C++ 出纳程序

这是一个出纳计划，我迫切需要指导。该程序计算一个项目的成本和为该项目支付的金额，并计算要返还给客户的剩余“变化”。然而，我有几个问题我还没有解决。第一个是，在某些金额上，exp:cost=0.25，paid=100.00，程序会在没有零钱的地方吐出一分钱。第二个问题是为cout提供的参数：只有零钱（二十美元、十美元、五美元、一美元）和硬币（四分之一、一角硬币、五分镍币、一便士）只有在零钱中存在时才能使用。没有0个20个10个等。。。 所需示例：成本=0.25，已付=100.00将是4二十（s）1十（s）1五（s） 4一（s）三个季度；一角硬币、五分镍币和便士不应显示，因为它们是0。我是否只剩下大量的If语句，或者我还能做些什么；也许换一下？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;
}

#包括
#包括
#包括
使用名称空间std；
int main（）
{
浮动采购、支付、变更；
零钱，一元，五元，十元，两元；
四分之一，五分之一，一角，一便士；
//程序说明
cout当您发出更改时，您不会从更改的值中减去它。例如，尝试使用输入跟踪您的代码购买：$0.25，付款：$100。当您到达该行时

nickel = (change%10) /5;

你已经用3个季度支付了差额，但是计算机不知道bc的零钱值仍然是9975.9975%10=5，5/5=1，所以你得到一个五分镍币。为了避免这一情况，每当你计算要返还的零钱金额时，就这样从总零钱中减去它

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

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

以下是我的代码工作版本：

#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;
}

#包括
#包括
#包括
使用名称空间std；
int main（）
{
浮动购买、支付；
零钱，一美元，五美元，十美元，两美元；
四分之一，五分之一，一角，一便士；
//程序说明
你能试着调试你的每一个问题吗？也许你想分割计算并编写一些自动化测试？这可能有助于你在不引入新错误的情况下一个一个地修复错误。你可能想用你的编程语言标记你的问题，以针对正确的读者。下面的阅读可能会也没有帮助。