이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
/*
Code by @marlov
*/
#include <iostream>
#include <fstream>
#include <string>
#include <sstream>
#include <vector>
#include <string>
#include <cmath>
#include <algorithm>
#include <iomanip>
#include <utility>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <stack>
#include <queue>
#include <iterator>
using namespace std;
typedef long long ll;
typedef pair<int,int> pi;
#define maxN 10000000
int dpfront[maxN],dpback[maxN];
int delivery(int N,int K,int L,int position[]){
int ci=0;
while(position[ci]==0){
ci++;
N--;
}
int pos[N];
for(int i=ci;i<N;i++){
pos[i-ci]=position[i];
}
for(int i=0;i<N;i++){
if(i<K) dpfront[i]=min(L,2*pos[i]);
else dpfront[i]=min(L,2*pos[i])+dpfront[i-K];
}
for(int i=0;i<N;i++){
if(i<K) dpback[i]=min(L,2*(L-pos[N-i-1]));
else dpback[i]=min(L,2*(L-pos[N-i-1]))+dpback[N-i-1+K];
}
int result=min(dpback[N],dpfront[N]);
for(int i=1;i<N-1;i++){
result=max(dpfront[i]+dpback[N-i-1],result);
}
return result;
}
/* stuff you should look for
* int overflow, array bounds
* special cases (n=1,n=0?)
* do smth instead of nothing and stay organized
*/
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