Submission #1254708

#TimeUsernameProblemLanguageResultExecution timeMemory
1254708fasterthanyouFestival (IOI25_festival)C++20
5 / 100
56 ms6068 KiB
#include <bits/stdc++.h> using namespace std; vector<int> max_coupons(int A, vector<int> P, vector<int> T) { using ll = long long; const ll BIG = 2000000000LL; // 2e9 threshold described above int N = (int)P.size(); array<vector<pair<int,int>>, 5> byT; // byT[t] holds {P, idx} for (int i = 0; i < N; ++i) { int t = T[i]; if (t < 1 || t > 4) continue; // defensive byT[t].push_back({P[i], i}); } for (int t = 1; t <= 4; ++t) { sort(byT[t].begin(), byT[t].end()); // sort by price ascending } array<int, 5> ptr{}; // next unused per T vector<int> order; order.reserve(N); auto affordable = [&](int t, ll x) -> bool { return ptr[t] < (int)byT[t].size() && (ll)byT[t][ptr[t]].first <= x; }; ll x = (ll)A; // Main greedy loop while (true) { if (x >= BIG) break; ll bestVal = -1; int bestT = -1; int bestIdx = -1; int bestPrice = -1; for (int t = 1; t <= 4; ++t) { if (!affordable(t, x)) continue; int price = byT[t][ptr[t]].first; int idx = byT[t][ptr[t]].second; // candidate next tokens; safe since x < 2e9 here ll val = (ll)t * (x - (ll)price); // tie-break: larger val, then larger T, then smaller price, then smaller idx if (val > bestVal || (val == bestVal && t > bestT) || (val == bestVal && t == bestT && price < bestPrice) || (val == bestVal && t == bestT && price == bestPrice && idx < bestIdx)) { bestVal = val; bestT = t; bestIdx = idx; bestPrice = price; } } if (bestT == -1) { // nothing affordable in any group return order; } // take it order.push_back(bestIdx); ++ptr[bestT]; ll nextX = bestVal; if (nextX >= BIG) { x = BIG; break; } else { x = nextX; } } // If we reached BIG, we can finish deterministically: if (x >= BIG) { // All remaining T>1 coupons are affordable and non-decreasing for (int t = 4; t >= 2; --t) { for (int k = ptr[t]; k < (int)byT[t].size(); ++k) { order.push_back(byT[t][k].second); } ptr[t] = (int)byT[t].size(); } // Then all T=1 by price ascending for (int k = ptr[1]; k < (int)byT[1].size(); ++k) { order.push_back(byT[1][k].second); } ptr[1] = (int)byT[1].size(); return order; } // Otherwise (no affordable coupons), we’re done return order; }
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