The challenge

A giant whale has decided your submarine is its next meal, and it’s much faster than you are. There’s nowhere to run!

Suddenly, a swarm of crabs (each in its own tiny submarine - it’s too deep for them otherwise) zooms in to rescue you! They seem to be preparing to blast a hole in the ocean floor; sensors indicate a massive underground cave system just beyond where they’re aiming!

The crab submarines all need to be aligned before they’ll have enough power to blast a large enough hole for your submarine to get through. However, it doesn’t look like they’ll be aligned before the whale catches you! Maybe you can help?

There’s one major catch - crab submarines can only move horizontally.

You quickly make a list of the horizontal position of each crab (your puzzle input). Crab submarines have limited fuel, so you need to find a way to make all of their horizontal positions match while requiring them to spend as little fuel as possible.

[…]

Determine the horizontal position that the crabs can align to using the least fuel possible. How much fuel must they spend to align to that position?

You can read the challenge more in depth, here.

Part 1

Full solution

auto input = stdin.byLineCopy().front.splitter(",").map!(to!long); // input
input.map!(a => input.map!(b => abs(a-b)).sum).minElement.writeln; // calculate costs

Part 2

The crabs don’t seem interested in your proposed solution. Perhaps you misunderstand crab engineering?

As it turns out, crab submarine engines don’t burn fuel at a constant rate. Instead, each change of 1 step in horizontal position costs 1 more unit of fuel than the last: the first step costs 1, the second step costs 2, the third step costs 3, and so on.

[…]

Determine the horizontal position that the crabs can align to using the least fuel possible so they can make you an escape route! How much fuel must they spend to align to that position?

Full solution

auto input = stdin.byLineCopy().front.splitter(",").map!(to!long).array; // input
iota(input.maxElement).map!(a => input                                   // iter 0 to max element
    .map!(b => abs(a-b)*(abs(a-b)+1)/2).sum).minElement.writeln;         // calculate costs