
(1)若售退数据为'1A','1C','2A','2B','3C','2B','1A',则此时售出座位有
(2)实现上述功能的部分Python程序如下,请在划线处填入合适的代码。
n = 40;cnt = 0
tick =[0]*n#0 ~n-1号座位初始为待售状态,每位用0表示。
lst = ['5C','2A','7A','6C','6C','4D','7D','4D','10C]
j=0
while j!= len(lst):
puts =lst[j]
st=""
①
for i in puts;
if "0" <= i<= "9":
②
else:
num = int(st)
sq = ord(i)-65 #此时i为puts中读取的字母,以此计算本次操作的座位所处的列号
pos=(num-1)* 4 + sq
tick[pos]=1-tick[pos]
for i in range(0,n,2):
if ③
cnt+=1
print("待售的连坐座位统计:",cnt,"对")

同类型试题

y = sin x, x∈R, y∈[–1,1],周期为2π,函数图像以 x = (π/2) + kπ 为对称轴
y = arcsin x, x∈[–1,1], y∈[–π/2,π/2]
sin x = 0 ←→ arcsin x = 0
sin x = 1/2 ←→ arcsin x = π/6
sin x = √2/2 ←→ arcsin x = π/4
sin x = 1 ←→ arcsin x = π/2


y = sin x, x∈R, y∈[–1,1],周期为2π,函数图像以 x = (π/2) + kπ 为对称轴
y = arcsin x, x∈[–1,1], y∈[–π/2,π/2]
sin x = 0 ←→ arcsin x = 0
sin x = 1/2 ←→ arcsin x = π/6
sin x = √2/2 ←→ arcsin x = π/4
sin x = 1 ←→ arcsin x = π/2

