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[? AMILAH: ?] Ready to cash out?

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Let's say you're owed back $0.32.

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The cashier hands you 32
pennies worth $0.01 each.

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Well, perhaps you might rather they
hand you three dimes and two pennies--

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also $0.32.

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Can you think of a way where the cashier
could hand you back even fewer coins?

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We're going to implement that.

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Central to this algorithm is the concept
of using the largest coin possible.

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The four available coins to us
are quarters, worth $0.25, dimes,

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worth $0.10, nickels, worth $0.05,
and pennies, worth $0.01 each.

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OK.

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So let's go back to our
example of being owed $0.32.

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So let's ask ourselves
first, can a quarter be used?

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Yes, because $0.32 is larger than $0.25.

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And so we're now owed $0.07,
and we've used one coin.

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Can we use another quarter?

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The answer is no because
$0.07 is less than $0.25,

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and so we proceed to
the next largest coin--

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a dime.

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Well, $0.07 is less than $0.10,
and so we can't use a dime.

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We go on, then, to a nickel,
and a nickel worth $0.05

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is less than the $0.07
that we're owed back.

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And so we can indeed use one more
coin, and now we're owed back $0.02.

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If we ask ourselves whether we can
use that same coin, the answer is no.

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The value that we're owed
back is less than a nickel.

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So then we go to the
last coin available.

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Can we use a penny?

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And the answer is that
yes, indeed, we'll

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be using two more pennies for
a total of four coins used.

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And so you'll see, if you were to run
this exact same problem with the staff

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solution and eventually yours, then
running the cash problem with $0.32

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would get you an output of four.

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All right.

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So what are our tasks for this problem?

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First, we want to prompt the
user for an amount of change

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that they're owed back.

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Then, we want to implement
an algorithm that

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always uses the largest coin possible.

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We want to keep track of the coins used,
and print that final number of coins.

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So let's talk about
prompting the user first.

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Whenever we prompt the
user, we have to make sure

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that we only accept values that
will make sense for our problem.

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In this case, you'll also find the
get float function in the CS50 library

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quite useful.

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Here's a snippet of code that will
ensure that the user provides you

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with a valid float.

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Here, I have a do-while loop that will
execute the body of my loop at least

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once, prompting the user using the
get float function found in the CS50

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library.

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After that, as long as
that float is invalid,

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then I'll continue to
re-prompt the user.

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If, however, that float
is valid, then I'll

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accept it and move on
to the rest of my code.

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So now we have a dollar
value from the user.

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However, that's stored
as a float, and we

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know that floats have
floating point in precision.

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So in order to avoid that, I
might convert those floats--

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the dollar values-- into integers--

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cents.

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So look up the round function.

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This might come in handy.

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Now, we can advance to some
pseudocode for the cash problem.

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After getting the amount in
dollars that the user is owed

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and converting that to
cents, then we can proceed

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with using the largest coin available.

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While quarters can be used, we can
increase the count of coins used

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and decrease the amount owed
by a quarter every time.

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Once quarters can no longer be
used, we'll move on to dimes,

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and then nickels, and
then pennies, finally

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printing the number of coins used.

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One way to implement this pseudocode
might be to use greater than and less

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than operators, as well as
addition and subtraction,

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but I also want to propose
another way of solving

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this problem using modulo math.

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The modulo operator returns
the remainder of two integers

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after division.

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Some examples-- 50 modulo 5 gives
you 0 because 5 is a factor of 50.

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50 modulo 10 also gives you 0.

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There's no remainder after
dividing those two numbers.

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Any number modded with
itself will also give you 0.

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Now, let's talk about a non-zero case.

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50 modulo 49 gives you one
because 49 goes into 50 once,

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and then the remainder
of that is just 1.

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53 modulo 50, now, gives
you a remainder of 3.

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Let's go back to our example
where the user is owed back $0.32.

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Our first question is to ask
whether or not we can use quarters.

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What I want you to do now is go
through the same example, filling

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in the change owed box and the coins
used box using the modulo and division

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operators.

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All right.

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Now, you have two proposed ways
of implementing this cash problem.

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We've prompted the user
for some amount of change.

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We have used the largest coin
possible, keeping track of those coins,

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so the last step is to
print that final number.

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In order to print the
value of a variable,

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I'll need to use a placeholder for
whatever variable type that is.

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In this case, to print an integer, I use
the placeholder for an integer-- %i--

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and then pass in that variable.

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All right.

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And now you can print
the final number of coins

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used to give the user their change.

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My name is [? Amilah, ?]
and this was cash.

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