Divide without divide c++

Question

There is a problem I am supposed to solve that is normally easy, but it has a catch.

There are 2 types of candy. One type weighs m1 kg and is sold for s1 Euro. A second type weighs m2 kg and is sold for s2 Euro. All numbers are integers.

The question is, which type of candy costs more per kg?

The catch is, you can't use divide operation at all, neither / nor %.

For example, if we have the numbers as m1=2, s1=17, m2=3, and s2=14 then the answer needs to be that the first candies are more expensive as 17/2=8.5 and 14/3=4.(3).

As I am a C++ student, I am restricted to use only that which has been taught so far in the class to determine the more expensive candy. The only thing we learned so far was + - / * % and if statement with else. Also == > < && ||.

You could just use `div` which returns the quotient and remainder in a structure. But I assume you'e not meant to do that. — mediocrevegetable1, Oct 18 '21 at 06:38
Assuming both values are positive, try repeated subtraction. — Peter, Oct 18 '21 at 06:40
And the only thing we learned so far was + - / * % and if function with else. And == > < &&, || — Sigmar, Oct 18 '21 at 06:43
No you can't as we haven't learned it yet. I know how to use it but the teacher says you need to do only if we learned it in class. — Sigmar, Oct 18 '21 at 06:49
Does this answer your question? [Convert Fraction to decimal number and sort them](https://stackoverflow.com/questions/62549344/convert-fraction-to-decimal-number-and-sort-them) — phuclv, Oct 18 '21 at 07:03
duplicate: [Comparing fractions](https://stackoverflow.com/q/23177522/995714) — phuclv, Oct 18 '21 at 07:05

jxh · Accepted Answer · 2021-10-18T19:57:12.813

1

Compare X ≡ s1 * m2 with Y ≡ s2 * m1. If X > Y, then s1 / m1 > s2 / m2.

No division is required to do the comparison.

_{^{The caveat to this solution is that s1, s2, m1, and m2 should all have the same sign, and m1 and m2 should be non-zero.}}

Let's assume all the values are positive integers (hence, greater than 0). Consequently m1 * m2 is positive as well. Let z be the number such that:

z + (s1 / m1) = s2 / m2

By multiplying by m1 * m2 on both sides, we get:

z' + (s1 * m2) = s2 * m1 ∵ z' ≡ z × (m1 * m2)

Since z and z' have the same sign, the relational order of s1 / m1 and s1 / m2 is the same as the relational order of s1 * m2 and s2 * m1.

edited Oct 18 '21 at 19:57

answered Oct 18 '21 at 06:49

jxh

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One should also check that `m1 > 0` and `m2 > 0`. – Stephan Lechner Oct 18 '21 at 06:56
@StephanLechner Yes, there are edge cases. Answer not fully specified since it appears to be homework. – jxh Oct 18 '21 at 07:01

Divide without divide c++

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