“The Trip” challenge from Programming Challenges [closed]

Question

Here is a problem from the book Programming Challenges by Skiena and Revilla:

The Trip

A number of students are members of a club that travels annually to exotic locations. Their destinations in the past have included Indianapolis, Phoenix, Nashville, Philadelphia, San Jose, and Atlanta. This spring they are planning a trip to Eindhoven. The group agrees in advance to share expenses equally, but it is not practical to have them share every expense as it occurs. So individuals in the group pay for particular things, like meals, hotels, taxi rides, plane tickets, etc. After the trip, each student's expenses are tallied and money is exchanged so that the net cost to each is the same, to within one cent. In the past, this money exchange has been tedious and time consuming. Your job is to compute, from a list of expenses, the minimum amount of money that must change hands in order to equalize (within a cent) all the students' costs.

The Input

Standard input will contain the information for several trips. The information for each trip consists of a line containing a positive integer, n, the number of students on the trip, followed by n lines of input, each containing the amount, in dollars and cents, spent by a student. There are no more than 1000 students and no student spent more than $10,000.00. A single line containing 0 follows the information for the last trip.

The Output

For each trip, output a line stating the total amount of money, in dollars and cents, that must be exchanged to equalize the students' costs.

Sample Input

3
10.00
20.00
30.00
4
15.00
15.01
3.00
3.01
0

Sample Output

$10.00 
$11.99

Here is my solution which gives correct answer for all inputs I have tried so far, but there must be some case when it does not work because the judge says so. Can anybody help me to understand why?

int main (int argc, char * argv[]) {
    int64_t a;
    float paid[1000];
    std::vector<float> results;
    while (std::cin >> a && a != 0) {

        int sum = 0;
        for (int i = 0; i < a; i++) {
            std::cin >> paid[i];
            paid[i] *= 100;
            sum += paid[i];
        }
        int result = 0;
        int avg = (std::ceil)((float)sum/a);

        int mod = sum % a;
        int c = 0;
        if (mod > 0) {
            for (int i = 0; i < a; i++) {
                if (c >= a - mod)
                    break;
                if (paid[i] < avg) {
                    paid[i]++;
                    c++;
                }
            }
            for (int i = 0; i < a; i++) {
                if (c >= a - mod)
                    break;
                if (paid[i] > avg) {
                    paid[i] += a - mod - c;
                    break;
                }
            }
        }

        for (int i = 0; i < a; i++) {
            if (paid[i] > avg) {
                result += (paid[i] - avg);
            }
        }

        results.push_back((float)result/100);
    }
    std::cout << std::fixed;
    std::cout << std::setprecision(2);
    for (std::vector<float>::iterator it = results.begin(); it != results.end(); ++it) {
        std::cout << "$" << *it << std::endl;
    }
    return 0;
}

Answer 1

Your algorithm seems to be correct though a bit overcomplicated. But the code is not correct. consider:

9999.1
9999.1
9999.0
9999.1

$0.05

And the correct answer is $0.07

This is due to lack of precision in float implementation. It does not mean you should use double, it mean you should check your casts and try to avoid using float at money computations at all.

Also, i suggest to check your inputs.

Answer 2

• Try not to use single-character variables (with the exception of loop counter variables, such as i). It makes it harder for others to read your code.

• This isn't quite ideal:

  int avg = (std::ceil)((float)sum/a);
  

  std::ceil's return type is actually a floating-point type (can be either float or double depending on the overload). For an int value, you just need to have sum / a.

• You're mixing in some C features:

  Casting

  Instead of this C-style cast:

  results.push_back((float)result/100);
  

  use this C++-style cast:

  results.push_back(static_cast<float>(result)/100);
  

  C-style arrays

  You have this C-style array:

  float paid[1000];
  

  but you also use an std::vector. You can just use an std::vector for everything here, unless you may need a different storage container. If you have C++11, you can instead use an std::array in place of static arrays.

• You don't need to keep flushing the buffer with std::endl in the last loop. You can instead output "\n", which will just give you a newline. This is explained in more detail here.