Answer to Question #267975 in C++ for Imman

Question

Write a program that contains a function that displays' Welcome to CS127-8L!" Then, it asks for a double value such as 25.36, 100.02 etc. The maximum is 1000. Getting the function displays the value in words. Please see the sample runs below.

Run 1

Welcome to CS127-8L!

Enter a double value: 255.36

That is two hundred twenty-five and thirty-six centavos!

Run 2

Welcome to CS127-8L!

Enter a double value: 800.20

That is eight hundred and twenty centavos!

Answer 1

#include <iostream>
#include <string>
#include <math.h>
using namespace std;


void convert(int value){


	string first[20] = {"zero", "one", "two", "three","four","five","six","seven","eight","nine","ten",
		"eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eightteen","nineteen"};


	string second[10] = {"", "ten", "twenty", "thirty","forty","fifty","sixty","seventy","eighty","ninety"};


	if(value<0)
	{
		cout<<" ";
		convert(-value);
	}
	else if(value>=1000)
	{
		convert(value/1000);
		cout<<" thousand";
		if(value % 1000)
		{
			if(value % 1000 < 100)
			{
				cout << " ";
			}
			cout << " " ;
			convert(value % 1000);
		}
	}
	else if(value >= 100)
	{
		convert(value / 100);
		cout<<" hundred";
		if(value % 100)
		{
			cout << " ";
			convert (value % 100);
		}
	}
	else if(value >= 20)
	{
		cout << second[value / 10];
		if(value % 10)
		{
			cout << "-";
			convert(value % 10);
		}
	}
	else
	{
		cout<<first[value];
	} 
}


int main(){




	double value=-1;
	
	cout<<"Welcome to CS127-8L!\n";
	while(value<0 ||value>1000){
		cout<<"Enter a double value: ";
		cin>>value;
	}


	int y  = int(value * 1000);
	int integralpart=y / 1000;
	int fractionalPart=(value-integralpart)*100;


	cout<<"That is ";
	convert(integralpart);
	if(fractionalPart>0){
		cout<<" and ";
		convert(fractionalPart);
		cout<<" centavos!";
	}
	
	cin>>value;
	return 0;
}