Add 2016 day 15 puzzles

2016/puzzle-15-01.cc

@@ -0,0 +1,64 @@
+//
+// Created by Matthew Gretton-Dann on 06/12/2021.
+//
+
+#include <iostream>
+#include <regex>
+#include <string>
+#include <vector>
+
+struct Circle
+{
+  explicit Circle(std::string const& s)
+  {
+    std::regex re{"Disc #\\d has (\\d+) positions; at time=0, it is at position (\\d+)."};
+    std::smatch m;
+    if (!std::regex_search(s, m, re)) {
+      abort();
+    }
+    circumference_ = std::stoul(m.str(1));
+    start_ = std::stoul(m.str(2));
+  }
+
+  auto circumference() const noexcept -> unsigned { return circumference_; }
+  auto start() const noexcept -> unsigned { return start_; }
+
+private:
+  unsigned circumference_;
+  unsigned start_;
+};
+
+auto main() -> int
+{
+  std::string line;
+  std::vector<Circle> circles;
+  while (std::getline(std::cin, line)) {
+    circles.emplace_back(line);
+  }
+
+  unsigned time{0};
+  unsigned adder{1};
+  unsigned offset{0};
+  /* For each circle we want the value of cirle.start() + offset + time to be 0 (mod number of
+   * positions).  Where the offset is how many circles we've dropped through (starting at 1), and
+   * time is the time released.
+   *
+   * Once we've worked out the time for the first circle, all future
+   * potential releases will be (t1 + N * pos1), for circle 2 it will be (t2 + N * pos1 * pos2).
+   * We use adder to keep track of this.
+   *
+   * We assume all circumferences are co-prime.
+   */
+  for (auto const& circle : circles) {
+    ++offset;
+    while ((circle.start() + offset + time) % circle.circumference() != 0) {
+      time += adder;
+    }
+    std::cout << "Circle start " << circle.start() << " Position " << circle.circumference()
+              << " release time " << time << '\n';
+    adder *= circle.circumference();
+  }
+
+  std::cout << "Release time: " << time << '\n';
+  return 0;
+}

2016/puzzle-15-02.cc

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