Submission #995012

#TimeUsernameProblemLanguageResultExecution timeMemory
995012caterpillowThe Big Prize (IOI17_prize)C++17
99.72 / 100
42 ms1988 KiB
#include "prize.h" #include <bits/stdc++.h> using namespace std; using ll = long long; using pl = pair<ll, ll>; using pi = pair<int, int>; #define vt vector #define f first #define s second #define pb push_back #define all(x) x.begin(), x.end() #define size(x) ((int) (x).size()) #define FOR(i, a, b) for (int i = (a); i < (b); i++) #define ROF(i, a, b) for (int i = (b) - 1; i >= (a); i--) #define F0R(i, b) FOR (i, 0, b) #define endl '\n' const ll INF = 1e18; const int inf = 1e9; template<template<typename> class Container, typename T> ostream& operator<<(ostream& os, Container<T> o) { os << "{"; int g = size(o); for (auto i : o) os << i << ((--g) == 0 ? "" : ", "); os << "}"; return os; } void _print() { cerr << "\n"; } template<typename T, typename ...V> void _print(T t, V... v) { cerr << t; if (sizeof...(v)) cerr << ", "; _print(v...); } #ifdef LOCAL #define cerr(x...) cerr << #x << " = "; _print(x); #else #define cerr(x...) #define cerr if (0) std::cerr #endif /* there are at most O(sqrt n) different types of prizes constraints suggest 2 * sqrt(n) * log2(n) we can differentiate levels of boxes based on sum of query results there are at most sqrt non-complete-trash prizes if we can isolate each "level" of prizes, we can work our way up to the top consider only the worst type of prize there are at most sqrt(n) connected components of prizes better than this we can bsearch for these components to find a l point, move it right while the # of better prizes to its left doesnt change to find a r point, move it right while the # of better prizes to its right doesnt change take these points, and squish the new array such that we only consider these indices now do this recursively unitl we end up with an array of length 1 */ map<int, pi> memo; pi _query(int i) { if (memo.count(i)) return memo[i]; vt<int> res = ask(i); return memo[i] = {res[0], res[1]}; } struct Array { vt<int> actual; void pb(int v) { actual.pb(v); } pi query(int i) { if (i < 0) return {0, 0}; return _query(actual[i]); } int operator[](int i) { return actual[i]; } int sz() { return size(actual); } }; /* to find a l point, move it right while the # of better prizes to its left doesnt change (so query res is equal or less than original) take these points, and squish the new array such that we only consider these indices now */ void solve(Array& a, int tot) { vt<pi> rngs; // inc, exc int n = a.sz(); int l = -1, r; while (l < n) { int lo = -1, hi = n; int bruh = a.query(l).f; if (bruh > tot) { solve(a, bruh); return; } while (hi - lo > 1) { int m = (lo + hi) / 2; auto [x, y] = a.query(m); if (x + y > tot) { solve(a, x + y); return; } if (m <= l || (x + y == tot && x <= bruh)) lo = m; else hi = m; } l = lo + 1; if (l == n) break; // reached end of arr int r = l + 1; // exc while (r < n) { auto [lhs, rhs] = a.query(r); if (lhs + rhs > tot) { solve(a, lhs + rhs); return; } if (lhs + rhs == tot) break; else r++; } rngs.pb({l, r}); l = r; } Array narr; for (auto [l, r] : rngs) { FOR (i, l, r) { narr.pb(a[i]); } } a = narr; } int find_best(int _n) { Array arr; int n = _n; F0R (i, n) { arr.pb(i); } while (arr.sz() > 1) { solve(arr, arr.query(0).f + arr.query(0).s); } auto [l, r] = _query(arr[0]); assert(l + r == 0); return arr[0]; }

Compilation message (stderr)

prize.cpp:44: warning: "cerr" redefined
   44 | #define cerr if (0) std::cerr
      | 
prize.cpp:43: note: this is the location of the previous definition
   43 | #define cerr(x...)
      | 
prize.cpp: In function 'void solve(Array&, int)':
prize.cpp:101:14: warning: unused variable 'r' [-Wunused-variable]
  101 |  int l = -1, r;
      |              ^
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...