Doubling a variable numerous times, and adding all iterations of the variable together

Question

EDITED FOR CLARITY'S SAKE. I APOLOGIZE FOR THE CONFUSION :3

Okay, so I'm following an online CS class, and we're supposed to write a program, in C, that will tell you how much money you'd have if you had a penny at the beginning of the month and doubled it every day.

Each day you would get double what you had yesterday PLUS everything from the previous days.

Example: You start with .01 and what to calculate a running total by day 3. So the first day is .01, second day is .02, third day is .04. On day 3 you would have 0.01+0.02+0.04 (.09).

The program intends to calculate this process over the duration of any given month (28 - 31 days).

I'm having a really hard time trying to implement this. I've got it doubling it, but I'm not sure how to the previously-calculated days together.

Here's my code:

#include <stdio.h>
#include <math.h>

int main(void) {

    /*days represents total days in months*/
    /*pens represents the number of pennies on the first day*/

    long long days;
    long long pens;

    do {
      printf("Enter the number of days in the month: ");
      scanf("%llu", &days);
    } while(days < 28 || days > 31);    

    printf("Enter the initial number of pennies: ");
    scanf("%llu", &pens);

    for (int i=0; i<= days-1; i++) {
      pens += pow(2,i);
      printf("You'll have $%llu\n", pens);
    }
}

edit2: Okay, so I think I fixed it thanks to all your awesome advice. I changed the last part to:

for (int i=0; i<= days-1; i++)
        {
            pens = pens + (pens * 2);

        }
        total = pens / 100;
        printf("You'll have $%.2f\n", total);

    } 

Though there is still a slight issue with the output (which, I'm thinking, is due to the data type I'm using?) It prints out:

You'll have $0.00 You'll have $0.00 You'll have $0.00 You'll have $0.00 You'll have $2.00 You'll have $7.00 You'll have $21.00 You'll have $65.00 You'll have $196.00 You'll have $590.00 You'll have $1771.00 You'll have $5314.00 You'll have $15943.00 You'll have $47829.00 You'll have $143489.00 You'll have $430467.00 You'll have $1291401.00 You'll have $3874204.00

etc.

Pretty good, but I'm betting it's not that accurate since the first few iterations are 0.00.

Answer 1

Well, let's translate what you have to do into pseudocode:

var daily_amount = 0.01
var total = daily_amount

iteration_over_days:
    daily_amount *= 2
    total += daily_amount

From there, all you need to do is translate to C.

Enjoy!

Instead of using the long datatype, you can also start at 1 and then at the end divide by 100: $0.01 -> $1

Answer 2

I think the following line needs change:

pens+=pow(2,i);

It should be:

pens = pens + (pens * 2)

The logic is that if you start with 1 penny on day 1, then it will be something like this

Day 1 : 1 + (1 * 2) = 3
Day 2 : 3 + (3 * 2) = 9
Day 3 : 9 + (9 * 2) = 27
and so on..

I hope this is the logic which you are expecting. By doing pow(2,i), you are just adding square of the day count to your total money each day which does't seem to be the expected logic based on your question.

Answer 3

This is the code you are looking at:

total = 0;
for (int i = 1; i <= days; i++) {
     total += pens;
     pens = pens*2;
}
printf("You'll have $%d\n", total);

Answer 4

Don't compute the power each time! Just keep track of a running count and double it every day.

Here's the core of the algorithm you need for total_days number of days:

typedef unsigned long long int my_uint;

my_uint initial_payment = 1;
my_uint total_pennies = 0;

for (my_uint day_payment = initial_payment, day = 0; days != total_days; ++day)
{
    total_pennies += day_payment;
    day_payment *= 2;
}

// now have total_pennies

---

Of course you could notice that 1 + q + q² + q³ + ... + q^N − 1 is the same as (q^N − 1) / (q − 1) and thus compute the answer in one single step (e.g. for q = 2 in your case and N number of days).