Question

I need to break down the amounts on their respective bill/coin.

The output is somewhat like this:

So far, here is my code: (I made the last few codes a comment one ‘cos the
errors come from there)

  { 
  int x,y;

  printf("Enter input: ");
  scanf("%d",&x);

  y=x/1000;
  printf("\n# of $1000 bill: %d",y);
  x = x%1000;

  y=x/500;      
  printf("\n# of 4500 bill: %d",y);    
  x = (x%500);

  y=x/200;
  printf("\n#. of $200 bill: %d",y);    
  x = (x%200);

  y=x/100;
  printf("\n# of $100 bill: %d",y);    
  x = (x%100);

  y=x/50;
  printf("\n# of $50 bill: %d",y);    
  x = (x%50);

  y=x/20;
  printf("\n# of $20 bill: %d",y);    
  x = (x%20);

  y=x/10;
  printf("\n#. of $10 coin: %d",y);    
  x = (x%10);

  y=x/5;
  printf("\n#. of $5 coin: %d",y);    
  x = (x%5);

  y=x/1;
  printf("\n# of $1 coin: %d",y);    
  x = (x%1);


  getch();
  return 0;
  }

I hope you’ll help me out with this. :/

Thanks!

Answer 1

It’s a simple heuristic problem. You already have the right idea. For a start, if you want to use int (and that’s probably a good idea in your case to avoid fp-precision headaches and the need to use a separate fmod), you’ll have to scale the floating-point input and your modulo/divisors by 100. Also to cut down on some redundancy, consider using a function like:

// returns number of units and subtracts unit_size * result
// from val
int units(int* val, int unit_size)
{
    int num = *val / unit_size;
    *val %= unit_size;
    return num;
}

printf("No. of P1000 bill: %d\n",units(&x, 1000 * 100) );
printf("No. of P500 bill: %d\n",units(&x, 500 * 100) );
printf("No. of P200 bill: %d\n",units(&x, 200 * 100) );
etc.

That should cut down on redundant code a little bit. Maybe not worth getting too fancy for it and I suspect this is homework. Complete solution:

// returns number of units and subtracts unit_size * result
// from val
int units(int* val, int unit_size)
{
    int num = *val / unit_size;
    *val %= unit_size;
    return num;
}

int main()
{
    printf("Enter input: ");

    float amount;
    scanf("%f",&amount);
    int x = (int)(amount * 100.0 + 0.5);

    printf("No. of P1000 bill: %d\n", units(&x, 1000 * 100) );
    printf("No. of P500 bill: %d\n", units(&x, 500 * 100) );
    printf("No. of P200 bill: %d\n", units(&x, 200 * 100) );
    printf("No. of P100 bill: %d\n", units(&x, 100 * 100) );
    printf("No. of P50 bill: %d\n", units(&x, 50 * 100) );
    printf("No. of P20 bill: %d\n", units(&x, 20 * 100) );
    printf("No. of P10 coin: %d\n", units(&x, 10 * 100) );
    printf("No. of P5 coin: %d\n", units(&x, 10 * 100) );
    printf("No. of P1 coin: %d\n", units(&x, 1 * 100) );
    printf("No. of 25 cents: %d\n", units(&x, 25) );
    printf("No. of 1 cent: %d\n", units(&x, 1) );

    return 0;
}

[Edit] If you have trouble understanding pointers, then just do it the way you wrote without using the units function but modify it accordingly to read in a float and multiply by 100 as in the example above.

[Edit] Requested:

int main()
{
    printf("Enter input: ");

    float amount;
    scanf("%f",&amount);
    int x = (int)(amount * 100.0 + 0.5); // x stores the user input in cents

    int y = x / 100000; // 1000 dollars is 100,000 cents
    printf("\nNo. of P1000 bill: %d",y);
    x = x % 100000;

    ...

    y=x / 25; // we're working with cents, so 25 = 25 cents
    printf("\nNo. of 25 cents: %d",y);    
    x = (x % 25);

    ...
}