class assignment 2 - mostly about loops and stuff

inflation.cpp

/*
 * Write a program that estimates the inflation rate for the past year. The program prompts the user for
 * the current and last year prices of a representative item (e.g., milk, bread, etc.). It then calculates and
 * prints the inflation rate (as a percent value) by analyzing the price difference. In addition, the program
 * prints out an estimate, based on the calculated inflation rate, of the price of the item one and two years
 * hence. Your program should allow the user to repeat the estimation as often as the user wishes. Use
 * functions to calculate the inflation rate and the price estimation.
 *
 */

#include <iostream>
#include <stdio.h>
using namespace std;

int main(int argc, const char *argv[])
{

   float
      currPrice = 0,
      prevPrice = 0,
      currInflation = 0,
      avrgInflation = 0;

   for (float i = 1; true; i++) {
      // get input
      cout << "enter the price of an item last year, followed by its price this year. Enter 0 for either value to quit: ";
      cin >> prevPrice;
      if(!prevPrice){ return 0; }
      cin >> currPrice;
      if(!currPrice){ return 0; }

      //calc current Inflation change
      currInflation = currPrice / prevPrice; // to get the change in percent

      // output the estimated price based on the inflation for that item
      printf("Based on an inflation rate of %.2f%, the price next year will be $%.2f and the price in two years will be $%.2f\n",
         (currInflation - 1) * 100, currPrice * currInflation, currPrice * currInflation * currInflation
      );

      // calculate running average of inflation values (just for fun)
      avrgInflation = (avrgInflation * ( ( i - 1) / i ) ) + (currInflation / i);
      // if this is not the first iteration, output the new values based on the average inflation rate
      if(i > 1){
         printf("Based on an average inflation rate of %.2f%%, the price next year will be $%.2f and the price in two years will be $%.2f\n",
            (avrgInflation - 1) * 100,
            currPrice * avrgInflation,
            currPrice * avrgInflation * avrgInflation
         );

      }
      // output a blank line to make it look nice
      cout << endl;
   }
   
   return 0;
}

perfect_numbers.cpp

#include <iostream>
#include <cmath>

bool isPerfect(int n){
   int ub = (int) ceil(sqrt(n));
   int sum = 1;
   for (int i = 2; i < ub; ++i) {
      if(n % i == 0){
         sum += i;
         sum += (n/i);

      }
   }
   if(ub*ub == n){
      sum += ub;
   }

   return n == sum;
}

int main(int argc, const char *argv[])
{
   
   using namespace std;
   for (int i = 0; i < 10000; i++) {
      if(isPerfect(i)){
         cout << i << endl;
      }
   }

   return 0;
}

primes.cpp

/*
 * PROBLEM STATEMENT:
 *
 * Write a program that inputs a lower and an upper range and then finds and prints all prime numbers in
 * that range. A prime number is a number such that one and itself are the only numbers that evenly
 * divide it. For example for the range [3, 20] the program should output the following: 3, 5 , 7 , 11, 13,
 * 17, 19.
 *
 * For this task I'll use the "Sieve of Eratosthenes" to calculate the primes
 * it is fairly straight forward and is explained below.
 *
 * input is given via command line arguments, their order does not matter
 *
 */

#include <stdio.h>
#include <stdlib.h>

using namespace std;

// has to be declared as int not void or else compiler complains when used in the short circuit below
// ("void value not ignored as it ought to be")
int die(const char * message){
   puts(message);
   exit(0);
   return 0; // will never get here, but needs a return val
}

// with some help from http://www.java2s.com/Code/C/Pointer/SwapfunctionExchangethevaluesbypointers.htm
void swap(int *a, int *b){
   int c = *a;
   *a = *b;
   *b = c;
}

int main(int argc, const char *argv[])
{
   // short circuits!
   argc == 3 or die("expecting exactly 2 arguments (upper and lower bounds).");

   int lowerBound, upperBound;

   lowerBound = atoi(argv[1]);
   upperBound = atoi(argv[2]);

   // eliminate any human error in reversing the arguments
   if (lowerBound > upperBound) swap(&upperBound, &lowerBound);


   // to calculate all the primes we will use the Sieve of Eratosthenes 
   // ( http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes )
   // unfortunately this requires getting all primes from 2 to max
   // this is really inefficient if the lower and upper bounds are both
   // really high, for example [10000,10020], but in most normal
   // cases I think it should be the best approach. I'm sure there's
   // some good way to figure out the optimal method for a given range,
   // but I'm not too concerned about that.
   int range[upperBound - 1];

   // fill the sieve with all ints from 2 to upperBound
   for(int i = 0; i < (upperBound - 1); ++i){
      range[i] = i + 2;
   }

   // this boolean keeps track of whether we've printed any prime to the
   // screen yet, so that the formatting ends up being nice, and there is
   // no trailing ", "
   bool haventPrintedYet = true;
   for(int i = 0; i < (upperBound - 1); ++i){
      // if the current entry is not 0, it means it is prime
      if(range[i]){
         // so we set up some temp variables and then loop through, filtering out
         // every number in the range that is a multiple of the number we have now
         int p = range[i];
         int j = i + p;
         while(j < (upperBound - 1)){
            range[j] = 0;
            j += p;
         }

         // as explained above, make sure the formatting is good.
         if(p >= lowerBound){
            printf(haventPrintedYet ? "[ %d" : ", %d", p);
            haventPrintedYet = false;
         }
      }
   }

   puts(" ]");
   return 0;
}

vending.cpp

/*
 * PROBLEM STATEMENT:
 *
 * A newly installed vending machine sells DVDs for $15. Inside every DVD is a coupon. 6 coupons
 * are needed to get one DVD for free. Write a program that inputs the number of dollars and outputs
 * how many DVDs a customer can collect after spending all cash money and redeeming as many
 * coupons as possible. The program should also output the number of leftover coupons.
 * 
 * For example, if customer “John” has 500 dollars then he can initially buy 33 DVDs. This gives him
 * 33 coupons. He can redeem 30 coupons for 5 additional DVDs. These 5 additional DVDs have 5
 * more coupons, so we now have a total of 8 coupons when added to the 3 remaining coupons from the
 * original purchase. This gives John enough to redeem one final DVD. As a result John has received
 * 39 DVDs and still has 3 leftover coupons.
 */

/*
 * First the basics:
 * inputs:
 *    dollars
 * outputs:
 *    number of DVDs
 *
 *
 * Algorithm:
 *    - first step is to get the initial number of DVDs based on the dollar amount
 *    - after this is done we need to iteratively calculate how many DVDs can
 *    be purchased using tokens
 *       - each iteration we spend coupons and get back a smaller number of coupons and some DVDs
 *       - we repeat this until we can no longer buy any DVDs with the number of coupons we have.
 */

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

using namespace std;

int main()
{
   const float DOLLAR_COST_OF_DVD = 15.00;
   const int COUPON_COST_OF_DVD = 6;

   float inputAmount;
   int
      DVDs = 0,
      coupons = 0;

   // get input from the user
   cout << "enter the amount of money to be spent: ";
   cin >> inputAmount;

   // calc the initial number of DVDs purchased / coupons obtained
   DVDs = (int)floor(inputAmount / DOLLAR_COST_OF_DVD);
   coupons = DVDs;

   int newDVDs = 0;
   while(coupons >= COUPON_COST_OF_DVD){
      newDVDs = coupons / COUPON_COST_OF_DVD; // this works because int division is truncated towards zero, not rounded
      coupons = coupons % COUPON_COST_OF_DVD; // the coupons that remain is the remainder of the int division
      coupons += newDVDs;
      DVDs += newDVDs;
   }

   printf("with $%.2f you can buy %d DVDs. You will have %d coupons left over\n", inputAmount, DVDs, coupons);

   return 0;
}

windchill.cpp

/*
 * During winter, meteorologists often report the so-called wind chill factor, which takes into account the
 * wind speed in addition to the temperature. The wind chill factor for a given temperature (in
 * Celsius) and wind speed may be approximated by the following formula:
 * 
 *                w = (9/200)(5.27sqrt(v) + 10.45 - 0.28v)(t - 33) + 33
 *
 * where
 *    v = wind speed in m/sec
 *    t = temperature in degrees Celsius: t <= 10
 *    w = wind chill index (in degrees Celsius)
 *
 * Write a program that prompts the user for the current temperature and wind speed and then outputs the
 * corresponding wind chill index. Note that the user should be given the choice between entering the
 * temperature in Celsius or Fahrenheit. Use functions for the wind chill calculation and the conversion
 * between Celsius and Fahrenheit (and vice versa). Your code should ensure that restrictions on the
 * temperature for calculating the wind chill are not violated.
 *
 * Gordon Bailey for Dr. Daniel Sinnig's COEN 243, fall 2011
 * 
 */

#include <math.h>
#include <string>
#include <iostream>
#include <cstdlib>
#include <stdio.h>

using namespace std;

float windchill(float v, float t)
{
   printf("windspeed: %.2f temp: %.2f\n\n", v, t);
   return (9.0/200.0)*((5.27*sqrt(v)) + 10.45 - (0.28*v))*(t - 33.0) + 33.0;
}

float farenheitToCelsius(float f){
   return 0.5555555555555555*(f-32);
}

int main(int argc, const char *argv[])
{
   float windspeed, temperature; // (temperature in C)
   string tempStr;

   // get inputs from user
   cout << "enter the wind speed (in metres / second): ";
   cin >> windspeed;

   cout << "enter the temperature (add a C or F suffix to specify units, Celcius is default): ";
   cin >> tempStr;

   // calculate temperature based on input value and unit
   // accepted temperatures are in the format "x", "xc", "xC", "xf", "xF"
   // where x is a numerical string, e.g "1", "2.34", "-5.67"
   
   // first we find the first occurence of any of the characters that
   // represents a unit. The default unit is 'c'.
   size_t pos = tempStr.find_first_of("fFcC");
   char unit = 'c';
   // if we could find a unit character then we extract it and then take the
   // substring up to, but not including, that character
   if (pos != string::npos){
      unit = tempStr[pos];
      tempStr = tempStr.substr(0, pos);
   }

   // we then convert the resultant string to a float value
   // there is no validation/error correction right now...
   temperature = atof(tempStr.c_str());
   // based on the unit we convert the temperature to Celsius if necessary
   if(unit == 'f' || unit == 'F'){
      temperature = farenheitToCelsius(temperature);
   }

   if(temperature > 10.0){
      printf("The temperature you entered (%.2f degrees C) is too warm (it should be less than 10.00 degrees C)\n\n", temperature);
      return 0;
   }

   printf("the windchill makes it feel like %.2f degrees.\n\n", windchill(windspeed, temperature));

   return 0;
}