CODEWITHSUNDEEP
c
Zetto
Zetto

Reputation: 61

'atm' simulator

I'm trying to solve this problem where I need to simulate a sort of atm cash out. The program will ask 3 types of dollar bills and store it in 3 different bays e.g. bay 1 = 20; bay 2 = 50; bay 3 = 100; After the program asks for the cash out amount and give a maximum of 3 options if it is avaiable.

For example: cash out amount: U$ 220

A) 2 x U$ 100 + 1 x U$ 20
B) 1 x U$ 100 + 2 x U$ 50 + 1 x U$ 20
C) 4 x U$ 50 + 1 x U$ 20

Im struggling in finding a way to solve the B & C options, i don't know if im tunnel visioning in the way that i 'solved' the A option because i feel like there must be a much simpler way of calculating it

a,b & c variables are a bubble sort of the bay values

 if(cash_out >= a){
      left = total % a; //total = cash_out;
      if(left == 0){
          cont_A = total / a;
      } else {
          cont_A = total / a;
          total = left;
          if(total >= b){
              left = total % b;
              cont_B = total / b;
              total = left;
              if(total != 0){
                  cont_C = total / c;
              }
          } else {
              cont_C = total / c;
          }
      }
 } else if(cash_out >= b){
      left = total % b;
      if(left == 0){
          cont_B = total / b;
          left = total % b;
          if(left == 0){
              cont_C = total / c;
          }
      } else {
          cont_B = total / b;           
          total = left;
          if(total >= c){
              cont_C = total / c;
          }
      }
 } else {
       cont_C = total / c;
 }

Just to be clear i'm not asking so someone will solve the whole thing for me, I just want to understand how I should be approaching the problem

The code is what i did for the A option

Upvotes: 0

Views: 194

Answers (1)

4386427
4386427

Reputation: 44339

As your problem involves fairly small numbers, you could start with a brute force method, i.e. simply trying all combinations in the possible range.

Like

#include <stdio.h>

int main(void) {
    int bay[3] = {100, 50, 20};
    int cash_out = 220;

    for (int x = cash_out/bay[0]; x >= 0; --x)
    {
        for (int y = cash_out/bay[1]; y >= 0; --y)
        {
            for (int z = cash_out/bay[2]; z >= 0; --z)
            {
                if (cash_out == (x * bay[0] + y*bay[1] + z*bay[2]))
                {
                    printf("%d %d %d\n", x, y, z);
                }
            }
        }
    }
    return 0;
}

This will print:

2 0 1
1 2 1
1 0 6
0 4 1
0 2 6
0 0 11

Of cause a brute force method like the above is not optimal. So your next steps are to optimize the code.

First step would be to get rid of the inner most for-loop as it is completely unnecessary. I'll leave that as an exercise for you (tip: z can be calculated directly).

Second step would be to limit the range of the second for-loop by taking into account the amount "used" by the first loop. Again, I'll leave that as an exercise for you.

Upvotes: 2

Related Questions