Programming Practice

[M_D_K]

Your first attempt was a total fail
It printed "FooBar" before 1 and your app stopped at 99 which is an off by one error. This make it a fail.

You second attempt much better you successfully did a recursive version and no off by one error. So 10 points for the recursive and I gonna assume that you also fixed the for loop version so total 18


Also other scores:
BOMERMAN: 16 points did both the for loop version and recursive gained the full 16.
MakazeGamer: 20 points and beat BOMBERMAN to completion.
M_D_K(thats me): 30 points since I made the challenge

OMFG EDIT!!!!2 BOMERMAN has 16 points total.

[KuramaYoko10]

Your first attempt was a total fail
It printed "FooBar" before 1 and your app stopped at 99 which is an off by one error. This make it a fail.

You second attempt much better you successfully did a recursive version and no off by one error. So 10 points for the recursive and I gonna assume that you also fixed the for loop version so total 18


Also other scores:
BOMERMAN: 20 points did both the for loop version and recursive gained the full 20.
MakazeGamer: 20 points and beat BOMBERMAN to completion.
M_D_K(thats me): 30 points since I made the challenge

Nice... I am happy I got better and a 18 out of 20!!

However I dont see those two errors in the first program attempt... I copied and paste the code I posted here in the forum and I got it all right - starting on 1 , and finishing on Bar (100 is a multiple of 5)!!

lol but it is ok... send more challenges =D

[M_D_K]

Sparda: 19 out of 20. So close but no cookie for you sparda.

[Ginto8]

OK then, with recursion:

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Code: Select all

#include <iostream>
#include <conio.h>

using namespace std;

void divisible(int number, int max)
{
     bool divBy3 = ((number % 3) == 0);
     bool divBy5 = ((number % 5) == 0);
     if(divBy3 && !divBy5)
          cout << "foo\n";
     else if(!divBy3 && divBy5)
          cout << "bar\n";
     else if(divBy3 && divBy5)
          cout << "foobar\n";
     else
          cout << number << endl;
     int nextNumber = number + 1;
     if(nextNumber <= max)
          divisible(nextNumber, max);
}

int main()
{
    divisible(1, 100);
    getch();
}

[M_D_K]

OK then, with recursion:

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Code: Select all

#include <iostream>
#include <conio.h>

using namespace std;

void divisible(int number, int max)
{
     bool divBy3 = ((number % 3) == 0);
     bool divBy5 = ((number % 5) == 0);
     if(divBy3 && !divBy5)
          cout << "foo\n";
     else if(!divBy3 && divBy5)
          cout << "bar\n";
     else if(divBy3 && divBy5)
          cout << "foobar\n";
     else
          cout << number << endl;
     int nextNumber = number + 1;
     if(nextNumber <= max)
          divisible(nextNumber, max);
}

int main()
{
    divisible(1, 100);
    getch();
}

Umm you did pretty bad. riddle me this batman do you ever see a foobar in your output? I doubt it. Since you put it at the bottom of an else if chain. no bonus points for you

[Arce]

MDK, i did yours since it's simple and easy...For those of you who're having trouble with it, I'd recommend acquainting yourselves with the modulus operator (% in most languages). It's very useful, and I use it several times in the editor.

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Recursion:

Java:

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// base case: O - Do nothing. 
// if you call foobar(100), it will print out starting at 1 and ending at 100.
public static void fooBar(int n) { 
     if(n!=0) {
          fooBar(n-1);

          if(n % 3 == 0 && n % 5 == 0)
               System.out.println("foobar");
          else if(n % 3 == 0)
               System.out.println("foo");
          else if(n % 5 == 0)
               System.out.println("bar");
     }

}

C++

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// base case: O - Do nothing. 
// if you call foobar(100), it will print out starting at 1 and ending at 100.
void fooBar(int n) {
     if(n!=0) {
          fooBar(n-1);

          if(n % 3 == 0 && n % 5 == 0)
              std::cout << "foobar\n";
          else if(n % 3 == 0)
               std::cout << "foo\n";
          else if(n % 5 == 0)
              std::cout << "bar\n";
     }
}

Blitz:

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Function fooBar(num) 
     if num >  0 Or num < 0 then
          if(num mod 3 = 0) and (num mod 5 = 0) then print "foobar"
          elseif (num mod 3 = 0) print "foo"
          elseif (num mod 5 = 0) print "bar"
     endif 
End Function

Iteration:

Java:

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public static void foobar() {
     for(int n =0;n<100;++n) {
           if(n % 3 == 0 && n % 5 == 0)
               System.out.println("foobar");
          else if(n % 3 == 0)
               System.out.println("foo");
          else if(n % 5 == 0)
               System.out.println("bar");
      }
}

C++

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void foobar() {
     for(int n =0;n<100;++n) {
           if(n % 3 == 0 && n % 5 == 0)
               std::cout << "foobar");
          else if(n % 3 == 0)
               std::cout << "foo";
          else if(n % 5 == 0)
               std::cout << "bar";
      }
}

Blitz:

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Function fooBar() 
     for num = 1 to 100
          if(num mod 3 = 0) and (num mod 5 = 0) then print "foobar"
          elseif (num mod 3 = 0) print "foo"
          elseif (num mod 5 = 0) print "bar"
     next 
End Function 

Also, has nobody else been able to do the coin one? I wasn't very specific, but what I wanted was this output:

10 cents = 0 quarters + 0 dimes + 0 nick1es + 1O pennies
10 cents = 0 quarters + 0 dimes + 1 nick1es + 5 pennies
10 cents = 0 quarters + 0 dimes + 2 nick1es + O pennies
10 cents = 0 quarters + 1 dimes + 0 nick1es + O pennies

For every correct combination of change to create the given amount. I'm NOT testing for efficiency, and no huge numbers will be tested. Again, I have some sample outputs if you need them...

Anyone want to try it? Doing it iteratively really presses your brain with compound loops, haha.

[KuramaYoko10]

Hey Arce.. have you tried my code yet?? ... it is there on the first page!!

BTW, can you pm me your solution for the coin problem written in C/C++ ??


thanks in advance

[Arce]

I'11 give your code a try nexttime I'm at schoo1. Current1y don't have a C++ dev environment setup! (working on it right now). Uninsta11ed my i11egit MVC++ too1s in favor of freeware so1utions (in the event that I create something worthwhi1e and don't want to get sued...)

Current1y have Jcreator for java and will upgrade to NetBeans when I get time. Will likely give codeblocks another try, or *shudder*Bloodshed DevC++...If I don't like either, I may revert to my legit MVC++ 6 (outdated, but I liked it...)

[Amarant]

Here's my attempt at M_D_K's challenge.

Since I have already seen all the other solutions I tried to solve it using a different approach.
- Not using % operator.
- No special code to print foobar when the number is a multiple of 3 and 5.

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Code: Select all

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

int main(int argc, char * argv[])
{
    int i, three = 1, five = 1;
    for(i = 1; i <= 100; i++, three++, five++)
    {
        if(three == 3){ printf("foo"); three = 0; }
        if(five == 5){ printf("bar"); five = 0; }
        if(five != 0 && three != 0) printf("%d",i);
        printf("\n");
    }
    return EXIT_SUCCESS;
}

[wearymemory]

I'm not that content with my code for a dirth of eloquence, but here's my solution to the coin handling problem in Java using nested for loops:

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Code: Select all

public class CoinHandler {
    
    public static void main(String[] args) {
        double val = 1.50;
        
        for(int quarters = 0; quarters * 25 <= val * 100; quarters++) {
            int newVal = (int) ((val * 100) - quarters * 25);
            for(int dimes = 0; dimes * 10 <= newVal; dimes++) {
                newVal -= dimes * 10;
                for(int nickels = 0; nickels * 5 <= newVal; nickels++) {
                    int pennies = (newVal - nickels * 5);
                    System.out.println("Quarters: " + quarters + ", Dimes: " + dimes + ", Nickles: " + nickels + ", Pennies:"+ pennies);
                }
            }
        } 
    }
}

I'd like to have a go at this using recursion tomorrow.
I'd also like to see yours, Arce.

[Ginto8]

OK then, with recursion:

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Code: Select all

#include <iostream>
#include <conio.h>

using namespace std;

void divisible(int number, int max)
{
     bool divBy3 = ((number % 3) == 0);
     bool divBy5 = ((number % 5) == 0);
     if(divBy3 && !divBy5)
          cout << "foo\n";
     else if(!divBy3 && divBy5)
          cout << "bar\n";
     else if(divBy3 && divBy5)
          cout << "foobar\n";
     else
          cout << number << endl;
     int nextNumber = number + 1;
     if(nextNumber <= max)
          divisible(nextNumber, max);
}

int main()
{
    divisible(1, 100);
    getch();
}

Umm you did pretty bad. riddle me this batman do you ever see a foobar in your output? I doubt it. Since you put it at the bottom of an else if chain. no bonus points for you

There are definitely foobar's in there. I put it at the bottom of an if else chain, but that doesn't matter because I was using && every time.

[KuramaYoko10]

OK then, with recursion:

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Code: Select all

#include <iostream>
#include <conio.h>

using namespace std;

void divisible(int number, int max)
{
     bool divBy3 = ((number % 3) == 0);
     bool divBy5 = ((number % 5) == 0);
     if(divBy3 && !divBy5)
          cout << "foo\n";
     else if(!divBy3 && divBy5)
          cout << "bar\n";
     else if(divBy3 && divBy5)
          cout << "foobar\n";
     else
          cout << number << endl;
     int nextNumber = number + 1;
     if(nextNumber <= max)
          divisible(nextNumber, max);
}

int main()
{
    divisible(1, 100);
    getch();
}

Umm you did pretty bad. riddle me this batman do you ever see a foobar in your output? I doubt it. Since you put it at the bottom of an else if chain. no bonus points for you

There are definitely foobar's in there. I put it at the bottom of an if else chain, but that doesn't matter because I was using && every time.

That should matter, since if it is in the bottom of the loop, it will check first for the first 'if' and if it that works it will skip the others 'else if' ...
But since Ginto8 have added a new argument in his 'if', denying other possibilities, it should work:

Code: Select all

 if(divBy3 && !divBy5)

[M_D_K]

OOOOPS I skimmed it since I was up for 30+ hrs

10 points for Ginto.

[Venom]

Hey Arce,

Very nice idea of yours ... I got pretty excited about it and with how I could improve my programming with those excercises!!

I have done the "coin handler" one, it works and shows every possibility, but it is not effective I guess, since I made it divided in several functions...
So, I ask you to look at my code and see if there is a way to put those 4 functions (checkQuarter, checkDime, checkNickle, checkPenny) in only one, using recursion maybe?

Ok here goes the code... bug free until now (I have fixed 2 since my first reply here)

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Code: Select all

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



void printPossibility();

//One function for each coin type check, since the loop is getting smaller and different for each coin value 
//going from 25 to 1
void checkPenny();
void checkNickle();
void checkDime();
void checkQuarter();

const int PENNY = 1;
const int NICKLE = 5;
const int DIME = 10;
const int QUARTER = 25;

//The input cents is the value the user have typed ... constant value
//The cents acquires the value of input, but it will be manipulated in the checking loop 
int inputCents;
int cents;
int pennyCoin = 0;
int nickleCoin = 0;
int dimeCoin = 0;
int quarterCoin = 0;

//It was necessary to set 'q' and 'qq' and so on, so that the 'q--' wouldnt interfere on the loop (for a = 0; a < qq ...)
int q = 0;
int qq = 0;

int d = 0;
int dd = 0;

int n = 0;
int nn = 0;

int p = 0;
int pp = 0;





void checkPenny()
{
	p = (int)(cents / PENNY);
	pp = p;

	if(pp >= 1)
	{
		for(int a = 0; a <= pp; a++)  //the equal sign in  'a<='  ...  fix bug 1
		{
			quarterCoin = q;
			dimeCoin = d;
			nickleCoin = n;
			pennyCoin = p;
			cents = inputCents - ((q*QUARTER) + (d*DIME) + (n*NICKLE) + (p*PENNY));

			if(cents == 0)
			{
				printPossibility();
			}

			p--;
		}
	}
}


void checkNickle()
{
	n = (int)(cents / NICKLE);
	nn = n;

	if(nn >= 1)
	{
		for(int a = 0; a < nn; a++)
		{
			quarterCoin = q;  //redefining the coin values for the coins checked before fix bug 2
			dimeCoin = d;
			nickleCoin = n;
			cents = inputCents - ((q*QUARTER) + (d*DIME) + (n*NICKLE) + (p*PENNY));

			if(cents == 0)
			{
				printPossibility();
			}
			else
			{
				checkPenny();
			}

			n--;
		}
	}
}


void checkDime()
{
	d = (int)(cents / DIME);
	dd = d;

	if(dd >= 1)
	{
		for(int a = 0; a < dd; a++)
		{
			quarterCoin = q;
			dimeCoin = d;
			cents = inputCents - ((q*QUARTER) + (d*DIME) + (n*NICKLE) + (p*PENNY));

			if(cents == 0)
			{
				printPossibility();
			}
			else
			{
				checkNickle();

				dimeCoin = d;
				cents = inputCents - (d * DIME);

				checkPenny();
			}

			d--;
		}
	}
}


void checkQuarter()
{
	q = (int)(cents / QUARTER);
	qq = q;

	if(qq >= 1)
	{
		for(int a = 0; a < qq; a++)
		{
			quarterCoin = q;
			cents = inputCents - ((q*QUARTER) + (d*DIME) + (n*NICKLE) + (p*PENNY));

			if(cents == 0)
			{
				printPossibility();
			}
			else
			{
				checkDime();

				quarterCoin = q;
				cents = inputCents - (q * QUARTER);

				checkNickle();

				quarterCoin = q;
				cents = inputCents - (q * QUARTER);

				checkPenny();
			}

			q--;
		}
	}
}


//The coins are checked from the highest (quarter) to the smallest (penny)
//*Try putting it in only one function
void handleCoins()
{
	checkQuarter();

	cents = inputCents;

	checkDime();
	
	cents = inputCents;

	checkNickle();

	cents = inputCents;

	checkPenny();
}


void printPossibility()
{
	printf("%d pennies , %d nickles , %d dimes , %d quarters \n\n", pennyCoin, nickleCoin, dimeCoin, quarterCoin);

	pennyCoin = 0;
	nickleCoin = 0;
	dimeCoin = 0;
	quarterCoin = 0;
}


int main(int argc, char *argv[])
{
	printf("Please enter an amount of cents that you want me to handle: \n");
	scanf("%d", &inputCents);
	system("CLS");
	

	while(inputCents > 0)
	{
		cents = inputCents;

		handleCoins();

		inputCents = 0;
	}
	
	system("PAUSE");
	return 0;
}

Hey Kurama...

I Remade your code with recursion (just learned with the foobar-Arce`s post up there...so i don`t know if that`s the best way ^^)

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Code: Select all

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

void Check(int b);

const int CoinVal[4] = {25,10,5,1};
int Coinqnt[4];

int InputCents;
int Cents;

int main(int argc, char *argv[])
{
   printf("Please enter an amount of cents that you want me to handle: ");
   scanf("%d", &InputCents);
   system("CLS");
   Cents = InputCents;
   if (InputCents > 0){
	   Check(0);
   }
   else{
	   printf("Invalid Number \n");
   }
   system("PAUSE");
   return 0;
}

void Check(int b)
{
  for(Coinqnt[b] = Cents / CoinVal[b]; Coinqnt[b] >= 0; Coinqnt[b]--)
  {
	 Cents = InputCents - ((Coinqnt[0]*CoinVal[0]) + (Coinqnt[1]*CoinVal[1]) + (Coinqnt[2]*CoinVal[2]) + (Coinqnt[3]*CoinVal[3]));
	
     if(Cents == 0)
     {
		 printf("%d cents = %d pennies , %d nickles , %d dimes , %d quarters \n\n", InputCents, Coinqnt[3], Coinqnt[2], Coinqnt[1], Coinqnt[0]);
     }
	 else if(b!=3)
     {
		 Check(b+1);
     }
  }
}

[KuramaYoko10]

Ok... I dont know if I should bring this topic up again, but I like the idea of it, and I will post a very simple challenge which I remember doing when I was learning the basics of C (which I still am... kinda lol)

You are requested to print on screen X numbers of the fibonacci sequence (starting from 0), being X the variable that the user inputs.
The fibonacci sequence is an interesting sequence of numbers in which you get the previous number, add to your current number to get the result of the next number of the sequence ... here is how it starts:

Code: Select all

0  1  1  2  3  5  8  13  ...

*more info at: http://en.wikipedia.org/wiki/Fibonacci_sequence


I have the code done, but I will try thinking on different ways of solving it, so it will be an exercise to me as well ... I will wait for some replys and then post my code here (and because i am not on my computer right now =P)