Basic program for calculating money-CodePudding

I have a basic assignment and can't get the program right. The assignment is to make a program that displays the minimum amount of banknotes and coins necessary to pay.

#include <iostream>
using namespace std;

int main()
{
    int pari;
    cin >> pari;
    switch (pari)
    {
        case 1: cout << pari/5000 << "*5000" << endl;
        break;
        case 2: cout << pari/1000 << "*1000" << endl;
        break;
        case 3: cout << pari/500 << "*500" << endl;
        break;
        case 4: cout << pari/100 << "*100" << endl;
        break;
        case 5: cout << pari/50 << "*50" << endl;
        break;
        case 6: cout << pari/10 << "*10" << endl;
        break;
        case 7: cout << pari/5 << "*5" << endl;
        break;
        case 8: cout << pari/2 << "*2" << endl;
        break;
        case 9: cout << pari/1 << "*1" << endl;
        break;
        default: cout << "WRONG";
    }
    return 0;
}

For example:

Input:

Output:

I tried with switch case, with if statements, but nothing works.

CodePudding user response：

To get the kind of output you have shown, use logic that looks more like this instead:

#include <iostream>
using namespace std;

int main()
{
    int pari;
    cin >> pari;

    cout << pari/5000 << "*5000" << endl;
    pari %= 5000;

    cout << pari/1000 << "*1000" << endl;
    pari %= 1000;

    cout << pari/500 << "*500" << endl;
    pari %= 500;

    cout << pari/100 << "*100" << endl;
    pari %= 100;

    cout << pari/50 << "*50" << endl;
    pari %= 50;

    cout << pari/10 << "*10" << endl;
    pari %= 10;

    cout << pari/5 << "*5" << endl;
    pari %= 5;

    cout << pari/2 << "*2" << endl;
    pari %= 2;

    cout << pari/1 << "*1" << endl;

    return 0;
}

Online Demo

Which can be simplified if you put the banknotes in an array and loop through it, eg:

#include <iostream>
using namespace std;

int main()
{
    const int bankNotes[] = {5000, 1000, 500, 100, 50, 10, 5, 2, 1};
    const int numBankNotes = sizeof(bankNotes)/sizeof(bankNotes[0]);

    int pari;
    cin >> pari;

    for (int i = 0; i < numBankNotes;   i) {
        cout << pari/bankNotes[i] << "*" << bankNotes[i] << endl;
        pari %= bankNotes[i];
    }

    return 0;
}

Online Demo

CodePudding user response：

I have written several versions for you here. Hopefully it will help you to understand the procedure.

Version 1

This is a navie version. This is how we would do it if we were doing it by hand.

int main()
{
  int input_value = 0;
  std::cin >> input_value;         // First we get the input.
                                   // We start with the highest value banknote.

  int       value                = input_value;
  int const number_of_5000_notes = value / 5000; // How many of these notes do 
                                                 // we need?
  value                          = value % 5000; // Now calculate the rest.

  int const number_of_1000_notes = value / 1000; // How many of these notes do 
                                                 // we need? 
  value                          = value % 1000; // Now calculate the rest.
  int const number_of_500_notes  = value / 500;
  value                          = value % 500;
  int const number_of_100_notes  = value / 100;
  value                          = value % 100;
  int const number_of_50_notes   = value / 50;
  value                          = value % 50;
  int const number_of_10_notes   = value / 10;
  value                          = value % 10;
  int const number_of_5_notes    = value / 5;
  value                          = value % 5;
  int const number_of_2_notes    = value / 2;
  value                          = value % 2;
  int const number_of_1_notes    = value;

  // At the end we write the output           
  std::cout << "Input: " << input_value << std::endl;
  std::cout << "Output:" << std::endl;
  std::cout << number_of_5000_notes << " x 5000" << std::endl;
  std::cout << number_of_1000_notes << " x 1000" << std::endl;
  std::cout << number_of_500_notes << " x 500" << std::endl;
  std::cout << number_of_100_notes << " x 100" << std::endl;
  std::cout << number_of_50_notes << " x 50" << std::endl;
  std::cout << number_of_10_notes << " x 10" << std::endl;
  std::cout << number_of_5_notes << " x 5" << std::endl;
  std::cout << number_of_2_notes << " x 2" << std::endl;
  std::cout << number_of_1_notes << " x 1" << std::endl;

  return 0;
}

Version 2

This is a more advanced version

int main()
{
  int value = 0;
  std::cin >> value; // Get input

  // Check input
  if (value == 0)
  {
    std::cout << "No value or 0 has been entered";
    return 0;
  }

  // Output on the fly
  std::cout << "Input: " << value << std::endl;
  std::cout << "Output:" << std::endl;

  // loop over a sorted list of banknotes.
  for (auto note_value_ent : {5000, 1000, 500, 100, 50, 10, 5, 2, 1})
  {
    int const number_of_notes = value / note_value_ent;
    value %= note_value_ent;
    std::cout << number_of_notes << " x " << note_value_ent << std::endl;
  }
  return 0;
}

Both versions give the same result (except in the case of an invalid entry).