copy pasting the rules from last year’s thread:
Rules: no spoilers.
The other rules are made up aswe go along.
Share code by link to a forge, home page, pastebin (Eric Wastl has one here) or code section in a comment.
I can’t sleep, so here’s 1-1 and 1-2, unfortunately I couldn’t think of any silly solutions this time, so it’s straightforward instead:
spoiler
#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>
#include <set>
#include <iterator>
int main() {
std::multiset<int> l, r;
int a, b;
while (std::cin >> a >> b) {
l.insert(a); r.insert(b);
}
std::vector<int> delta;
std::transform(l.begin(), l.end(), r.begin(), std::back_inserter(delta),
[](int x, int y) { return std::abs(x-y); }
);
std::cout << std::accumulate(delta.begin(), delta.end(), 0) << std::endl;
}
spoiler
#include <iostream>
#include <numeric>
#include <set>
int main() {
std::multiset<int> l, r;
int a, b;
while (std::cin >> a >> b) {
l.insert(a); r.insert(b);
}
std::cout << std::accumulate(l.begin(), l.end(), 0, [&r](int acc, int x) {
return acc + x * r.count(x);
}) << std::endl;
}
2-1: I have quickly run out of hecks to give. This is the sort of problem that gives prolog programmers feelings of smug superiority.
spoiler
#include <string>
#include <iostream>
#include <sstream>
int main() {
int safe = 0;
std::string s;
while (std::getline(std::cin, s)) {
std::istringstream iss(s);
int a, b, c;
if (!(iss >> a >> b)) {
safe++; continue;
}
if (a == b || std::abs(a-b) > 3) continue;
bool increasing = b > a;
while (iss >> c) {
if (b == c || std::abs(b-c) > 3) goto structuredprogrammingisfornerds;
switch (increasing) {
case false:
if (c < b) { b = c; continue; }
goto structuredprogrammingisfornerds;
case true:
if(c > b) { b = c; continue; }
goto structuredprogrammingisfornerds;
}
}
safe++;
structuredprogrammingisfornerds:;
}
std::cout << safe << std::endl;
}
As usual the second part has punished me for my cowboy code, so I’ll have to take a different more annoying tack (maybe tomorrow). Or you know I could just double down on the haphazard approach…
I decided to double down on 2-2, since bad code is one of life’s little pleasures. Where we’re going we won’t need big-oh notation
spoiler
#include <string>
#include <iostream>
#include <sstream>
#include <vector>
#include <iterator>
template <typename It>
bool seemslegit(It begin, It end) {
if (std::distance(begin, end) == 1) {
return true;
}
int a = *begin++;
int b = *begin++;
if (a == b || std::abs(a-b) > 3) return false;;
bool increasing = b > a;
while (begin != end) {
int c = *begin++;
if (b == c || std::abs(b-c) > 3) return false;;
switch (increasing) {
case false:
if (c < b) { b = c; continue; }
return false;
case true:
if(c > b) { b = c; continue; }
return false;
}
}
return true;
}
template <typename It>
void debug(It begin, It end) {
bool legit = seemslegit(begin, end);
while (begin != end) {
std::cout << *begin++ << " ";
}
//std::cout << ": " << std::boolalpha << legit << std::endl;
}
int main() {
int safe = 0;
std::string s;
while (std::getline(std::cin, s)) {
std::istringstream iss(s);
std::vector<int> report((std::istream_iterator<int>(iss)),
std::istream_iterator<int>());
debug(report.begin(), report.end());
if (seemslegit(report.begin(), report.end())) {
safe++;
std::cout << "\n\n";
continue;
}
for (int i = 0; i < report.size(); ++i) {
auto report2 = report;
auto it = report2.erase(report2.begin()+i);
debug(report2.begin(), report2.end());
if (seemslegit(report2.begin(), report2.end())) {
safe++;
break;
}
}
std::cout << "\n\n";
}
std::cout << safe << std::endl;
}
Commentary
Doing this “efficiently” should be possible. since you only need ~2-ish look-back you should be able to score reports in O(n) time. One complication is you might get the direction wrong, need to consider that erasing one of the first two elements could change the direction. But that requires thinking, and shoving all the permutations into a function with ungodly amounts of copying does not.
re: 2-2
yeah that’s what I ended up thinking. Just try the brute force and if it’s too slow, maybe I’ll try be smarter about it.
Ah thanks for reminding me that time progresses and that it is now December.
I have done 1.1 through 2.2 and have nothing interesting to say about them.
It’s that time of the year again. Last year was tough for me, i got laid off in the middle of dec and it kinda killed the vibe. I’ll see how long I keep up this year. My historical backlog is growing but I’ve made peace with it.
Day 3
3-2
I expect much wailing and gnashing of teeth regarding the parsing, which of course is utterly trivial if you know a bit if regex.
I got bit by the input being more than one line. Embarrasing.
I wonder if any input starts with a “don’t()” or if it’s too early for Erik to pull such trickery.
For 3: I made dart one-liners for both. Pasting the juicy parts.
3:1
RegExp(r"mul\((\d*),(\d*)\)").allMatches(input).fold<int>( 0, (p, e) => p + e.groups([1, 2]).fold(1, (p, f) => p * int.parse(f!))));
3:2
My original solution found do, don’t and mul entries, then stepped through them to get the solve. I decided to try regex my way through it. What I realised was that you want to ignore strings starting with don’t() and ending at the first do(). Some amount of trial and error later, I figured out the (ecma*) regex to do it, which I am proud of:
RegExp(r"(?:don\'t\(\)(?:.(?<!do\(\)))*do\(\))|(?:mul\((\d*),(\d*)\))") .allMatches(input) .fold<int>( 0, (p, e) => p + (e.group(0)![0] != 'd' ? e.groups([1, 2]).fold<int>(1, (p, f) => p * int.parse(f!)) : 0))
*ecma balls
