제출 #1129176

#제출 시각아이디문제언어결과실행 시간메모리
1129176jackofall718Snail (NOI18_snail)C++20
37 / 100
1 ms584 KiB
#include <bits/stdc++.h> using namespace std; typedef long long ll; int main() { ios::sync_with_stdio(false); cin.tie(nullptr); ll H; int N; cin >> H >> N; // well height, phases per day vector<ll> P(N); for (int i = 0; i < N; i++) { cin >> P[i]; } // 1) Build prefix array for one day with "clipping at 0" // prefix[i] = height after i-th phase in the day // (starting from height=0 at phase 0) // If negative, clip to 0 vector<ll> prefix(N); { ll curr = 0; for (int i = 0; i < N; i++) { curr += P[i]; if (curr < 0) curr = 0; // cannot go below 0 prefix[i] = curr; } } // Check if snail exits on day 0 (the very first day): for (int i = 0; i < N; i++) { if (prefix[i] >= H) { cout << 0 << " " << i << "\n"; return 0; } } // Net after one full day: ll net = prefix[N - 1]; // If after one full day the snail is at 0 or has not progressed, it will never get out: if (net <= 0) { cout << -1 << " " << -1 << "\n"; return 0; } // 2) If net > 0, we see how many full days we might skip. // We want day D such that D * net is close to H but < H, // then we simulate the partial day D. // To be safe, we do day = (H-1) // net, because if day*net is already >= H, // then the snail would have exited earlier. ll day = (H - 1) / net; // 0-based day index ll heightSoFar = day * net; // snail's height at the start of "day" day. // Now simulate day "day" phase-by-phase: ll curr = heightSoFar; for (int i = 0; i < N; i++) { curr += P[i]; if (curr < 0) curr = 0; if (curr >= H) { cout << day << " " << i << "\n"; return 0; } } // If not out yet, it might mean we need day+1: // Because maybe heightSoFar + net < H, or the snail only gets out on the next day’s partial phases. day++; curr = day * net; // start of day (day+1 in 0-based) for (int i = 0; i < N; i++) { curr += P[i]; if (curr < 0) curr = 0; if (curr >= H) { cout << day << " " << i << "\n"; return 0; } } // If it still hasn't exited, theoretically we could keep going day+2, day+3, ... // But often the above logic is enough if the day stepping is chosen carefully. // For safety, you might loop a few more days, but typically the day calculation should suffice. // If for some reason we never exit: cout << -1 << " " << -1 << "\n"; return 0; }
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