67 lines
1.8 KiB
C++
67 lines
1.8 KiB
C++
//
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// Created by Matthew Gretton-Dann on 06/12/2021.
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//
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#include <iostream>
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#include <regex>
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#include <string>
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#include <vector>
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struct Circle
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{
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explicit Circle(std::string const& s)
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{
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std::regex re{"Disc #\\d has (\\d+) positions; at time=0, it is at position (\\d+)."};
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std::smatch m;
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if (!std::regex_search(s, m, re)) {
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abort();
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}
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circumference_ = std::stoul(m.str(1));
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start_ = std::stoul(m.str(2));
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}
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Circle(unsigned circumference, unsigned start) : circumference_(circumference), start_(start) {}
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auto circumference() const noexcept -> unsigned { return circumference_; }
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auto start() const noexcept -> unsigned { return start_; }
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private:
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unsigned circumference_;
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unsigned start_;
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};
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auto main() -> int
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{
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std::string line;
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std::vector<Circle> circles;
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while (std::getline(std::cin, line)) {
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circles.emplace_back(line);
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}
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circles.emplace_back(11, 0);
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unsigned time{0};
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unsigned adder{1};
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unsigned offset{0};
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/* For each circle we want the value of cirle.start() + offset + time to be 0 (mod number of
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* positions). Where the offset is how many circles we've dropped through (starting at 1), and
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* time is the time released.
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*
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* Once we've worked out the time for the first circle, all future
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* potential releases will be (t1 + N * pos1), for circle 2 it will be (t2 + N * pos1 * pos2).
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* We use adder to keep track of this.
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*
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* We assume all circumferences are co-prime.
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*/
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for (auto const& circle : circles) {
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++offset;
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while ((circle.start() + offset + time) % circle.circumference() != 0) {
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time += adder;
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}
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std::cout << "Circle start " << circle.start() << " Position " << circle.circumference()
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<< " release time " << time << '\n';
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adder *= circle.circumference();
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}
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std::cout << "Release time: " << time << '\n';
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return 0;
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} |