Coin Change Maker Python Program










1















I am in a beginner programming course. We must do an exercise where we make a change maker program. The input has to be between 0-99 and must be represented in quarters, dimes, nickles, and pennies when the input is divided down between the four. I wrote a code that involved loops and whiles, but he wants something more easy and a smaller code. He gave me this as a way of helping me along:



c=int(input('Please enter an amount between 0-99:'))
print(c//25)
print(c%25)


He told us that this was basically all we needed and just needed to add in the dimes, nickles, and pennies. I try it multiple ways with the dimes, nickles, and pennies, but I cannot get the output right. Whenever I enter '99', I get 3 for quarters, 2 for dimes, 1 for nickles, and 0 for pennies. If anyone would be able to help me, that would be wonderful!










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  • There's an interesting wikipedia page on this algorithm: en.wikipedia.org/wiki/Change-making_problem.

    – NightShadeQueen
    Aug 31 '15 at 18:30






  • 1





    The tricky part is to know that you by taking as many quarters as possible without exceeding the sum, then as many dimes and then nickels and last pennies will actually result in the minimal amount of coins. And of course you should present the minimal amount or you could just anser to use c pennies.

    – skyking
    Aug 31 '15 at 18:32















1















I am in a beginner programming course. We must do an exercise where we make a change maker program. The input has to be between 0-99 and must be represented in quarters, dimes, nickles, and pennies when the input is divided down between the four. I wrote a code that involved loops and whiles, but he wants something more easy and a smaller code. He gave me this as a way of helping me along:



c=int(input('Please enter an amount between 0-99:'))
print(c//25)
print(c%25)


He told us that this was basically all we needed and just needed to add in the dimes, nickles, and pennies. I try it multiple ways with the dimes, nickles, and pennies, but I cannot get the output right. Whenever I enter '99', I get 3 for quarters, 2 for dimes, 1 for nickles, and 0 for pennies. If anyone would be able to help me, that would be wonderful!










share|improve this question
























  • There's an interesting wikipedia page on this algorithm: en.wikipedia.org/wiki/Change-making_problem.

    – NightShadeQueen
    Aug 31 '15 at 18:30






  • 1





    The tricky part is to know that you by taking as many quarters as possible without exceeding the sum, then as many dimes and then nickels and last pennies will actually result in the minimal amount of coins. And of course you should present the minimal amount or you could just anser to use c pennies.

    – skyking
    Aug 31 '15 at 18:32













1












1








1


3






I am in a beginner programming course. We must do an exercise where we make a change maker program. The input has to be between 0-99 and must be represented in quarters, dimes, nickles, and pennies when the input is divided down between the four. I wrote a code that involved loops and whiles, but he wants something more easy and a smaller code. He gave me this as a way of helping me along:



c=int(input('Please enter an amount between 0-99:'))
print(c//25)
print(c%25)


He told us that this was basically all we needed and just needed to add in the dimes, nickles, and pennies. I try it multiple ways with the dimes, nickles, and pennies, but I cannot get the output right. Whenever I enter '99', I get 3 for quarters, 2 for dimes, 1 for nickles, and 0 for pennies. If anyone would be able to help me, that would be wonderful!










share|improve this question
















I am in a beginner programming course. We must do an exercise where we make a change maker program. The input has to be between 0-99 and must be represented in quarters, dimes, nickles, and pennies when the input is divided down between the four. I wrote a code that involved loops and whiles, but he wants something more easy and a smaller code. He gave me this as a way of helping me along:



c=int(input('Please enter an amount between 0-99:'))
print(c//25)
print(c%25)


He told us that this was basically all we needed and just needed to add in the dimes, nickles, and pennies. I try it multiple ways with the dimes, nickles, and pennies, but I cannot get the output right. Whenever I enter '99', I get 3 for quarters, 2 for dimes, 1 for nickles, and 0 for pennies. If anyone would be able to help me, that would be wonderful!







python






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edited Aug 31 '15 at 18:10









Eric Renouf

11.1k32349




11.1k32349










asked Aug 31 '15 at 18:06









bulsona15bulsona15

813




813












  • There's an interesting wikipedia page on this algorithm: en.wikipedia.org/wiki/Change-making_problem.

    – NightShadeQueen
    Aug 31 '15 at 18:30






  • 1





    The tricky part is to know that you by taking as many quarters as possible without exceeding the sum, then as many dimes and then nickels and last pennies will actually result in the minimal amount of coins. And of course you should present the minimal amount or you could just anser to use c pennies.

    – skyking
    Aug 31 '15 at 18:32

















  • There's an interesting wikipedia page on this algorithm: en.wikipedia.org/wiki/Change-making_problem.

    – NightShadeQueen
    Aug 31 '15 at 18:30






  • 1





    The tricky part is to know that you by taking as many quarters as possible without exceeding the sum, then as many dimes and then nickels and last pennies will actually result in the minimal amount of coins. And of course you should present the minimal amount or you could just anser to use c pennies.

    – skyking
    Aug 31 '15 at 18:32
















There's an interesting wikipedia page on this algorithm: en.wikipedia.org/wiki/Change-making_problem.

– NightShadeQueen
Aug 31 '15 at 18:30





There's an interesting wikipedia page on this algorithm: en.wikipedia.org/wiki/Change-making_problem.

– NightShadeQueen
Aug 31 '15 at 18:30




1




1





The tricky part is to know that you by taking as many quarters as possible without exceeding the sum, then as many dimes and then nickels and last pennies will actually result in the minimal amount of coins. And of course you should present the minimal amount or you could just anser to use c pennies.

– skyking
Aug 31 '15 at 18:32





The tricky part is to know that you by taking as many quarters as possible without exceeding the sum, then as many dimes and then nickels and last pennies will actually result in the minimal amount of coins. And of course you should present the minimal amount or you could just anser to use c pennies.

– skyking
Aug 31 '15 at 18:32












3 Answers
3






active

oldest

votes


















3














I'm now sure about what you want to achieve. Using the modulo operator you could easily find out how many quarters, dimes, nickles and pennies.



Let's just say you input 99.



c=int(input('Please enter an amount between 0-99:'))
print(c//25, "quarters")
c = c%25
print(c//10, "dimes")
c = c%10
print(c//5, "nickles")
c = c%5
print(c//1, "pennies")


this would print out:



3 quarters
2 dimes
0 nickles
4 pennies





share|improve this answer























  • Thank you so much for your response! I appreciate the help! I can see where I am going wrong and I now know how to fix it!

    – bulsona15
    Aug 31 '15 at 18:32











  • no problem, just make sur you understand the % and // operators. that's the whole point of this exercice.

    – Saimu
    Aug 31 '15 at 18:34






  • 2





    There's also divmod, which does both of those operations at the same time.

    – NightShadeQueen
    Aug 31 '15 at 18:35


















2














The actual trick is knowing that because each coin is worth at least twice of the next smaller denomination, you can use a greedy algorithm. The rest is just implementation detail.



Here's a slightly DRY'er (but possibly, uh, more confusing) implementation. All I'm really doing differently is using a list to store my results, and taking advantage of tuple unpacking and divmod. Also, this is a little easier to extend in the future: All I need to do to support $1 bills is to change coins to [100, 25, 10, 5, 1]. And so on.



coins = [25,10,5,1] #values of possible coins, in descending order
results = [0]*len(coins) #doing this and not appends to make tuple unpacking work
initial_change = int(input('Change to make: ')) #use raw_input for python2
remaining_change = initial_change
for index, coin in enumerate(coins):
results[index], remaining_change = divmod(remaining_change, coin)
print("In order to make change for %d cents:" % initial_change)
for amount, coin in zip(results, coins):
print(" %d %d cent piece(s)" % (amount, coin))


Gives you:



Change to make: 99
In order to make change for 99 cents:
3 25 cent piece(s)
2 10 cent piece(s)
0 5 cent piece(s)
4 1 cent piece(s)





share|improve this answer
































    1














    n = int(input("Enter a number between 0-99"))
    q = n // 25
    n %= 25
    d = n // 10
    n %= 10
    ni = n // 5
    n %= 5
    c = n % 5
    print(str(q) +" " + str(d) +" " + str(ni) + " " + str(c))


    I hope this helps? Something like this but don't just copy it. Everytime you divide by 25 10 5 you must lose that part because it's already counted.At the end print what ever you want :).






    share|improve this answer























    • Thank you for your help! I am definitely trying to make things harder than they really are! I appreciate the help!

      – bulsona15
      Aug 31 '15 at 18:33











    • No it's ok. If you just started I understand :) .you are welcome ;)

      – Hybr1d
      Aug 31 '15 at 18:47










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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    I'm now sure about what you want to achieve. Using the modulo operator you could easily find out how many quarters, dimes, nickles and pennies.



    Let's just say you input 99.



    c=int(input('Please enter an amount between 0-99:'))
    print(c//25, "quarters")
    c = c%25
    print(c//10, "dimes")
    c = c%10
    print(c//5, "nickles")
    c = c%5
    print(c//1, "pennies")


    this would print out:



    3 quarters
    2 dimes
    0 nickles
    4 pennies





    share|improve this answer























    • Thank you so much for your response! I appreciate the help! I can see where I am going wrong and I now know how to fix it!

      – bulsona15
      Aug 31 '15 at 18:32











    • no problem, just make sur you understand the % and // operators. that's the whole point of this exercice.

      – Saimu
      Aug 31 '15 at 18:34






    • 2





      There's also divmod, which does both of those operations at the same time.

      – NightShadeQueen
      Aug 31 '15 at 18:35















    3














    I'm now sure about what you want to achieve. Using the modulo operator you could easily find out how many quarters, dimes, nickles and pennies.



    Let's just say you input 99.



    c=int(input('Please enter an amount between 0-99:'))
    print(c//25, "quarters")
    c = c%25
    print(c//10, "dimes")
    c = c%10
    print(c//5, "nickles")
    c = c%5
    print(c//1, "pennies")


    this would print out:



    3 quarters
    2 dimes
    0 nickles
    4 pennies





    share|improve this answer























    • Thank you so much for your response! I appreciate the help! I can see where I am going wrong and I now know how to fix it!

      – bulsona15
      Aug 31 '15 at 18:32











    • no problem, just make sur you understand the % and // operators. that's the whole point of this exercice.

      – Saimu
      Aug 31 '15 at 18:34






    • 2





      There's also divmod, which does both of those operations at the same time.

      – NightShadeQueen
      Aug 31 '15 at 18:35













    3












    3








    3







    I'm now sure about what you want to achieve. Using the modulo operator you could easily find out how many quarters, dimes, nickles and pennies.



    Let's just say you input 99.



    c=int(input('Please enter an amount between 0-99:'))
    print(c//25, "quarters")
    c = c%25
    print(c//10, "dimes")
    c = c%10
    print(c//5, "nickles")
    c = c%5
    print(c//1, "pennies")


    this would print out:



    3 quarters
    2 dimes
    0 nickles
    4 pennies





    share|improve this answer













    I'm now sure about what you want to achieve. Using the modulo operator you could easily find out how many quarters, dimes, nickles and pennies.



    Let's just say you input 99.



    c=int(input('Please enter an amount between 0-99:'))
    print(c//25, "quarters")
    c = c%25
    print(c//10, "dimes")
    c = c%10
    print(c//5, "nickles")
    c = c%5
    print(c//1, "pennies")


    this would print out:



    3 quarters
    2 dimes
    0 nickles
    4 pennies






    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Aug 31 '15 at 18:26









    SaimuSaimu

    244139




    244139












    • Thank you so much for your response! I appreciate the help! I can see where I am going wrong and I now know how to fix it!

      – bulsona15
      Aug 31 '15 at 18:32











    • no problem, just make sur you understand the % and // operators. that's the whole point of this exercice.

      – Saimu
      Aug 31 '15 at 18:34






    • 2





      There's also divmod, which does both of those operations at the same time.

      – NightShadeQueen
      Aug 31 '15 at 18:35

















    • Thank you so much for your response! I appreciate the help! I can see where I am going wrong and I now know how to fix it!

      – bulsona15
      Aug 31 '15 at 18:32











    • no problem, just make sur you understand the % and // operators. that's the whole point of this exercice.

      – Saimu
      Aug 31 '15 at 18:34






    • 2





      There's also divmod, which does both of those operations at the same time.

      – NightShadeQueen
      Aug 31 '15 at 18:35
















    Thank you so much for your response! I appreciate the help! I can see where I am going wrong and I now know how to fix it!

    – bulsona15
    Aug 31 '15 at 18:32





    Thank you so much for your response! I appreciate the help! I can see where I am going wrong and I now know how to fix it!

    – bulsona15
    Aug 31 '15 at 18:32













    no problem, just make sur you understand the % and // operators. that's the whole point of this exercice.

    – Saimu
    Aug 31 '15 at 18:34





    no problem, just make sur you understand the % and // operators. that's the whole point of this exercice.

    – Saimu
    Aug 31 '15 at 18:34




    2




    2





    There's also divmod, which does both of those operations at the same time.

    – NightShadeQueen
    Aug 31 '15 at 18:35





    There's also divmod, which does both of those operations at the same time.

    – NightShadeQueen
    Aug 31 '15 at 18:35













    2














    The actual trick is knowing that because each coin is worth at least twice of the next smaller denomination, you can use a greedy algorithm. The rest is just implementation detail.



    Here's a slightly DRY'er (but possibly, uh, more confusing) implementation. All I'm really doing differently is using a list to store my results, and taking advantage of tuple unpacking and divmod. Also, this is a little easier to extend in the future: All I need to do to support $1 bills is to change coins to [100, 25, 10, 5, 1]. And so on.



    coins = [25,10,5,1] #values of possible coins, in descending order
    results = [0]*len(coins) #doing this and not appends to make tuple unpacking work
    initial_change = int(input('Change to make: ')) #use raw_input for python2
    remaining_change = initial_change
    for index, coin in enumerate(coins):
    results[index], remaining_change = divmod(remaining_change, coin)
    print("In order to make change for %d cents:" % initial_change)
    for amount, coin in zip(results, coins):
    print(" %d %d cent piece(s)" % (amount, coin))


    Gives you:



    Change to make: 99
    In order to make change for 99 cents:
    3 25 cent piece(s)
    2 10 cent piece(s)
    0 5 cent piece(s)
    4 1 cent piece(s)





    share|improve this answer





























      2














      The actual trick is knowing that because each coin is worth at least twice of the next smaller denomination, you can use a greedy algorithm. The rest is just implementation detail.



      Here's a slightly DRY'er (but possibly, uh, more confusing) implementation. All I'm really doing differently is using a list to store my results, and taking advantage of tuple unpacking and divmod. Also, this is a little easier to extend in the future: All I need to do to support $1 bills is to change coins to [100, 25, 10, 5, 1]. And so on.



      coins = [25,10,5,1] #values of possible coins, in descending order
      results = [0]*len(coins) #doing this and not appends to make tuple unpacking work
      initial_change = int(input('Change to make: ')) #use raw_input for python2
      remaining_change = initial_change
      for index, coin in enumerate(coins):
      results[index], remaining_change = divmod(remaining_change, coin)
      print("In order to make change for %d cents:" % initial_change)
      for amount, coin in zip(results, coins):
      print(" %d %d cent piece(s)" % (amount, coin))


      Gives you:



      Change to make: 99
      In order to make change for 99 cents:
      3 25 cent piece(s)
      2 10 cent piece(s)
      0 5 cent piece(s)
      4 1 cent piece(s)





      share|improve this answer



























        2












        2








        2







        The actual trick is knowing that because each coin is worth at least twice of the next smaller denomination, you can use a greedy algorithm. The rest is just implementation detail.



        Here's a slightly DRY'er (but possibly, uh, more confusing) implementation. All I'm really doing differently is using a list to store my results, and taking advantage of tuple unpacking and divmod. Also, this is a little easier to extend in the future: All I need to do to support $1 bills is to change coins to [100, 25, 10, 5, 1]. And so on.



        coins = [25,10,5,1] #values of possible coins, in descending order
        results = [0]*len(coins) #doing this and not appends to make tuple unpacking work
        initial_change = int(input('Change to make: ')) #use raw_input for python2
        remaining_change = initial_change
        for index, coin in enumerate(coins):
        results[index], remaining_change = divmod(remaining_change, coin)
        print("In order to make change for %d cents:" % initial_change)
        for amount, coin in zip(results, coins):
        print(" %d %d cent piece(s)" % (amount, coin))


        Gives you:



        Change to make: 99
        In order to make change for 99 cents:
        3 25 cent piece(s)
        2 10 cent piece(s)
        0 5 cent piece(s)
        4 1 cent piece(s)





        share|improve this answer















        The actual trick is knowing that because each coin is worth at least twice of the next smaller denomination, you can use a greedy algorithm. The rest is just implementation detail.



        Here's a slightly DRY'er (but possibly, uh, more confusing) implementation. All I'm really doing differently is using a list to store my results, and taking advantage of tuple unpacking and divmod. Also, this is a little easier to extend in the future: All I need to do to support $1 bills is to change coins to [100, 25, 10, 5, 1]. And so on.



        coins = [25,10,5,1] #values of possible coins, in descending order
        results = [0]*len(coins) #doing this and not appends to make tuple unpacking work
        initial_change = int(input('Change to make: ')) #use raw_input for python2
        remaining_change = initial_change
        for index, coin in enumerate(coins):
        results[index], remaining_change = divmod(remaining_change, coin)
        print("In order to make change for %d cents:" % initial_change)
        for amount, coin in zip(results, coins):
        print(" %d %d cent piece(s)" % (amount, coin))


        Gives you:



        Change to make: 99
        In order to make change for 99 cents:
        3 25 cent piece(s)
        2 10 cent piece(s)
        0 5 cent piece(s)
        4 1 cent piece(s)






        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Aug 31 '15 at 19:21

























        answered Aug 31 '15 at 19:14









        NightShadeQueenNightShadeQueen

        2,44331532




        2,44331532





















            1














            n = int(input("Enter a number between 0-99"))
            q = n // 25
            n %= 25
            d = n // 10
            n %= 10
            ni = n // 5
            n %= 5
            c = n % 5
            print(str(q) +" " + str(d) +" " + str(ni) + " " + str(c))


            I hope this helps? Something like this but don't just copy it. Everytime you divide by 25 10 5 you must lose that part because it's already counted.At the end print what ever you want :).






            share|improve this answer























            • Thank you for your help! I am definitely trying to make things harder than they really are! I appreciate the help!

              – bulsona15
              Aug 31 '15 at 18:33











            • No it's ok. If you just started I understand :) .you are welcome ;)

              – Hybr1d
              Aug 31 '15 at 18:47















            1














            n = int(input("Enter a number between 0-99"))
            q = n // 25
            n %= 25
            d = n // 10
            n %= 10
            ni = n // 5
            n %= 5
            c = n % 5
            print(str(q) +" " + str(d) +" " + str(ni) + " " + str(c))


            I hope this helps? Something like this but don't just copy it. Everytime you divide by 25 10 5 you must lose that part because it's already counted.At the end print what ever you want :).






            share|improve this answer























            • Thank you for your help! I am definitely trying to make things harder than they really are! I appreciate the help!

              – bulsona15
              Aug 31 '15 at 18:33











            • No it's ok. If you just started I understand :) .you are welcome ;)

              – Hybr1d
              Aug 31 '15 at 18:47













            1












            1








            1







            n = int(input("Enter a number between 0-99"))
            q = n // 25
            n %= 25
            d = n // 10
            n %= 10
            ni = n // 5
            n %= 5
            c = n % 5
            print(str(q) +" " + str(d) +" " + str(ni) + " " + str(c))


            I hope this helps? Something like this but don't just copy it. Everytime you divide by 25 10 5 you must lose that part because it's already counted.At the end print what ever you want :).






            share|improve this answer













            n = int(input("Enter a number between 0-99"))
            q = n // 25
            n %= 25
            d = n // 10
            n %= 10
            ni = n // 5
            n %= 5
            c = n % 5
            print(str(q) +" " + str(d) +" " + str(ni) + " " + str(c))


            I hope this helps? Something like this but don't just copy it. Everytime you divide by 25 10 5 you must lose that part because it's already counted.At the end print what ever you want :).







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Aug 31 '15 at 18:26









            Hybr1dHybr1d

            85110




            85110












            • Thank you for your help! I am definitely trying to make things harder than they really are! I appreciate the help!

              – bulsona15
              Aug 31 '15 at 18:33











            • No it's ok. If you just started I understand :) .you are welcome ;)

              – Hybr1d
              Aug 31 '15 at 18:47

















            • Thank you for your help! I am definitely trying to make things harder than they really are! I appreciate the help!

              – bulsona15
              Aug 31 '15 at 18:33











            • No it's ok. If you just started I understand :) .you are welcome ;)

              – Hybr1d
              Aug 31 '15 at 18:47
















            Thank you for your help! I am definitely trying to make things harder than they really are! I appreciate the help!

            – bulsona15
            Aug 31 '15 at 18:33





            Thank you for your help! I am definitely trying to make things harder than they really are! I appreciate the help!

            – bulsona15
            Aug 31 '15 at 18:33













            No it's ok. If you just started I understand :) .you are welcome ;)

            – Hybr1d
            Aug 31 '15 at 18:47





            No it's ok. If you just started I understand :) .you are welcome ;)

            – Hybr1d
            Aug 31 '15 at 18:47

















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