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I've got Pset 1, 'Greedy' working, I just think I can refactor it better. How can I get rid of ints a, b, c and d?

a = amount /25; amount = amount % 25;

b = amount /10; amount = amount % 10;

c = amount /5; amount = amount % 5;

d = amount /1; amount = amount & 1;

coins = (a+b+c+d);

where amount is the rounded amount inputted.

Thanks

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  • Two side questions. What is the purpose of amount = amount & 1; at the end? The & operator is a bit shift operator. I can't figure out what you were doing here, or is it a typo? Fortunately, it has no effect on the program since amount isn't used after this. Second, what happens when you divide any number by 1? ;-)
    – Cliff B
    Dec 13, 2017 at 23:57

2 Answers 2

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Use multiple while loops to solve it.

e.g. for dimes.....

//Checking for dime usage
while (amount >= 10)
{
    amount = amount - 10;
    coins = coins + 1;
 printf("amount = %d\ncoins = %i\n",amount,coins);
}

or pennies..

//Checking for penny usage
while (amount > 0)
{
    amount = amount - 1;
    coins = coins + 1;
    printf("amount = %d\ncoins = %i\n",amount,coins);
}

If this was helpful, Please vote up and mark as accepted by clicking.

Regards, Nathan

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  • While loops will work, the technique that ExpertShaun used is far more efficient. The improvement, as @curiouskiwi suggested, is to use a single var, coins, to accumulate the coins, eliminating the need for a,b,c and d. Consider this. Say that the amount is $1,000,000. A loop would have to execute 4 million times to get the answer. With ExpertShaun's technique, the program executes exactly 8 calculations, no matter what the starting amount of cash. It could be simplified to 7, but I'll leave that for anyone else to figure out for themselves. ;-) (don't post it here.)
    – Cliff B
    Dec 13, 2017 at 23:45
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You could simply use coins as the variable that holds the coin value and add to it with each equation.

coins = 0;
coins += amount/25; ...
coins += amount/10 ....   etc etc.

Then, at the end, coins will already be the total.

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