Calculate possible combinations in C - c

Question

I've started studying C and I'm trying to practice it developing a small application. Please, could you give any tips about what to do here?
I want to buy shoes from three different brands (brandA=50; brandB=100; brandC=150). I need to spend 2000 dollars on it and buy exactly 20 shoes.
How could I write a program to display all possible combinations?
E.g. brandA (10 shoes), brandB (0 shoe), brandC(10 shoes);
brandA(1 shoe), brandB (3 shoes), brandC (11 shoes), etc.
Please, I don't want the full code now but tips about how to do it.
I really appreciate any help. Tks!
I've updated my post to include a code. Does this code make any sense?
int main(void) {
int brandA=50, brandB=100, brandC=150, ba, bb, bc;
for(ba=0;ba<=20;ba++) {
for(bb=0;bb<=20;bb++) {
for(bc=0;bc<=20;bc++) {
if(ba+bb+bc==20 && (ba*brandA)+(bb*brandB)+(bc*brandC)==2000 {
printf("You can buy %d brandA, %d brandB, %d brandC", ba,bb,bc);
}
}
}
}
return 0;
}

Answer 1

For first you have to have an algorithm...
Take all zero shoes brandA=0; brandB=0; brandC=0
Check total quantity= 0+0+0 = 0
If it is not 20 pcs - pass it
If it equal 20 pcs ( for example: brandA=5; brandB=5; brandC=10) - check total price.
If total price equal 2000 - show it, if not - pass it.
Increment brandA value till 20
repeat steps 2-6
Increment brandB value till 20
repeat steps 2-8
Increment brandC value till 20
repeat steps 2-10
Note: you can use 3 included 'for' cycles :)

Answer 2

If you are a beginner I suggest to start with backtracking and recursion.
Even if backtracking is a costly technique it's great for a beginner to see how recursion can provide a simple yet powerful solution to a problem.
Here are some resources for you to start: http://web.cse.ohio-state.edu/~gurari/course/cis680/cis680Ch19.html
And if you are serious about programming you should also read some books about algorithms and data structures since you will rely heavily on these basic fundamentals:
1. The Algorithm Design Manual
2. The Pragmatic Programmer: From Journeyman to Master

Answer 3

Now that you have it working, I feel no guilt in suggesting a potential refinement. Rather than a pure brute force triple loop, as others have mentioned, you can use the relationship between A, B & C to eliminate the third loop. Remember in C, there are generally many ways to approach any given problem and many ways to handle the output. As long as the logic and syntax are correct, then the only difference will be in the efficiency of the algorithms. Here is an example of eliminating the third loop:
#include <stdio.h>
int main (void) {
int ca = 50;
int cb = 100;
int cc = 150;
int budget = 2000;
int pairs = 20;
int a, b, c, cost;
for (a = 0; a < budget/ca; a++)
for (b = 0; b < budget/cb; b++)
{
c = pairs - (a + b);
if ((cost = a * ca + b * cb + c * cc) != budget)
continue;
printf ("\n a (%2d) * %3d = %4d"
"\n b (%2d) * %3d = %4d"
"\n c (%2d) * %3d = %4d\n",
a, ca, a * ca, b, cb, b * cb, c, cc, c * cc);
printf (" ===================\n (%d) %d\n", pairs, budget);
}
return 0;
}
Compiling
Since you are new to C, when you compile your code, make sure you always compile with warnings enabled. The compiler warnings are there for a reason, and there are very, very few circumstances where you can rely on code that doesn't compile without warnings. At minimum, you will want to compile with -Wall -Wextra enabled (gcc). You can also include -pedantic if you want to check your code against virtually all possible warnings. For example, to compile the code above:
$ gcc -Wall -Wextra -pedantic -o bin/shoes shoes.c
If you want to add optimizations to the fullest extent, you can add:
-Ofast (-03 with gcc < 4.6)
Output
$ ./bin/shoes
a ( 1) * 50 = 50
b (18) * 100 = 1800
c ( 1) * 150 = 150
===================
(20) 2000
a ( 2) * 50 = 100
b (16) * 100 = 1600
c ( 2) * 150 = 300
===================
(20) 2000
a ( 3) * 50 = 150
b (14) * 100 = 1400
c ( 3) * 150 = 450
===================
(20) 2000
<..snip..>
a ( 9) * 50 = 450
b ( 2) * 100 = 200
c ( 9) * 150 = 1350
===================
(20) 2000
a (10) * 50 = 500
b ( 0) * 100 = 0
c (10) * 150 = 1500
===================
(20) 2000
Good luck learning C. There is no other comparable language that gives you the precise low-level control that you have in C. (assembler excluded) But that precise control doesn't come for free. There is a learning curve involved and there is a bit more to cover in C before you will lose that "fish out of water" feeling and feel comfortable with the language. The benefit of learning C, with the low-level access it provides, is it will greatly improve your understanding of how programming works. That knowledge is applicable to all other programming languages (no matter how hard the other languages work to hide the details from you). C is time well spent.