제출 #219566

#제출 시각아이디문제언어결과실행 시간메모리
219566mode149256선물상자 (IOI15_boxes)C++14
0 / 100
5 ms384 KiB
/*input */ #include <bits/stdc++.h> #include "boxes.h" using namespace std; typedef long long ll; typedef long double ld; typedef complex<ld> cd; typedef pair<int, int> pi; typedef pair<ll, ll> pl; typedef pair<ld, ld> pd; typedef vector<int> vi; typedef vector<vi> vii; typedef vector<ld> vd; typedef vector<ll> vl; typedef vector<vl> vll; typedef vector<pi> vpi; typedef vector<vpi> vpii; typedef vector<pl> vpl; typedef vector<cd> vcd; typedef vector<pd> vpd; typedef vector<bool> vb; typedef vector<vb> vbb; typedef std::string str; typedef std::vector<str> vs; #define x first #define y second #define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n" const int MOD = 1000000007; const ll INF = std::numeric_limits<ll>::max(); const int MX = 100101; const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L; template<typename T> pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); } template<typename T> pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); } template<typename T> T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); } template<typename T> T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); } template<typename T> void print(vector<T> vec, string name = "") { cout << name; for (auto u : vec) cout << u << ' '; cout << '\n'; } long long delivery(int N, int K, int L, int p[]) { vl pos(N); for (int i = 0; i < N; ++i) pos[i] = p[i]; vl isKaires(N + 1); vl isDesines(N + 1); isKaires[0] = 0; // isDesines[N] = 0; auto dis = [&](int i) { i--; return pos[i] + min(pos[i], L - pos[i]); }; { for (int i = 1; i <= N; ++i) isKaires[i] = isKaires[max(0, i - K)] + dis(i); for (int i = 0; i < N; ++i) pos[i] = (L - pos[i]) % L; for (int i = 1; i <= N; ++i) isDesines[i] = isDesines[max(0, i - K)] + dis(i); } ll ats = 1e17; for (int i = 0; i <= N; ++i) { ats = min(ats, isKaires[i] + isDesines[N - i]); // if (i) ats = min(ats, isKaires[i - 1] + isDesines[N - i]); } // printf("%lld\n", ats); return ats; }
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