이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include<bits/stdc++.h>
using namespace std;
#define REP(i, n) for(int i = 0; i < n; i++)
#define FOR(i, a, b) for(int i = a; i <= b; i++)
#define ST first
#define ND second
ostream& operator<<(ostream &out, string str) {
for(char c : str) out << c;
return out;
}
template<class L, class R> ostream& operator<<(ostream &out, pair<L, R> p) {
return out << "(" << p.ST << ", " << p.ND << ")";
}
template<class T> auto operator<<(ostream &out, T a) -> decltype(a.begin(), out) {
out << "{";
for(auto it = a.begin(); it != a.end(); it = next(it))
out << (it != a.begin() ? ", " : "") << *it;
return out << "}";
}
void dump() { cerr << "\n"; }
template<class T, class... Ts> void dump(T a, Ts... x) {
cerr << a << ", ";
dump(x...);
}
#ifdef DEBUG
# define debug(...) cerr << "[" #__VA_ARGS__ "]: ", dump(__VA_ARGS__)
#else
# define debug(...) false
#endif
template<class T> int size(T && a) { return (int) a.size(); }
using LL = long long;
using PII = pair<int, int>;
#include "boxes.h"
long long delivery(int N, int K, int L, int p[]) {
sort(p, p + N);
vector<LL> left(K), right(K);
REP(i, N)
left[(i + 1) % K] += min(2 * p[i], L);
LL ans = left[N % K];
for(int i = N - 1; i >= 0; i--) {
left[(i + 1) % K] -= min(2 * p[i], L);
right[i % K] += min(2 * (L - p[i]), L);
ans = min(ans, left[i % K] + right[i % K]);
}
return ans;
}
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