이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/*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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