제출 #752912

#제출 시각아이디문제언어결과실행 시간메모리
752912I_love_Hoang_Yen서열 (APIO23_sequence)C++17
7 / 100
189 ms31664 KiB
#include "sequence.h" #include <bits/stdc++.h> #define SZ(s) ((int) ((s).size())) using namespace std; vector<int> medians(const vector<int> values) { int n = SZ(values); assert(n > 0); if (n % 2 == 0) { return {values[n/2 - 1], values[n / 2]}; } else { return {values[n/2]}; } } int sub1(const vector<int>& a) { int n = SZ(a); int res = 0; for (int l = 0; l < n; ++l) { for (int r = l; r < n; ++r) { vector<int> b(a.begin() + l, a.begin() + r + 1); unordered_map<int, int> cnt; for (int val : b) cnt[val] += 1; sort(b.begin(), b.end()); auto meds = medians(b); for (int med : meds) { res = max(res, cnt[med]); } } } return res; } #include <ext/pb_ds/assoc_container.hpp> // Common file #include <ext/pb_ds/tree_policy.hpp> // Including tree_order_statistics_node_update using namespace __gnu_pbds; typedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set; using namespace __gnu_pbds; int sub2(const vector<int>& a) { int n = SZ(a); int res = 0; for (int l = 0; l < n; ++l) { unordered_map<int, int> cnt; ordered_set values; for (int r = l; r < n; ++r) { cnt[a[r]]++; values.insert(a[r]); int k = SZ(values); if (k % 2 == 0) { res = max(res, cnt[*values.find_by_order(k / 2 - 1)]); res = max(res, cnt[*values.find_by_order(k / 2)]); } else { res = max(res, cnt[*values.find_by_order(k / 2)]); } } } return res; } bool is_sub_3(const std::vector<int> a) { auto mit = max_element(a.begin(), a.end()); return std::is_sorted(a.begin(), mit) && std::is_sorted(mit, a.end(), std::greater<int>()); } int len(const std::pair<int,int>& p) { return p.second - p.first + 1; } bool can_be_median(int cnt_less, int cnt_equal, int cnt_greater) { return cnt_equal + cnt_less >= cnt_greater; } int sub3(const vector<int>& a) { int n = SZ(a); unordered_map<int, vector<pair<int,int>>> pos; int l = 0; while (l < n) { int r = l; while (r < n && a[l] == a[r]) ++r; pos[a[l]].emplace_back(l, r-1); l = r; } int res = 0; for (const auto& [val, lrs] : pos) { // only 1 segment for (const auto& lr : lrs) res = max(res, len(lr)); // 2 segments if (SZ(lrs) < 2) continue; assert(SZ(lrs) == 2); int cnt_equal = len(lrs[0]) + len(lrs[1]); int cnt_greater = len({lrs[0].second + 1, lrs[1].first - 1}); int cnt_less = len({0, lrs[0].first - 1}) + len({lrs[1].second + 1, n-1}); if (can_be_median(cnt_less, cnt_equal, cnt_greater)) { res = max(res, cnt_equal); } } return res; } int sub4(const vector<int>& a) { int n = SZ(a); vector<vector<int>> cnt(3, vector<int> (n, 0)); for (int val = 0; val < 3; ++val) { for (int i = 0; i < n; ++i) { if (i) cnt[val][i] = cnt[val][i-1]; cnt[val][i] += (val+1) == a[i]; } } for (int l = 0; l < n; ++l) { } } int sequence(int n, std::vector<int> a) { if (is_sub_3(a)) return sub3(a); if (*max_element(a.begin(), a.end()) <= 3) return sub4(a); if (n <= 2000) return sub2(a); return 0; }

컴파일 시 표준 에러 (stderr) 메시지

sequence.cpp: In function 'int sub4(const std::vector<int>&)':
sequence.cpp:118:1: warning: no return statement in function returning non-void [-Wreturn-type]
  118 | }
      | ^
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