I'm trying to write a program to calculate compound interest, but I'm not sure how to use the pow() function. My assignment is this:
Dvijesh makes his first $1,025.75 deposit into an IRA earning 4.125% compounded annually on his 24th birthday and his last $1,025.75 deposit on his 35th birthday. With no additional deposits, the money in the IRA continues to earn 4.125% interest compounded monthly until Dvijesh retires on his 65th birthday. Write a program to find out the amount of money that Dvijesh has in his IRA on his 35th and 65th birthday? How much interest did Dvijesh earn on his 35th and 65th birthday?
FV = PMT{(1+i)^n-1 / i}
A=P(1+i)^n
FV:Future Value; PMT:Periodic Payment; i:Rate Per Period; n:Number Of
Payments; A:Future Amount; P:Principal
Right now, I'm trying to calculate the first formula, and this is what I have so far:
#include <iostream>
#include <cmath>
using namespace std;
int main()
{
double futureValue, periodic_payment, rate_per_period;
int n; // number of payments
cout << "Enter the periodic payment, rate per period, and time of investment: ";
cin >> periodic_payment >> rate_per_period >> n;
// Future value of 35th birthday
futureValue = periodic_payment * pow((1 + rate_per_period, n) * 0.01) / (rate_per_period));
return 0;
}
We wrote something similar in my C++ class, but the formula was different. I'm not sure how to write FV = PMT{(1+i)^n-1 / i} in C++.
pow() (and std::pow()) takes 2 input arguments, but you are passing in only 1 value - the result of this calculation:
(1 + rate_per_period, n) * 0.01
Because of your use of the comma operator, (1 + rate_per_period, n) returns n, thus the above is effectively just this simpler calculation:
n * 0.01
That is the sole value you are passing to pow(). But that is not what it wants.
The formula (1+i)^n-1 / i does not translate into this code, as you have written it:
pow((1 + rate_per_period, n) * 0.01)
It actually translates into code more like this instead:
pow(1 + rate_per_period, n - 1 / i)
or:
pow(1 + rate_per_period, n - 1) / i
or:
pow(1 + rate_per_period, n) - 1 / i
Depending on how you read the formula. However, those solutions are all wrong, because the formulas you have shown are written incorrectly to begin with! So, you are translating them into C++ code incorrectly.
The correct formula is:
Related
Full problem: (from https://codeforces.com/problemset/problem/466/A)
Ann has recently started commuting by subway. We know that a one ride subway ticket costs a rubles. Besides, Ann found out that she can buy a special ticket for m rides (she can buy it several times). It costs b rubles. Ann did the math; she will need to use subway n times. Help Ann, tell her what is the minimum sum of money she will have to spend to make n rides?
Input
The single line contains four space-separated integers n, m, a, b (1 ≤ n, m, a, b ≤ 1000) — the number of rides Ann has planned, the number of rides covered by the m ride ticket, the price of a one ride ticket and the price of an m ride ticket.
Output
Print a single integer — the minimum sum in rubles that Ann will need to spend.
Solution from Codeforces (https://codeforces.com/contest/466/submission/7784793):
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
int n, m, a, b;
cin >> n >> m >> a >> b;
if (m * a <= b)
cout << n * a << "\n";
else
cout << (n/m) * b + min((n%m) * a, b) << "\n";
return 0;
}
Although I understand that the conditional statements check if there should be a special ride ticket bought or not, but how is the expression (m * a <= b) derived, and how it is useful in checking if there should be special tickets bought or not? Additionally, i understand how (n/m) * b) is derived, but min((n%m) * a, b) really confuses me.
if (m * a <= b) covers the nonsensical case (in the real world) where the special ticket is at least as expensive as regular tickets for the same number of journeys. Eg a=1, b=10, m=5. Here m * a <= b so it's cheaper to buy the regular tickets.
Having established that it might be worthwhile to buy the regular tickets you should buy (n/m) of them at a cost of (n/m) * b rubles. The remaining journeys could be paid for with one special ticket costing b rubles or with (n%m) regular tickets costing (n%m) * a rubles. Whichever of those is cheaper gives the formula min((n%m) * a, b).
I'm looking at figuring out this program for a lab, I have been working on it for a while now, but cannot for the life of me figure out how this works.
The objective is to display an active equation as it decrements from from the initial value.
For example:
20000.00 + 83.33 - 150.00 = 19933.33
19933.33 + 83.06 - 150.00 = 19866.39
19866.39 + 82.78 - 150.00 = 19799.17
19799.17 + 82.50 - 150.00 = 19731.66
19731.66 + 82.22 - 150.00 = 19663.88
19663.88 + 81.93 - 150.00 = 19595.81
19595.81 + 81.65 - 150.00 = 19527.46
And so forth. I have to display that as an output on the screen. But am not sure how to keep a decrementing total like that and how to display it as an active equation like that in cout form.
The number on the far left is an initial loan that a user inputs, the number in the middle is the interest rate which is calculate using p*r/p (initial loan * interestrate(user will input this as well)/ initial loan). And the number on the right just before the equal sign is a payment which the usual will enter.
The goal is to get it to perform 10 iterations or once the initial loan is fully paid off whichever comes first.
Here is a little guidance
Finance
First you got your basics wrong. If this is finance and it looks this way :-), p*r/p is crap.
The second column in your plot is not a rate, neither is it an interest rate, it is an interest.
P is the loan ammount
r is an annual interest rate
The interest is calculated using P times r/12 since the payments you show are monthly in case r is entered mathematically (e.g. 0.05 for 5 %) or P*r/1200 in case r is counterconventional entered as percentage.
C++
The input of the parameters could be done e.g.
double P, r, q;
std::cout << "Enter P, q and r:\t";
std::cin >> P >> r >> q;
you will need to have the numbers printed fixed precision, this can by done with
std::cout << std::fixed << std::setprecision(2)
one last hint: The needed include files will be
#include <iostream>
#include <iomanip>
last you will need a loop have a look for for-loops or do-while loops.
This should help you to get your homework a good start.
I please check this problem I'm creating a Time Base app but I'm having problem getting to work around the modulus oper (%) I want the remainder of 50%60 which I'm expecting to output 10 but it just give me the Lhvalues instead i.e 50. How do I go about it.
Here is a part review of the code.
void setM(int m){
if ((m+min)>59){
hour+=((min+m)/60);
min=0;
min=(min+m)%60;
}
else min+=m;
}
In the code m is passed in as 50 and min is passed in as 10
How do I get the output to be 10 for min in this equation min=(min+m)%60; without reversing the equation i.e
60%(min+m)
in C++ expression a % b returns remainder of division of a by b (if they are positive. For negative numbers sign of result is implementation defined).
you should do : 60 % 50 if you want to divide by 50
Or, if you want to get mins, i think you don't need to make min=0.
When you do 50 % 60, you get a remiainder of 50 since 50 cannot be divided by 60.
To get around this error, you can try doing do something like 70 % 60 to get the correct value as a result, since you do not want to use 60 % 50
This would follow the following logic:
Find the difference between 60 and min + m after min is set to zero if min + mis less than 60. Store it in a variable var initially set to zero.
check if the result is negative; if it is, then set it to positive by multiplying it by -1
When you do the operation, do min = ((min + m) + var) % 60; instead.
***Note: As I am unfamiliar with a Time Base App and what its purpose is, this solution may or may not be required, hence please inform me in the comments before downvoting if I there is anything wrong with my answer. Thanks!
It looks like you are trying to convert an integral number of minutes to an hour/minute pair. That would look more like this instead:
void setM(int m)
{
hour = m / 60;
min = m % 60;
}
If you are trying to add an integral number of minutes to an existing hour/minute pair, it would look more like this:
void addM(int m)
{
int value = (hour * 60) + min;
value += m;
hour = value / 60;
min = value % 60;
}
Or
void addM(int m)
{
setM(((hour * 60) + min) + m);
}
I'm new to C++ and was wondering if i was on the right track? I'm kind of confused about this but was hoping for possibly some helpful hints on things i am missing/ have wrong....i know its not completely finished i still need to do the breakdown of the dollars,quarters....etc
Question: A cash register uses an automated coin machine to help make change. We assume that a clerk is handed money to pay for purchases. For change, the clerk returns to the customer any paper money and directs the coin machine to distribute any changes less then $1. In this exercise, you are to simulate the action of the clerk and the machine.
At the cash register, we need access to the purchase price and the payment. The change, which is the difference between the payment and the purchase prices, is a real number. The whole part represents the change in dollars and the fractional part is the change in cents that is returned in quarters, dimes, nickels, and pennies. For instance, with a payment of $10 to cover purchases of $3.08, the required change is $6.92. The clerk hand out $6 and the coin machine distributes 3 quarters, 1 dime, 1 nickel, and 2 pennies for the 92 cents.
92 = 3(25) + 1(10) + 1(5) + 2
Use real-number objects that identify the purchase price (price), the amount of payment (payment), and the change (change). The main program computes the amount of change (coinChange) and partitions it into dollars (dollars), quarters (quarters), dimes (dimes), nickels (nickels), and pennies (pennies).
You must declare constants for quarters (quarters), dimes (dimes), nickels (nickels), and pennies (pennies). You must use compound operators in the calculations. You must use setreal(w,p) and setw(n) for the output.
What I have done so far:
// Me
// A BRIEF PROGRAM DESCRIPTION FOR CHAPTER 2, HOMEWORK 4
// COMMENT THE PREPROCESSOR
#include <iostream.h>
// COMMENT THE PREPROCESSOR STATEMENT
#include "textlib.h"
int main( )
{
// COMMENT THE CONSTANTS
const int QUARTER_AMOUNT = 25;
const int DIME_AMOUNT = 10;
// COMMENT THE OBJECTS
double price;
double payment;
double change;
int numofDollars;
int numofQuarters;
int numofDimes;
int numofNickles;
int numofPennies;
int coinChange;
cout << "Enter the purchase total: ";
cin >> price;
cout << "Enter the payment: $";
cin >> payment;
// COMMENT THE CALCULATION
change = payment - price;
numofDollars = int(change);
coinChange = (int((change / numofDollars) * 100));
numofQuarters = coinChange / 25;
coinChange = coinChange / (numofQuarters * 25);
numofDimes = coinChange / 10;
numofNickles = coinChange / 5;
numofPennies = coinChange / 1;
// OUTPUT THE INFORMATION
return 0;
}
Yes, you are on the right track. Your general structure is sound. These sorts of homework assignments almost always have a structure like this:
int main () {
// read in the data
...
// Do the math
...
// Write out the data
...
}
You do have some math errors. Try stepping through the code with a pencil and paper, pretending that you are the computer. Also, try stepping through the code with your debugger, examining the variables after each line. Compare what actually happened to what you expected.
I have only been working on c++ for about a month. I am not really understanding how it works, however I need to write a program for school. I used a void function and it seems to be working so far,but I have no idea what to do next I am lost at line 44, I am not sure how to make it work, is there a way to take a value from a certain string? If the value is in both strings how would I determine which value? Here is my assignment:
A parking garage charges a $2.00 minimum fee to park for up to three hours. The garage charges an additional $0.50 per hour for each hour or part thereof in excess of three hours. The maximum charge for any given 24-hour period is $10.00. People who park their cars for longer than 24 hours will pay $8.00 per day.
Write a program that calculates and prints the parking charges. The inputs to your program are the date and time when a car enters the parking garage, and the date and time when the same car leaves the parking garage. Both inputs are in the format of YY/MM/DD hh:mm
Here's the code I've written so far:
#include <iostream>
#include <fstream>
#include <iomanip>
#include <string>
#include <cmath>
#include <algorithm>
#include <sstream>
using namespace std;
stringstream ss;
string enter_date;
string enter_time;
string exit_date;
string exit_time;
int calculatecharge;
int num;
int i;
int year;
int month;
int ddmmyyChar;
int dayStr;
string line;
int x;
void atk()
{
getline (cin,line); // This is the line entered by the user
stringstream ss1(line); // Use stringstream to interpret that line
ss >> enter_date >> enter_time;
stringstream ss2(enter_date); // Use stringstream to interpret date
string year, month, day;
getline (ss2, year, '/');
}
int main()
{
cout << "Please enter the date and time the car is entering "<< endl
<< "the parking garage in the following format: YY/MM/DD hh:mm"<< endl;
atk();
cout << "Please enter the date and time the car is exiting "<< endl
<< "the parking garage in the following format: YY/MM/DD hh:mm"<< endl;
atk();
if (hr - hr < 3)
cout<<"Parking fee due: $2.00" << endl;
Write a program that calculates and prints the parking charges.
This is the goal of our program. Basically, this is the output.
The inputs to your program are the date and time when a car enters the
parking garage, and the date and time when the same car leaves the
parking garage. Both inputs are in the format of YY/MM/DD hh:mm
So, we want some way of translating the date format entered as a string into a time difference. You could store the time in an int, which represents the amount of minutes elapsed during the parking period (I choose minutes since this is the smallest time period inputted). The challenge here is the parsing of the string into this integer.
You could write a function like this:
int parseDate( std::string dateStr )
{
// Format: YY/MM/DD hh:mm
int year = atoi( dateStr.substr( 0, 2 ).c_str() );
int month = atoi( dateStr.substr( 3, 2 ).c_str() );
int day = atoi( dateStr.substr( 6, 2 ).c_str() );
int hour = atoi( dateStr.substr( 9, 2 ).c_str() );
int min = atoi( dateStr.substr( 12, 2 ).c_str() );
// Now calculate no. of mins and return this
int totalMins = 0;
totalMins += ( year * 365 * 24 * 60 ); // Warning: may not be accurate enough
totalMins += ( month * 30 * 24 * 60 ); // in terms of leap years and the fact
totalMins += ( day * 24 * 60 ); // that some months have 31 days
totalMins += ( hour * 60 );
totalMins += ( min );
return totalMins;
}
Careful! My function here is just an illustration, and does not take into account subtleties like leap years and varying month length. You will probably need to improve on it. The important thing is to recognise that it attempts to take a string and return the number of minutes that have elapsed since year '00. This means we simply have to subtract two integers from the two date strings to find the elapsed time:
int startTime = parseDate( startDateString );
int endTime = parseDate( endDateString );
int elapsedTime = endTime - startTime; // elapsedTime is no. of minutes parked
This is probably the hardest part of the problem, once you have this worked out, the rest should be more straightforward. I will give you a few more tips:
A parking garage charges a $2.00 minimum fee to park for up to three
hours.
Basically just a flat rate: No matter what, the output variable that describes the cost should be equal to at least 2.00.
The garage charges an additional $0.50 per hour for each hour
or part thereof in excess of three hours.
Work out the amount of hours elapsed past three hours - subtract 180 from elapsedTime. If this is greater than 0, then divide it by 60 and store the result in a float (since it is not necessarily an integer result), called, say, excessHours. Use excessHours = floor( excessHours ) + 1; to round this number up. Now multiply this by 0.5; this is the extra cost. (Try to understand why this works mathematically).
The maximum charge for any
given 24-hour period is $10.00. People who park their cars for longer
than 24 hours will pay $8.00 per day.
I will leave this up to you to work out, since this is homework after all. Hopefully I have provided enough here for you to get the gist of what needs to be done. There are many possible approaches to this problem too, this is just one, and may or may not be "the best".
First of all as both the string for the date and the string for the time are continuous(contain no spaces), you do not need to use stringstream to parse the line. You can read the date and time just like this:
cin >> enter_date >> enter_time;
cin >> exit_date >> exit_time;
Now what you need is to convert these strings to actual dates and times. As the format of both is fixed you can write something like this:
void parse_date(const string& date_string, int& y, int& m, int& d) {
y = (date_string[0] - '0')*10 + date_string[1] - '0'; // YY/../..
m = (date_string[3] - '0')*10 + date_string[4] - '0'; // ../mm/..
d = (date_string[6] - '0')*10 + date_string[7] - '0'; // ../../dd
}
Of course this code is somewhat ugly and can be written in better way but I believe this way it is easier to understand. Having this function it should be obvious how to write this one:
void parse_time(const string& time_string, int& h, int &m);
Now that you have the date and time parsed, you need to implement a method that subtracts two dates. IN fact what you care about is the number of hours that elapsed from the enter date time to the exit date time rounded up. What I suggest here is that you convert the dates to number of days from some initial moment(say 00/01/01) and then subtract the two values. Then convert both times to number of minutes since 00:00 and again subtract them. This is not language specific so I believe these tips should be enough. Again using some built-in libraries this can be done easier but I don't think this is the idea of your assignment.
After you have the number of hours rounded up all you need to do is actually to implement the rules in the statement. This will only take a few ifs and is in fact quite easy.
Hope this helps. I am purposefully not giving more detailed explanations so that there is something left for you to think about. After all this is homework and is meant to make you think how to do it.
You don't need to perform separate input reading and parsing operations. You could pass the necessary variables by reference and read the input directly into the variables using stringstream. I would use a structure to store the date and time and overload operator- with an algorithm for subtracting two date/time values. Here's how I would do it:
#include <iostream>
#include <sstream>
using namespace std;
struct DateTime {
// Variables for each part of the date and time
int year, month, day, hour, minute;
// Naive date and time subtraction algorithm
int operator-(const DateTime& rval) const {
// Total minutes for left side of operator
int lvalMinutes = 525600 * year
+ 43200 * month
+ 1440 * day
+ 60 * hour
+ minute;
// Total minutes for right side of operator
int rvalMinutes = 525600 * rval.year
+ 43200 * rval.month
+ 1440 * rval.day
+ 60 * rval.hour
+ rval.minute;
// Subtract the total minutes to determine the difference between
// the two DateTime's and return the result.
return lvalMinutes - rvalMinutes;
}
};
bool inputDateTime(DateTime& dt) {
// A string used to store user input.
string line;
// A dummy variable for handling separator characters like "/" and ":".
char dummy;
// Read the user's input.
getline(cin, line);
stringstream lineStream(line);
// Parse the input and store each value into the correct variables.
lineStream >> dt.year >> dummy >> dt.month >> dummy >> dt.day >> dummy
>> dt.hour >> dummy >> dt.minute;
// If input was invalid, print an error and return false to signal failure
// to the caller. Otherwise, return true to indicate success.
if(!lineStream) {
cerr << "You entered an invalid date/time value." << endl;
return false;
} else
return true;
}
From here, in the main() function, declare two DateTime structures, one for the entry time and one for the exit time. Then read in both DateTime's, subtract the entry time from the exit time, and use the result to generate the correct output.