Force sum of weights in vector to be 100 - c++

Question

I need to populate a vector so that it holds a sum of weights, where the sum must be 100. In other words, the number of items is equal to the divisor, and its values are the quotient, to ensure (force) the sum of the vector to equal 100.
Something like this: 100/3=3.333333...
vector[0]=33.33
vector[1]=33.34
vector[2]=33.33
The sum of this needs to be exactly 100 (some sort of selective rounding?)
Another example: 100/6 = 16.66666667
vector[0]=16.67
vector[1]=16.67
vector[2]=16.66
vector[3]=16.67
vector[4]=16.67
vector[5]=16.66
I've seen something like this done in grocery stores where something on sale might be 3 for $11, so the register displays the prices like 3.67, 3.66, and so on.
The values must add up to exactly 100 though I was thinking of doing this with an epsilon but that wouldn't work.
const int divisor = 6;
const int dividend = 10;
std::vector<double> myVec;
myVec.resize(6);
for (int i = 0; i < divisor; ++i)
{
...some magic that I don't know how to do
}
EDIT: The client wants the values stored (and displayed) in values fixed at two decimal places to visually see they add to 100.

Answer 1

Like the comments say, store money in terms of cents.
#include <vector>
#include <iostream>
#include <iomanip>
std::vector<int> function(int divisor, int total) {
std::vector<int> myVec(divisor);
for (int i = 0; i < divisor; ++i) {
myVec[i] = total/divisor; //rounding down
if (i < total%divisor) //for each leftover
myVec[i] += 1; //add one of the leftovers
}
return myVec;
}
void print_dollars(int cents) {
std::cout << (cents/100) << '.';
std::cout << std::setw(2) << std::setfill('0') << (cents%100) << ' ';
}
int main() {
std::vector<int> r = function(6, 10000);
int sum=0;
for(int i=0; i<r.size(); ++i) {
print_dollars(r[i]);
sum += r[i];
}
std::cout << '\n';
print_dollars(sum);
}
//16.67 16.67 16.67 16.67 16.66 16.66
//100.00
When you divide 100 by 6, you get 16, with 4 leftover. This will put those 4 leftover in each of the first four slots of the vector. Proof of compilation: http://ideone.com/jrInai

Answer 2

There's no one "correct" way to do this. A place to start would be to add up the contents of the vector, and find the difference between 100 and that result you obtained. How you'd fold that in to the individual items would inherently be a heuristic. There are a couple of routes you could take:
Add the difference you found divided by the number of elements in the vector to each element in the vector. This has the advantage that it'll affect an individual value by the smallest amount possible in order to achieve your constraint.
You might want to just add the difference to the first or last element in the vector. This has the advantage that the fewest number of elements in the vector are modified.
You might want to list a separate rounding error element in the vector, which will just be the difference. This gives the most "correct" answer, but might not be what your users want.
Only you can decide what kind of heuristic to use based on the application you're building.
It should be noted that using floating point numbers (e.g. float, double, and long double) may result in errors when storing money values -- you should use fixed point decimal arithmetic for such calculations, because that's how money calculations are done in "the real world". Because floating point uses the base 2 number system internally (on most systems), there will be small rounding errors induced in the conversion from decimal to binary and back. You'll likely have no problem with small values, but if the dollar value is large you'll start seeing problems with the number of digits of precision available in a double.

Answer 3

You can divide into whatever is left as you go, subtracting the last value from the remaining amount.
const int divisor = 6;
const int dividend = 10;
std::vector<double> myVec;
myVec.reserve(6);
double remain = 100.0;
for (int i = divisor; i >= 1; --i)
{
double val = remain / (double)i;
remain -= val;
myVec.push_back(val);
}

Answer 4

In your example,
100/6=16.67 (rounded)
Then you just multiply it by 6-1=5 and get 83.35
And now you know that in order to make the sum to be exactly 100, you need to make the price of the last element to be equal to
100 - 83.35 = 16.65