how many weeks to become a millionaire?

[xiikryssiix]

i have to create a program that calculates how many weeks it will take to become a millionaire given an allowance of pennies.

for example...
1st week - 1 cent
2nd week - 2 cents
3rd week - 4 cents
4th week - 8 cents
5th week - 16 cents
so forth, so on...

i sort of need help getting started. so far, i only have a basic equation to begin such program..

(2 ^ n - 1)

in which n is the number pennies of that week.

the total goal is 1,000,000 in dollars, which is 100,000,000 pennies.

i am unsure how to set anything up. so far, i only have...

#include <iostream>
#include <iomanip>
using namespace std;

#include <cmath>

int main()
{
   
   total = ((2 * pow(n)) - 1);
       
    
    
    cout << fixed;
    cout << setprecision(2);    
    
    cout << "In " << weeks << " weeks, I'll be a millionaire." << endl;  
    
    cin.ignore();
    cin.get();
    return 0;
}

can anyone leave me some suggestions how to get going with this program?

[helios]

It involves base 2 logarithms.

[jsmith]

Here is an algorithm:


initialize total_amount_earned to 0.
initialize week_number to 1.
while the total_amount_earned is less than $1,000,000 do:
add to total_amount_earned the amount earned in week "week_number".
increment week_number.


Now when the loop terminates, what will the value of total_amount_earned and week_number be?
For example, total_amount_earned = 1, but week_number = 2, right? (because we increment
week_number). So you might need to output an "adjusted" week_number.

[EDIT: This is just one way to do it. There are other ways as well that don't require loops.]

[tition]

Hehe can't we just do:

std::cout<< "Number of weeks:"<<10;

Sure it would give you a bad grade, but also the satisfaction you annoyed your teacher, who gave you such a lovely problem... ;)

[smilodon]

10 weeks to become a millionaire, awesome. Though you could figure out whatever binary number(consisting of 1 followed by all zeroes) is closest to 100mil, and then the number of bits in that number is your answer.

note this post was edited.

[Kiana]

Based on the allowance that you posted..

#include <winnt.h>
#include <sal.h>
#include <stdio.h>

#define PENNIES_NEEDED 100000000

int __cdecl main(__in int argc, __in_ecount_z(argc) char* argv[], __in_z char* envp[])
{
	UNREFERENCED_PARAMETER(argc);
	UNREFERENCED_PARAMETER(argv);
	UNREFERENCED_PARAMETER(envp);

	UINT	i = 0, j = 1;

	for(; j < PENNIES_NEEDED; i++)
		j *= 2;

	printf("it will take %i weeks for you to become a millionaire", i);
	getchar();
	return EXIT_SUCCESS;
}

:) You can use cout instead of printf, I'm just more of a C person.

[helios]

Christ.
Please tell me how you learnt C so I can discourage people from taking that path, Kiana.

[xiikryssiix]

finally got it to work. thanks for all the suggestions. i did change my equation to having 2 logs within it. im finally getting the output i wanted.

THANKS a million for taking the time to help me out.