收银员计划

Question

这是一个我急需指导的收银员计划。该程序获取物品的成本和为该物品支付的金额，并计算要返还给客户的剩余“零钱”。但是，我遇到了一些我无法解决的问题。第一个是在一定的金额上，exp: cost= 0.25,paid= 100.00，程序会吐出一个 nickle，而应该没有 nickle 用于改变。第二个问题是为 cout 提供的参数：只有个别美元（二十、十、五、一）和硬币（25美分硬币、10美分硬币、5 美分硬币、1 美分硬币）只有当它们出现在零钱中时才可以 cout。没有 0 20 0 10 等等... 所需示例：成本 = 0.25，已支付 = 100.00 将是 4 二十（s） 1 十（s） 1 五（s） 4 个 3 个季度；硬币、镍和便士不应该显示，因为它们是 0。我是否只剩下大量的 If 语句，或者还有其他我可以做的事情吗？也许切换？我不确定，因为我还是个新手。

源代码：

#include <iostream>
#include <iomanip>
#include <cmath>

using namespace std;

int main()
{
    float purchase, payment, change;
    int new_change, one_dollar, five_dollar, ten_dollar, twen_dollar;
    int quarter, nickel, dime, penny;

    // Program explanation
    cout << "This program calculates the change of a payment\n";
    cout << "towards a purchase." << endl << endl;

    // user prompt to input item cost and
    cout << "- - Enter the following Information - -" << endl;
    cout << endl << "Cost of Item: $";
    cin >> purchase;
    cout << endl << endl << "Payment: $";
    cin >> payment;

    // calculating the balance that is owed back to customer
    change = (payment - purchase) * 100;

    // this assignment used so that we can use modulo
    new_change = ceil(change);

    // Following equations used to retrieve respective bill
    // amounts owed back to purchaser
    twen_dollar = new_change / 2000;
    ten_dollar = (new_change % 2000) / 1000;
    five_dollar = (new_change % 1000) / 500;
    one_dollar = (new_change % 500) / 100;

    // Following equations used to retrieve respective coin
    // amounts owed back to purchaser
    quarter = (new_change % 100) / 25;
    dime = (new_change % 25) / 10;
    nickel = (new_change % 10) / 5;
    penny = new_change % 5;



    cout << endl << change << "  " << new_change << "   " << twen_dollar << " 
    twenty " << ten_dollar << " ten ";
    cout << five_dollar << " five " << one_dollar << " one " << quarter;
    cout << " quarter " << dime << " dime " << nickel << " nickel ";
    cout << penny << " penny ";

    return 0;
}

cout 格式不是最终格式，我在 cout 上显示了“change”和“new_change”，因为在“new_change = change”分配期间我遇到了数字减少的另一个问题； ceil(change) 似乎已经修复了它。欢迎任何建议和批评，只要它们具有建设性。感谢您的宝贵时间。

Answer 1

当您分发零钱时，您并没有从零钱的价值中减去它。例如，尝试使用输入 purchase:$0.25 和 payment:$100 来跟踪您的代码。当你到达线路时

nickel = (change%10) /5;

您已经使用 3 个季度支付了差额，但是计算机不知道 bc 的零钱价值仍然是 9975。9975%10 = 5, 5/5 = 1 所以你得到一个镍币。为了避免这种情况，每当您计算要回馈的更改量时，请像这样从总更改中减去它。

ten_dollar = change/1000;
change -= 1000*ten_dollar;

five_dollar = change /500;
change -= 500*five_dollar;

这是我的代码的工作版本：

#include <iostream>
#include <iomanip>
#include <cmath>

using namespace std;

int main()
{
    float purchase, payment;
    int change, one_dollar, five_dollar, ten_dollar, twen_dollar;
    int quarter, nickel, dime, penny;

    // Program explanation
    cout << "This program calculates the change of a payment\n";
    cout << "towards a purchase." << endl << endl;

    // user prompt to input item cost and
    cout << "- - Enter the following Information - -" << endl;
    cout << endl << "Cost of Item: $";
    cin >> purchase;
    cout << endl << endl << "Payment: $";
    cin >> payment;

    // calculating the balance that is owed back to customer
    change = (payment - purchase)*100;

    cout<<"$"<<(float)change/100<<endl;
    // this assignment used so that we can use modulo
    //don't need this, change is an int at this point
    //new_change = ceil(change);

    // Following equations used to retrieve respective bill
    // amounts owed back to purchaser
    twen_dollar = change / 2000;
    change -= 2000*twen_dollar;

    ten_dollar = change / 1000;
    change -= 1000*ten_dollar;

    five_dollar = change / 500;
    change -= 500*five_dollar;

    one_dollar = change / 100;
    change -= 100*one_dollar;

    // Following equations used to retrieve respective coin
    // amounts owed back to purchaser
    quarter = change / 25;
    change -= 25*quarter;

    dime = change / 10;
    change -= 10*dime;

    nickel = change / 5;
    change -= 5*nickel;

    penny = change;



    cout << twen_dollar << "twenty " << ten_dollar << " ten ";
    cout << five_dollar << " five " << one_dollar << " one " << quarter;
    cout << " quarter " << dime << " dime " << nickel << " nickel ";
    cout << penny << " penny ";

    return 0;
}