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Problem solution · C++

Stern Brocot Tree

Stern Brocot Tree: a C++ solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to NyaanNyaan Competitive Programming Library.

Technique
Binary search
Source
NyaanNyaan Competitive Programming Library
Length
62 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Binary search

For Stern Brocot Tree, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 62 lines of C++ from the credited upstream file yosupo-stern-brocot-tree.test.cpp.
  • The implementation keeps its working state in language-native values and containers.
  • 1 loop block detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from NyaanNyaan Competitive Programming Library by NyaanNyaan and is used under the CC0-1.0 licence.

Open problem ↗Original source ↗
Full codeStern Brocot Tree · C++C++
Use this to learn the idea, then write your own version.
#define PROBLEM "https://judge.yosupo.jp/problem/stern_brocot_tree"//#include "../../template/template.hpp"//#include "../../math/stern-brocot-tree.hpp" using namespace Nyaan; using SBT = SternBrocotTreeNode<ll>; void q() {  ins(cmd);  if (cmd == "DECODE_PATH") {    ini(n);    SBT f;    rep(_, n) {      char c;      in(c);      inl(x);      if (c == 'R') f.go_right(x);      if (c == 'L') f.go_left(x);    }    out(f.x, f.y);  } else if (cmd == "ENCODE_PATH") {    inl(x, y);    SBT f{x, y};    cout << f.seq.size() << " ";    each(s, f.seq) {      cout << (s > 0 ? 'R' : 'L') << " ";      cout << abs(s) << " ";    }    cout << "\n";  } else if (cmd == "LCA") {    inl(x1, y1, x2, y2);    SBT f{x1, y1}, g{x2, y2};    SBT h = SBT::lca(f, g);    out(h.x, h.y);  } else if (cmd == "ANCESTOR") {    inl(k, x, y);    SBT f{x, y};     ll l = f.depth() - k;    if (l < 0) {      out(-1);    } else {      bool b = f.go_parent(l);      assert(b == true);      out(f.x, f.y);    }  } else {    inl(x, y);    SBT f{x, y};    out(f.lower_bound(), f.upper_bound());  }} void Nyaan::solve() {  int t = 1;  in(t);  while (t--) q();}

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