Submission #422485

#	Time	Username	Problem	Language	Result	Execution time	Memory
422485		MarcoMeijer	Boxes with souvenirs (IOI15_boxes)	C++14	20 / 100	1 ms	332 KiB

boxes

#include "boxes.h"
#include <bits/stdc++.h>
using namespace std;

// macros
typedef long long ll;
typedef long double ld;
typedef pair<int, int> ii;
typedef pair<ll, ll> lll;
typedef tuple<int, int, int> iii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef vector<iii> viii;
typedef vector<ll> vll;
typedef vector<lll> vlll;
#define REP(a,b,c) for(int a=int(b); a<int(c); a++)
#define RE(a,c) REP(a,0,c)
#define RE1(a,c) REP(a,1,c+1)
#define REI(a,b,c) REP(a,b,c+1)
#define REV(a,b,c) for(int a=int(c-1); a>=int(b); a--)
#define FOR(a,b) for(auto& a : b)
#define all(a) a.begin(), a.end()
#define INF 1e9
#define EPS 1e-9
#define pb push_back
#define popb pop_back
#define fi first
#define se second
#define sz size()
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());

// output
template<class T1, class T2> void OUT(const pair<T1,T2>& x);
template<class T> void OUT(const vector<T>& x);
template<class T> void OUT(const T& x) {cout << x;}
template<class H, class... T> void OUT(const H& h, const T&... t) {OUT(h); OUT(t...); }
template<class T1, class T2> void OUT(const pair<T1,T2>& x) {OUT(x.fi,' ',x.se);}
template<class T> void OUT(const vector<T>& x) {RE(i,x.size()) OUT(i==0?"":" ",x[i]);}
template<class... T> void OUTL(const T&... t) {OUT(t..., "\n"); }
template<class H> void OUTLS(const H& h) {OUTL(h); }
template<class H, class... T> void OUTLS(const H& h, const T&... t) {OUT(h,' '); OUTLS(t...); }

ll delivery(int N, int K, int L, int P[]) {
    ll n=N, k=K, l=L;
    
    ll ans = 1e18;
    vll a, b; a.resize(n+1,0), b.resize(n+1,0);
    vll p;
    RE(i,n) p.pb(P[i]);

    RE(j,2) {
        ll res = 0;
        RE(i,n) {
            res += p[i];
            if((i%k) == 0) {
                res += p[i];
            } else {
                res -= p[i-1];
                res += p[i] - p[i-1];
            }

            (j ? b : a)[i+1] = res;
        }
        reverse(all(p));
        RE(i,n) p[i] = L - p[i];
    }

    RE(i,n+1)
        ans = min(ans, a[i] + b[n-i]);

    RE(i,n+1-k)
        ans = min(ans, a[i] + b[n-i-k] + l);

    return ans;
}

• Subtask 1: 10 / 10
• Subtask 2: 10 / 10
• Subtask 3: 0 / 15
• Subtask 4: 0 / 15
• Subtask 5: 0 / 20
• Subtask 6: 0 / 30