
        elif beans ==1:
            return 1

对于这个问题有这么多的代码，我深表歉意。但是，我想向那些查看它的人展示它正在测试什么，因此如果他们愿意，他们可以复制所有代码以自己运行它。

“新” 我的代码赢得了 96% 的时间，但我喜欢它赢得 100% 的时间，这不可能吗？

4

0 回答 0