It completely depends the datatype that you have assigned for the variables 'a' , 'b' , and 'c'. Check the compiler that you are using for the size of the datatype in bytes. Add them and thus you will get the answer.
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b+b+b+c+c+c+c =3b+4c
2b + 2c or 2(b + c)
And how does this relate to coins?
A+c= 2a+b
Because there is no way to define the divisors, the equations cannot be evaluated.