大家一起學python-day2 -布林型別,基礎運算和基礎流程
阿新 • • 發佈:2018-12-29
''' #布林型 #1.0,空元組,列表,字典,字串,none都屬於flase print(bool(0))#整型 print(bool([]))#列表 print(bool(()))#元組 print(bool({}))#字典和集合 print(bool(''))#字串 print(bool(None))#None #2.非零,非空列表,字典,字串,元組 print(bool(1))#整形 print(bool([1,1]))#列表 print(bool({'name':'word'}))#字典 print(bool({1,2,3,4,5,6,7,7})) print(bool('sdad'))#字串 print(bool((1,2,3,4,)))#元組 #注:對於字典和集合(兩個都是大括號{}),空{}的type是字典型別! print(type({})) ''' # 3.運算子 a = 10 b = 20 print(a + b) # 加法運算子 print(a - b) # 減法運算子 print(a * b) # 乘法運算子 print(a / b) # 除法運算子 print(a ** b) # 冪運算子 print(9 // 4) # 地板除,向下取整數 print(9 % 4) # 取餘數#可用這個判斷是否是奇數或者偶數 # 4.複合運算 # += --> a=a+b # *= --> a=a*b # -= --> a=a-b # 5.比較運算子 print(10 > 0) # 大於 print(10 != 20) # 不等於 print(10 < 20) # 小於 print(10 == 10) # 等於 print(10 >= 10) # 大於等於 print(10 <= 10) # 小於等於 # 6.邏輯運算子 b = 10 > 4 # true b1 = 10 > 2 # true b2 = 2 > 3 # false b3 = 10 < 5 # false print(b and b1 and b2) print(b or b1 or b2) print(b1 or b2 or b3) print(b2 or b3) print(b) print(not b) a = 10 b = 'aa' print(a and b) # 當a為真,需要看後面的是否也為真,所以返回後面b print(a or b) # 有一個真的就可以,所以當a為真的時候,不用往後看,直接輸入a #7.邏輯運算的優先順序 #not > and >or #8.import random money = 100 while True: print('1代表剪刀') print('2代表石頭') print('3代表布') player = input("請您輸入剪刀或者石頭或者布") if player.isdigit() == False: print("輸入錯誤!請重新輸入") continue else: player = int(player) computer = random.randrange(1, 3) print("電腦出的是", computer) if player == computer: print("平局,請重新再來") elif player == 1 and computer == 3 or player == 2 and computer == 1 or player == 3 and computer == 2: print("您贏了") money += 10 else: print("你輸了") money -= 10 print("您還有", money, '錢') if money == 0: print('太菜了!沒錢了!')
# a = 1 # while a < 10: # print(" "*(10-a),'*'*a) # print() # a += 1 ''' #求和 a = 1 b = 0 while a <= 100: b += a a += 1 print(b) #求偶數 c = 1 d = 0 while c <= 100: if c % 2 == 0: d += c c += 1 print(d) ''' # 求和平均值 # a = 1 # b = 0 # c = 0 # while a <= 5: # print("請輸入您的第", a,"個數:") # c = input() # c = int(c) # b = b + c # a += 1 # print("平均值:", b / 5) # print("和:", b) # 逆序輸出 #第一種方法 a = int(123/100)#百位 b = int((123-a*100)/10)#十位 c = int(123-int((123/100))*100-int((123-a*100)/10)*10)#個位 d = c*100 +b*10+a print(d) #第二種方法 a = '123' b = a[2]+a[1]+a[0] print(b)