Question

这是我非常需要指导的收银员计划。该程序会计算一件商品的成本和该商品的付款金额，并计算出将要返还给客户的剩余“零钱”。但是，我遇到了一些无法解决的问题。首先是，在某些金额上，例如：成本= 0.25，已付款= 100.00，该程序会吐出一个零散的地方，该零钱应该没有零散的变化。第二个问题是cout的参数：只有零钱（包括零钱，二十，十，五，一）和硬币（四分之一，一角硬币，镍币，几美分）才具有cout。没有0二十0十等... 所需的示例：成本= 0.25，已支付= 100.00将是4个20（s）1个10（s）1个5（s） 4个季度3个季度；角钱，镍币和便士硬币不应该显示，因为它们为0。我是否只剩下大量的If语句，或者还有其他可以做的事情；也许切换？我不确定，因为我仍然是新手。

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的格式不是最终的，我在cout上显示了“ change”和“ new_change”，因为在“ new_change = change”分配期间，我遇到了另一个问题，即数字减少了； ceil（change）似乎已经修复了它。只要有建设性，任何建议和批评都将受到欢迎。谢谢您的宝贵时间。

Answer 1

发出变更时，并不是从变更值中减去变更。例如，尝试使用输入Purchase：$ 0.25和Payment：$ 100来跟踪代码。当你上线

nickel = (change%10) /5;

您已经用3个季度支付了差额，但是计算机不知道bc的变化值仍然是9975。9975％10 = 5、5 / 5 = 1，所以您得到了1镍。为了避免在每次计算要返还的更改量时避免这种情况，请从总更改中减去该金额。

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;
}

收银员计划

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