Problems with terminating loop at specific time - c++

Question

Having trouble with my loop. The program is supposed to resemble a print out of a
lotto ticket. The user enters in how many sets of lotto numbers he/she wants. Each line is labeled alphabetically, yet if someone wants more then ten lines (letter J) of lotto numbers the program is supposed to start back over at A again. My problem is that if anyone enters 10 (or any interval of ten) "mega" gets printed like so:
"Mega" should only be printed again if there is another line of lotto numbers.
In "int main()" inside the "for()" is my attempt to remedy this problem.
#include <iostream> //I/O
#include <iomanip> //setw
#include <ctime> //seeding srand
#include <string> //size
#define RAND(a,b) (a+rand()% (b-a+1))
#define die(errmsg) {cerr << errmsg << endl; exit(1);}
using namespace std;
/*
Author: Zachary Stow
Date: July/20/15
Homework #5
Objective: To design a program that imitates the print out
of a lottery ticket.
*/
//********************************fillup()********************************
void fillup(int lotto[], int n, int from, int to)
{
void bubble_sort(int x[], int n);
for(int i = 0; i < n; i++)
{
lotto[i] = RAND(from,to);
}
bubble_sort(lotto,5);
}
//*****************************bubble_sort()******************************
void bubble_sort(int x[], int n)
{
for(int i = 0; i < n-1; i++)
{
int temp;
for(int j=i+1; j<n ; j++)
{
if(x[i] > x[j])
{
temp = x[i];
x[i] = x[j];
x[j] = temp;
}
}
}
}
//********************************print()*********************************
void print(int x[], int n)
{
for(int i = 0; i < n; i++)
{
cout << setfill('0') << setw(2) << x[i] <<" ";
}
cout <<" ";
cout << setfill('0') << setw(2) << RAND(1,46);
cout << endl;
}
//********************************isNumber********************************
bool isNumber(string str)
{
for(int i = 0; i < str.size(); i++)
{
if(!isdigit(str[i]))return(false);
}
return(true);
}
//**********************************check*********************************
void check(int argc, char **argv)
{
bool isNumber(string);
if(argc != 2)die("usage: megaMillion number_tickets");
string num_tickets = argv[1];
if(!isNumber(num_tickets))die("Not a digit."); //removed num_tickets for now
int num;
num = atoi(num_tickets.c_str());
if(num <= 0)die("Zero or negative number."); //doesnt work
}
//*********************************printmega()****************************
void printmega(int letter)
{
if(letter == 65)
{
cout << endl;
cout <<" Mega" << endl; //10 you get a mega
}
}
//*********************************main()*********************************
int main(int argc, char **argv)
{
void fillup(int x[], int n, int from, int to);
void print(int x[], int n);
void check(int argc, char **argv);
void printmega(int letter);
check(argc, argv);
srand(time(NULL));
cout <<" Mega" << endl;
int letter = 65;
for(int i = 0; i < atoi(argv[1]); i++)
{
if(i == atoi(argc[1]))cout << "Hi"; //my attempt to stop the loop from printing
//only mega after J
cout <<(char)letter++; //when theres no more lines
cout <<" ";
if(letter == 75)letter = 65;
int lotto[5];
fillup(lotto,5,1,56);
print(lotto,5);
printmega(letter);
}
return(0);
}

Answer 1

For this problem I would choose to use two nested loops, here is a minimal example:
#include <iostream>
int main(int argc, char** argv)
{
double aValues[] = {1, 2, 3, 4, 5, 6, 7, 8, 6, 2};
int nTotalLines = sizeof(aValues)/sizeof(double);
int nLinesPerBlock = 5;
int nLineNumber = 0;
for (int i = 0; i <= nTotalLines/nLinesPerBlock
&& nLineNumber < nTotalLines; ++i)
{
std::cout << "Start of block..." << std::endl;
for (int j = 0; j < nLinesPerBlock && nLineNumber < nTotalLines; ++j)
{
std::cout << " Number: " << aValues[nLineNumber++] << std::endl;
}
}
return 0;
}
You will need to adapt this code to fit your problem. It is not necessary to use nested loops but to me it seems more logical (you cycle one outer loop for each block and one inner loop for each line).
Another tip: Make use of variables so you don't need to keep typing things like atoi(argv[1]), instead create a variable int name = atoi(argv[1]); and use name wherever you need it.