Submission #1023558

#TimeUsernameProblemLanguageResultExecution timeMemory
1023558vjudge1Wish (LMIO19_noras)C++17
93 / 100
83 ms15048 KiB
#include <bits/stdc++.h> using namespace std; const double eps = 1e-13; void solve() { int n; long long r; cin >> n >> r; vector<long long> a(n), b(n), c(n), d(n); for (int i=0; i<n; i++) cin >> a[i] >> b[i] >> c[i] >> d[i]; vector<pair<long long, bool>> evs; for (int i=0; i<n; i++) { // p(t) = (a[i] + t * dx, b[i] + t * dy) // cari t sehingga (a[i] + t * dx)^2 + (b[i] + t * dy)^2 = r^2 // => a[i]^2 + t^2*dx^2 + 2*t*a[i]*dx + b[i]^2 + t^2*dy^2 + 2*t*b[i]*dy = r^2 // => t^2 * (dx^2 + dy^2) + t * (2*a[i]*dx + 2*b[i]*dy) + a[i]^2 + b[i]^2 - r^2 = 0 // => t^2 * A + t * B + C = 0 // pakai rumus akar persamaan kuadrat? long long dx = c[i] - a[i], dy = d[i] - b[i]; long long A = dx * dx + dy * dy, B = 2 * (a[i] * dx + b[i] * dy), C = a[i] * a[i] + b[i] * b[i] - r * r; long long D = B * B - 4 * A * C; if (D >= 0) { long double sd = sqrt(D); long double t1 = (-B + sd) / (2 * A); long double t2 = (-B - sd) / (2 * A); long long llt1 = ceil(min(t1, t2)); long long llt2 = floor(max(t1, t2)); evs.emplace_back(llt1, 0); evs.emplace_back(llt2, 1); } } sort(evs.begin(), evs.end()); int ans = 0, cnt = 0; for (int i=0; i<(int)evs.size(); i++) { auto &[t, rem] = evs[i]; if (t >= 0) ans = max(ans, cnt); if (rem) { cnt--; } else { cnt++; } } cout << ans << '\n'; } int main() { ios_base::sync_with_stdio(0); cin.tie(0); int tcs = 1; // cin >> tcs; while (tcs--) { solve(); } return 0; }
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