"The Trip" Programming Challenges - c++

Question

Here's the problem.
A group of students are members of a club that travels annually to different 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 share every expense as it occurs. Thus individuals in the group pay for particular things, such as meals, hotels, taxi rides, and plane tickets. 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 one cent) all the students' costs.
Input
Standard input will contain the information for several trips. Each trip consists of a line containing a positive integer n denoting the number of students on the trip. This is followed by n lines of input, each containing the amount spent by a student in dollars and cents. 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.
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
My code works for some test cases, but fails at others. I think it's because of a precision error in the float. However, I can't find the error.
For example,
Input:
4
9999.1
9999.1
9999.0
9999.1
Output:
$0.06
However, the output should be $0.07
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#define MAX 1000
using namespace std;
float money[MAX];
int main(){
int numOfStudents;
int i; // loop counter
double average; // of the costs
double negDiff, posDiff; // |amount-average|
double minDiff;
float total; // all the costs added together
while(scanf("%d", &numOfStudents) == 1){
if(numOfStudents == 0){
break;
}
memset(money, 0, sizeof(money));
total = 0;
for(i = 0; i < numOfStudents; i++){ // scan for the cost of each student - input into array
double m;
scanf("%lf", &m);
money[i] = m;
total += m;
}
average = total/numOfStudents;
negDiff = 0;
posDiff = 0;
for(i = 0; i < numOfStudents; i++){ // find the difference between average and each cost -> add together
if(money[i] > average){
posDiff += (long) ((money[i] - average) * 100.0) / 100.0;
}
else{
negDiff += (long) ((average - money[i]) * 100.0) / 100.0;
}
}
minDiff = 0;
if(posDiff > negDiff){ // find the minimum value for all to equal
minDiff = negDiff;
}
else{
minDiff = posDiff;
}
printf("$%.2lf\n", minDiff);
}
return 0;
}

Answer 1

If you are chopping the positive and negative differences upto 2 decimal places, chop the average also. This gives the result you are expecting (simplified working code)
#include <stdio.h>
int main() {
int n, i;
double average, negDiff, posDiff, minDiff, total;
while (scanf("%d", &n) == 1 && n != 0) {
double money[n];
total = 0.0;
for (i = 0; i < n; i++) {
scanf("%lf", &money[i]);
total += money[i];
}
average = (long) ((total / n) * 100.0) / 100.0;
negDiff = posDiff = 0.0;
for (i = 0; i < n; i++) {
if (money[i] > average)
posDiff += (long) ((money[i] - average) * 100.0) / 100.0;
else
negDiff += (long) ((average - money[i]) * 100.0) / 100.0;
}
minDiff = posDiff > negDiff ? negDiff : posDiff;
printf("$%.2lf\n", minDiff);
}
return 0;
}
Input
3
10.00 20.00 30.00
4
15.00 15.01 3.00 3.01
4
9999.1 9999.1 9999.0 9999.1
0
Output
$10.00
$11.99
$0.07
See http://ideone.com/cxhvlL

Answer 2

Well... you might be right. You really shouldn't be reading the money in as floats.
scanf ("%d.%d", &dollars, &pennies);
int m = dollars * 100 + pennies;
As for solving the problem in general, just make sure you stick to integral division
int average = total / numStudents;
int leftover = total % numStudents;
// numStudents - leftover students need to have paid $average
// leftover students need to have paid $average + 1
int paid = 0;
int recieved = 0;
for (int i = 0; i < numStudents; ++i)
{
// Since we start with the lowest amount owed, money array needs to have been
// sorted such that the student who paid the least comes first.
owed_money = average;
if (i > numStudents - leftover)
owed_money += 1;
if (money[i] < owed_money)
paid += owed_money - money[i];
else if (money[i] > owed_money)
recieved += money[i] - owed_money;
}
assert (paid == recieved);
Maybe there's a better way to do it? In any case, it's a hard problem, but I can promise your solution shouldn't contain any floating point arithmetic.

Answer 3

Calculate the average only down to cents rest needs to be dropped. That means we will be putting x cents in the side pot and since x is going to be less than n, in the end even if you distribute it only few will get and they will be atmost one cent more than others. So they are all with in the cents.
Now with that said. Here is the simple code for that
int main( )
{
int n;
cin >> n;
while( n ) {
vector<double> v;
while( n-- ) {
double x;
cin >> x;
v.push_back( x );
}
double avg = accumulate( begin(v), end(v), 0.0 ) / v.size();
avg = ((int)(avg*100))/100.00;
double exchange = 0;
for ( auto x : v ) { if ( x < avg ){ exchange += avg - x; } }
cout << exchange << endl;
cin >> n;
}
return 0;
}
and for this input
3
10.00
20.00
30.00
4
15.00
15.01
3.00
3.01
4
9999.1
9999.1
9999.0
9999.1
0
output is
10
11.99
0.07