C++收银员代码

Question

问题:

Giving change. Implement a program that directs a cashier how to give change. The program has two inputs: the amount due and the amount received from the customer. Display the dollars, quarters, dimes, nickels, and pennies that the customer should receive in return.

我目前拥有的:

#include <iostream>
using namespace std;


int main()
{
double amount_due;
double amount_recieved;

cout << "Enter amount due: " << endl;
cin >> amount_due;
cout << "Enter amount received: ";
cin >> amount_recieved;

int change = amount_recieved - amount_due;
int dollar = 100;
int quarters = 25;
int dimes = 10;
int nickels = 5;
int pennies = 1;



//Return change in full dollars

cout << "dollars: " << change % 100  << endl;


//Return change in quarters

cout << "quarters:  " << (change % 100) % 25 << endl;

//Return change in dimes

cout << "dimes: " << ((change % 100) % 25) % 10 << endl;

// Return change in nickels

cout << "nickels: " << (((change % 100) % 25) % 10) % 5 << endl;

//Return change in pennies

cout << "pennies:  " << ((((change % 100) % 25) % 10) % 5) % 1 << endl;



system("pause");
return 0;

}

我知道还有其他一些答案，但它们可能更高级，无法在我的代码中使用，我做错了什么？

Answer 1

你想做的和收银员一样。

首先确保零钱表示为整便士。

然后提供足够的美元，直到剩下的零钱少于一美元。然后转到 25 美分硬币，然后是 10 美分硬币、5 分硬币和 1 美分硬币。

所以对于美元的情况，伪代码是:

dollars = change / 100        # get integral number of dollars
change = change % 100         # and reduce change-left-to-go accordingly
print dollars, " dollars"

然后按照降低值(value)的顺序将该逻辑应用于其他硬币类型应该是一件简单的事情。