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c# - 每个客户练习的 C# 票分配

问题描述

我需要你的帮助来做这个练习。我需要为每个客户分配票。如果票多于客户，则票将添加到第一个客户，依此类推。

例子：

Enter ticket number: 10

Enter customer number: 5

结果：

customer#1 ticket#1 ticket#6
customer#2 ticket#2 ticket#7
customer#3 ticket#3 ticket#8
customer#4 ticket#4 ticket#9
customer#5 ticket#5 ticket#10

到目前为止，这是我的代码，它只能满足客户的第一个循环，但后续的票是我的问题。

List<int> customerNumberList = new List<int>();
List<int> ticketNumberList = new List<int>();

Console.Write("Enter Numer of Tickets: ");
int ticketCount = int.Parse(Console.ReadLine());

Console.Write("Enter Number of Customer: ");
int customerCount = int.Parse(Console.ReadLine());

for(int i = 1; i <= ticketCount; i++)
{
    ticketNumberList.Add(i);
}

for(int i = 1; i <= customerCount; i++)
{
    customerNumberList.Add(i);
}

if(customerNumberList.Count == 1)
{
    Console.WriteLine("Customer#1");

    for (int i = 0; i < ticketNumberList.Count; i++)
    {
        Console.WriteLine("Ticket#" + ticketNumberList[i]);
    }
}
else
{
    for (int i = 0; i < customerNumberList.Count; i++)
    {
        Console.WriteLine("Customer#" + customerNumberList[i]);

        for(int j = 0; j <ticketNumberList.Count; j++)
        {
            if(customerNumberList[i] == ticketNumberList[j])
            {
                Console.WriteLine("Ticket#" + ticketNumberList[j]);
            }
        }
    }
}

感谢大家

标签： c#console

解决方案

伪代码：

Ask for `int ticketCount`
Ask for `int customerCount`

customersWithTickets = List<int>[customerCount] (so an array of customers, each element is the list of the tickets of the customer)

set each element of customersWithTickets to a `new List<int>()`

int remainingTickets = ticketCount
int currentTicketNumber = 1

while remainingTickets != 0
    for each customersWithTickets
        add to current customersWithTickets the value of currentTicketNumber
        increment currentTicketNumber
        decrement remainingTickets

        if remainingTickets == 0 then break the for cycle
    end for
end while

for each customersWithTickets i = [0..customersWithTickets.Length[
    print customer#, without going to new line

    for each ticket of the current customersWithTickets j = [0..customersWithTickets[i].Count[
        print ticket# (`customersWithTickets[i][j]`), without going to new line
    end for

    print new line
end for

print the customers, each one with its ticket.

您显然可以走另一种方式（我们将这种方式称为“作弊”方式）：您实际上不需要存储客户的单票来打印它们。您可以简单地注意到，如果有 13 张票，有 5 个客户，则每个客户有 13 / 5 = 2 张票（其中 / 是整数除法），加上 13 mod 5 = 3（mod 是除法的余数，%在 C# 中）前 3 位客户有一张额外的票。对于门票的确切数量，它更容易：

the customer 1 will have 3 tickets: 1, 6, 11 
the customer 2 will have 3 tickets: 2, 7, 12
the customer 3 will have 3 tickets: 3, 8, 13
the customer 4 will have 2 tickets: 4, 9
the customer 5 will have 2 tickets: 5, 10

我希望清楚每个客户都会有以下形式的票：

the customer x will have n tickets (calculated as above): 
    (x + 0 * num of customers), 
    (x + 1 * num of customers), 
    (x + 2 * num of customers), 
    ...

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