이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "boxes.h"
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
int N,K,L;
int clst(int x){
if(x>L/2) x=L-x;
return (x)*2;
}
long long delivery(signed _n, signed _k, signed _l, signed p[]) {
N=_n;
K=_k;
L=_l;
vector<int> a(N);
vector<int> marked(N);
for (int i = 0; i < N; i++) a[i]=p[i];
sort(a.begin(),a.end());
int sumFULL=1e9;
int sumNOTFULL=0;
int lmt=N;
vector<pair<int,int>> res;
vector<int> res2;
vector<int> res3;
for (int i = 0; i < N; i++) { res.push_back({abs(L-2*a[i]),i}); }
sort(all(res));
for (int i = 0; i < sz(res); i++)
{
if(i<K) res3.push_back(res[i].second);
else {
res3.push_back(res[i].second);
res2.push_back(res[i].second);
}
}
sort(all(res2));
sort(all(res3));
vector<int> prf1(sz(res2));
lmt=sz(a);
for (int i = 0; i < sz(a); i++) { if(a[i]>L/2) { lmt=i; break; } }
for (int i = 0; i < lmt; i++) {
if(i<K) prf1[i]=a[i]*2;
else prf1[i]=a[i]*2+prf1[i-K];
}
for (int i = lmt; i < sz(res2); i--) {
if(i<lmt+K) prf1[i]=(L-a[i])*2;
else prf1[i]=(L-a[i])*2+prf1[i-K];
}
for (int i = max(0LL,lmt-K); i <= min(sz(a)-1,K+lmt); i++) sumFULL=min(sumFULL, L+prf1[i]+prf1[i+K]);
for (int i = lmt-1; i >= 0; i-=K) sumNOTFULL+=a[i]*2;
for (int i = lmt; i < sz(a); i+=K) sumNOTFULL+=(L-a[i])*2;
return min(sumFULL,sumNOTFULL);
}
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