A small company is writing its cheques by hand and wants to change
over to computer generated cheques.

Your boss has asked you, the new programmer, to produce a prototype
cheque writer that will accept the payees name and the amount payable
and display a cheque on the screen. The amount should be broken up
into ten thousand, thousand, hundreds and pounds. If the amount in any
of the categories is 0 then the word "zero" is put in that place.

The name of the company is MODCON Ltd.

Input the payee name: Joe
Input the amount payable: EUR 12045.67
*******************************************************
MODCON
Promises to pay Joe EUR 12045.67
1 ten_thousand
2 thousand
zero hundred
45 euro
67 cent
*******************************************************

Note that only one name is used. We cannot yet input first and
second names together.
Use only what you have learned to date Use any compiler you like to
test your code.

Try to get the cheque in the middle of the screen using <iomanip>
Identify the Inputs Outputs and constants.

Identify the data objects. Note that cheque in NOT an object as far as
we are concerned but is a collection of smaller objects that can be
seen on the screen in the example.

List the actions to be performed on those objects.

Write the algorithm consisting of the list of actions ordered
correctly and any control structures used.
Translate the algorithm into C++ code.

You need only test with positive numbers. Your code need not take
negative numbers into account.

#include <iostream.h>
#include <math.h> /* You need this to use the fmod dunction */

int main()
{
double amt,amt_over;
double field_val10k,field_val1k,field_valhun,field_val10,field_val1,field_valcent;

cout << "Enter amount to be paid:"<<endl;
cin >> amt;

/*Work out the Ten thousands field */
amt_over=fmod(amt,10000);
field_val10k=(amt-amt_over)/10000;

/*Work out the thousands field */
amt_over=fmod(amt,1000);
field_val1k=((amt-amt_over)/1000)-(field_val10k*10);

/*Work out the hundreds field */
amt_over=fmod(amt,100);
field_valhun=((amt-amt_over)/100)-((field_val10k*100)+(field_val1k*10));

/*Work out the tens field */
amt_over=fmod(amt,10);
field_val10=((amt-amt_over)/10)-((field_val10k*1000)+(field_val1k*100)+(field_valhun*10));

/*Work out the single unit field */
amt_over=fmod(amt,1);
field_val1=(amt-amt_over)-((field_val10k*10000)+(field_val1k*1000)+(field_valhun*100)+(field_val10*10));

/*Cent field is the easiest!!*/
field_valcent=(fmod(amt,1)*100);


cout << "#######################################"<<endl;
cout << "Ten thousands:\t" <<field_val10k<<endl;
cout << "Thousands:\t" <<field_val1k<<endl;
cout << "Hundreds:\t" <<field_valhun<<endl;
cout << "Tens:\t\t" <<field_val10<<endl;
cout << "Ones:\t\t" <<field_val1<<endl;
cout << "Cents:\t\t"<<field_valcent<<endl;
cout << "#######################################"<<endl;


return 0;
}

>however would this way return something like '4.5678?'
No, integers can't hold a precision, so everything past the radix is truncated. You would have 4. But you would also have known this if you had bothered to try it before asking this question.

#include<iostream>
#include<iomanip>
#include<string>


using namespace std;

int main()
{

string name;
double amount;
string prompt;
prompt = "Input payee name here like: ";
cout << prompt;
cin >> name;
cin.get();

prompt = "Input amount payable: ";
cout << prompt;
cin >> amount;
cin.get();

int ten_thousand;
int thousand;
int hundread;
int euro;
double cent;

cout << endl;
cout << endl;



// THE CHEQUE
cout << "*******************************************";
cout << endl << "\t"<< setw(20) << "MODCON" << endl;

cout << "Promises to pay " << "\t" << name << setw(20) << "EUR" << amount <<endl;

//Calculate amount of thousands.
int thousand = (int) (amount / 1000);
cout << "Thousands " << setw(20) << thousand;

cout << endl << "**********************************************";



cin.get();

return 0;
}

e.g.... lets say the amount is 35857.56

ten_thousand = (int) (amount / 10000);
ten_thousand comes up as 3.

thousand = (int) (amount / 1000) - (ten_thousand * 10);
thousand comes up as 5