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Coffee inTime
Asked:2020-09-25 02:11:04 +0000 UTC2020-09-25 02:11:04 +0000 UTC

最低总票价

  • 772

如何解决以下问题，是否需要穷举枚举或递归？我在下面附上了失败的代码。

n人在地铁上。众所周知，每个人都需要进行m次旅行。地铁票分为三种：

  • 一张单程票，花费一个卢布。

  • 一个人的无限票，费用为b卢布。

  • k人的团体无限票，费用为c卢布。

输入数据格式：

  • 第一行有 3 个整数n、m和k ( 1 ≤ n , m , k ≤ 1000)

  • 第二行有 3 个整数a、b和c (1 ≤ a , b , c ≤ 1000)

输出格式：打印单个整数 - 门票的最低总成本。

例子：

входные данные
7 4 5
2 7 20
выходные данные
34

失败的代码：

n,m,k = 7,4,5
a,b,c = 2,7,20
 
cost_1 = a
cost_2 = b/m
cost_3 = (c/m)/k
 
costs = {"cost_1": cost_1, "cost_2": cost_2, "cost_3": cost_3}
 
min_cost = None
ticket = ''
for key in costs:
    if min_cost is None or costs[key] < min_cost:
        min_cost = costs[key]
        ticket = key
 
 
result = 0
if ticket == "cost_1":
   result = cost_1 * m * n
 
elif ticket == "cost_2":
    result = b * n
 
elif ticket == "cost_3":
    if n % k == 0:
        result = n/k * c
    else:
        temp = n//k
        temp_2 = n - (temp * k)
 
        min_second_cost = None
        second_ticket = ''
        for key in costs:
            if key == ticket:
                continue
            else:
                min_second_cost = costs[key]
                second_ticket = key
 
 
        temp_result = 0
        if second_ticket == "cost_1":
            temp_result = cost_1 * m * temp_2
        elif second_ticket == "cost_2":
            temp_result = b * temp_2
 
        result = temp_result + temp * k
 
print(result)
python

1 个回答

  1. Best Answer
    def foo(n, m, k, a, b, c):
    cost0 = n * m * a
    cost1 = n * b
    cost2 = n // k * c
    print(cost0, cost1, cost2)
    if cost2 != 0:
        cost2 += foo(n % k, m, k, a, b, c)
    else:
        return min(cost0, cost1)
    return min(cost0, cost1, cost2)
    • 4

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