Loops and Penny Calculation

[Amehd]

I have a project due friday.
I have to type a c++ program that calculate how much a person would earn over a period of time if his or her salary is one penny the first day and two pennies the second day, and continues to double each day. i have typed this program but i cannot get it to work. do i need to fix my loop? is the loop supposed to be for or while?

//This program will calculate how much a person 
//would earn over a period of time if his or her salary
//is one penny the first day and two pennies the second 
//day, and continues to double each day

//Amehd Gutierrez

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

int main()
{
	int days;				//number of days
	double pennies = 0.01;	//pennies = $0.01
	double total = 0.0;		//Accumalator, initilized with $0.01

	//Get the number of days
	cout << "For how many days will the pay double? ";
	cin >> days;

	cout << endl << "Day             Total Pay \n";
	cout << "-------------------------------- \n";
	
	//Get the double of 0.01 for each day and accumalate the total.
	for ( int count = 1; count <=days; count++)
	{
		
		cout << count << "\t\t$\t" << (pow(2, pennies)) << endl;
	
	}

	//Displays the total 
	cout << setprecision(2) << fixed << showpoint;
	cout << "-------------------------------- \n";
	cout << "Total           $\t" << total << endl;
	
	return 0;
}

[Moschops]

At which point in that code are you actually changing the value of total?

[Amehd]

see thats the problem. i dont know where to go from there. i want the program to add up. so for example i put 10 days...... day1 .01 day 2 .02 day 3 .04. and then get the total

[hamsterman]

You could simply work out (or find) a formula to calculate the sum of a geometric progression, but the task was probably meant to make you use loops.

for...;...;...) part is OK. The problem is what you put in it. Now you just print 2^0.01 many times. What you need to do is to add something to total on every cycle. That something should double after every cycle. That something is pennies.

[Amehd]

double? as in * 2 or ^2

[moorecm]

I sounds like 2ⁿ. It's kind of funny that they only gave 2 terms, though.

Here, find the pattern:
1, 2, etc.

That kind of makes me chuckle.

[hamsterman]

sure you could keep adding 2ⁿ for n := 0 to days-1, but you need to understand the faster way too.

this is a geometric progression : b_n = b_n-1*2
let's say the task is to fill an array with 1, 2, 4...
the solution clearly (from the formula above) would be

arr[0] = 1;
for(int i = 1; i < array_size; i++)
   arr[i] = arr[i-1]*2;

but you don't need to store them anywhere. you don't need that array. still, new_a = last_a*2. then swap them and calculate another new_a. thanks to how computers work, you can write a = a*2; (or a *= 2;). then you just have add up n of those.

[rocketboy9000]

Or, you could do some more math, and find this solution:

#include "stdio.h"
#include "math.h"
int main(){
	double days;
	printf("Number of days: ");
	scanf("%lf",&days);
	printf("You will be paid $%.2lf",(pow(2,days)-1)/100);
}