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Wonderglade(奇境)

# You need to collect several items.# But, the burl wants the gems!# Pick up all appearing items EXCEPT gems.while True:    item = hero.findNearestItem()    if item:        # If item.type isn't equal to "gem":        if item.type != "gem":            # Move to the item's position.            hero.moveXY(item.pos.x, item.pos.y)

逻辑之路

注意：不是真的情况有多种写法，编写你认为可以的代码。

# 从巫师那得到两个秘密的真假值hero.moveXY(14, 24)secretA = hero.findNearestFriend().getSecretA()secretB = hero.findNearestFriend().getSecretB()# 如果 secretA 和 secretB 都为真，走上面的路；否则，走下面。# 查看提示，学会写逻辑表达式。secretC = secretA and secretBif secretC:    hero.moveXY(20, 33)else:    hero.moveXY(20, 15)hero.moveXY(26, 24)# 如果 secretA 和 secretB 中有一个为真，走上面。secretD = secretA or secretBif secretD:    hero.moveXY(32, 33)else:    hero.moveXY(32, 15)hero.moveXY(38, 24)# 如果 secretB 不是真的，走上面。secretE = secretB is not True#secretE = not secretBif secretE:    hero.moveXY(44, 33)else:    hero.moveXY(44, 15)hero.moveXY(50,24)

有用的对手

# 这片金币地中暗藏了致命的毒药。# 兽人正在进攻，而苦力尝试偷你的金币！while True:    enemy = hero.findNearestEnemy()    if enemy:        # 只在敌人类型不是 "peon" 的时候攻击。        if enemy.type != "peon":            hero.attack(enemy)    item = hero.findNearestItem()    if item:        # 只在物品的类型不是 "poison" 的时候收集。        if item.type != "poison":            hero.moveXY(item.pos.x, item.pos.y)        pass

Burlbole Grove（树精树精）

# Don't attack the burls!# Functions can return a value.# When a function is called, it will be equal to the value the function returns.def shouldAttack(target):	# return False if no target    if not target:        return False	# return False if target.type == "burl"    if target.type == "burl":        return False    # Otherwise, return True    return Truewhile True:    enemy = hero.findNearestEnemy()    # Here we use shouldAttack() to decide if we should attack!    # heroShouldAttack will be assigned the same value that shouldAttack() returns!    heroShouldAttack = shouldAttack(enemy)    if heroShouldAttack:    	hero.attack(enemy)

Cursed Wonderglade（被诅咒的梦境）

# Wonderglade has changed since our last visit.# Ogres cursed it and we should defeat them.# The burl still is collecting gems, so don't touch them.# Also don't attack the burl.while True:    # Find the nearest item.    # Collect it (if it exists) only if its type isn't "gem".    item = hero.findNearestItem()    if item and item.type != "gem":        hero.moveXY(item.pos.x,item.pos.y)    # Find the nearest enemy.    # Attack it if it exists and its type isn't "burl".    enemy = hero.findNearestEnemy()    if enemy and enemy.type != "burl":        hero.attack(enemy)

逻辑结论

# 移动到 Eszter 身边，从他那得到三个密码值hero.moveXY(24, 16)secretA = hero.findNearestFriend().getSecretA()secretB = hero.findNearestFriend().getSecretB()secretC = hero.findNearestFriend().getSecretC()# 如果 A 和 B 都为真，或者 C 为真，对 Tamas 说 "TRUE"。否则，说 "FALSE"。# 记得用上括号好好整理你的逻辑。tam = (secretA and secretB) or secretChero.moveXY(19, 26)hero.say(tam)# 如果 A 或 B 为真，并且 C 为真，对 Zsofi 说 "TRUE" 。否则，说 "FALSE"。zso = (secretA or secretB) and secretChero.moveXY(26, 36)hero.say(zso)# Say "TRUE" to Istvan if A OR C is true, AND if B OR C is true. Otherwise, say "FALSE."ist = (secretA or secretC) and (secretB or secretC)hero.moveXY(37, 34)hero.say(ist)# 如果 A 和 B都为真，或者 B 为真且 C 不为真，对 Csilla 说 "TRUE" 。否则，说 "FALSE"。csi = (secretA and secretB) or (secretB and secretC)hero.moveXY(40,22)hero.say(csi)

逻辑圈

# 移动到巫师的旁边，获得他的密码hero.moveXY(20, 24)secretA = hero.findNearestFriend().getSecretA()secretB = hero.findNearestFriend().getSecretB()secretC = hero.findNearestFriend().getSecretC()# If ALL three values are true, take the high path.# Otherwise, take the low path. Save the 4th value.secretD = secretA and secretB and secretCif secretD:    hero.moveXY(30, 33)else:    hero.moveXY(30, 15)# If ANY of the three values are true, take the left path.# Otherwise, go right. Save the 5th value.secretE = secretA or secretB or secretCif secretE:    hero.moveXY(20, 24)else:    hero.moveXY(40, 24)# If ALL five values are true, take the high path.# Otherwise, take the low path.secretF = secretA and secretB and secretC and secretD and secretEif secretF:    hero.moveXY(30, 33)else:    hero.moveXY(30, 15)

返回荆棘农场

函数中定义的参数不仅仅只有一个，也可以定义多个哦！

# 这个函数 “maybeBuildTrap” 定义了两个参数def maybeBuildTrap(x, y):    # 使用x和y作为移动的坐标。    hero.moveXY(x, y)    enemy = hero.findNearestEnemy()    if enemy:        pass        # 使用 buildXY 在特定 x 和 y 处建造 "fire-trap".        hero.buildXY("fire-trap", x, y)while True:    # 这会调用 maybeBuildTrap，并使用上方入口的坐标。    maybeBuildTrap(43, 50)        # 下面在左侧入口使用maybeBuildTrap！    maybeBuildTrap(25, 33)    # 在底部入口处使用“maybeBuildTrap” ！    maybeBuildTrap(43, 20)