제출 #1036290

#제출 시각아이디문제언어결과실행 시간메모리
1036290trucmaiA Difficult(y) Choice (BOI21_books)C++17
45 / 100
5 ms8276 KiB
#include "books.h" #include <bits/stdc++.h> #ifdef LOCAL #include "/home/trcmai/code/tools.h" #define debug(x...) \ cerr << "\e[91m" << __func__ << ":" << __LINE__ << " [" << #x << "] = ["; \ _print(x); \ cerr << "\e[39m" << endl; #else #define debug(x...) #endif using namespace std; #define all(a) a.begin(), a.end() #define ll long long #define endl '\n' const int N = 1e6 + 6, LOG = 27, MOD = 1e9 + 7; const ll INF = 1e18; ll a[N]; ll ask(int i) { if (a[i] != -1) return a[i]; else return a[i] = skim(i); } void solve(int N, int K, long long A, int S) { memset(a, -1, sizeof(a)); int l = 1, r = N - K + 1; // Check xem ton tai doan lien tiep nao thoa man khong while (l <= r) { int m = (r + l) >> 1; ll sum = 0; for (int i = m; i <= m + K - 1; ++i) { sum += ask(i); } if (sum >= A && sum <= 2 * A) { vector<int> res; for (int i = m; i <= m + K - 1; ++i) res.emplace_back(i); answer(res); } if (sum < A) l = m + 1; if (sum > 2 * A) r = m - 1; } // Neu khong ton tai, check xem doan [1,k - 1] co ghep duoc voi vi tri ith thoa man khong l = K, r = N - K + 1; int pos = -1; while (l <= r) { int m = (r + l) >> 1; ll val = ask(m); if (val >= A) { pos = m; r = m - 1; } else l = m + 1; } if (pos != -1) { vector<int> res; ll sum = ask(pos); for (int i = 1; i <= K - 1; ++i) { sum += ask(i); } if (sum >= A && sum <= 2 * A) { for (int i = 1; i <= K - 1; ++i) res.emplace_back(i); res.emplace_back(pos); answer(res); } } impossible(); }
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...