Submission #1212339

#	Time	Username	Problem	Language	Result	Execution time	Memory
1212339		Lemser	Boxes with souvenirs (IOI15_boxes)	C++20	20 / 100	0 ms	328 KiB

boxes

#include "boxes.h"
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx2")
#pragma GCC target("popcnt")
using namespace std;
 
using ll = long long;
using ull = unsigned long long;
using lld = long double;
using ii = pair<int,int>;
using pll = pair<ll, ll>;
 
using vi = vector<int>;
using vll = vector<ll>;
using vii = vector<ii>;
using vpll = vector<pll>;
using vlld = vector<lld>;
 
#define all(x) x.begin(),x.end()
#define lsb(x) x&(-x)
#define gcd(a,b) __gcd(a,b)
#define sz(x) (int)x.size()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define fls cout.flush()
 
#define fore(i, l, r) for (auto i = l; i < r; i++)
#define fo(i, n) fore (i, 0, n)
#define forex(i, r, l) for (auto i = r-1; i >= l; i--)
#define ffo(i, n) forex (i, n, 0)
 
bool cmin(ll &a, ll b) { if (b < a) { a=b; return 1; } return 0; }
bool cmax(ll &a, ll b) { if (b > a) { a=b; return 1; } return 0; }

const ll INF = 1e18;

long long delivery(int N, int K, int L, int p[]) {
    ll n = N, k = K, l = L;
    deque<ll> A, B;
    fo (i, n) {
        if (p[i] <= l/2) A.pb(p[i]);
        else B.pb(p[i]);
    }
    reverse(all(B));
    ll ans = 0;
    while (A.size() >= k) {
        ans += A[k-1] * 2ll;
        ll cnt = k;
        while (cnt--) A.pop_front();
    }
    while (B.size() >= k) {
        ans += (l - B[k-1]) * 2ll;
        ll cnt = k;
        while (cnt--) B.pop_front();
    }
    if (A.size()+B.size() == 0) return ans;
    if (A.size() == 0) return ans + (l - B.back()) * 2ll;
    if (B.size() == 0) return ans + (A.back()) * 2ll;
    ll nans = (A.size() > 0 ? A.back()*2ll : 0ll) + (B.size() > 0 ? (l - B.back())*2ll : 0ll);
    ll otans = min(l + (A.size()+B.size() > k ? (l - B[A.size()+B.size()-k-1]) * 2ll : 0), l + (A.size()+B.size() > k ? (A[A.size()+B.size()-k-1]) * 2ll : 0));;
    otans = min(otans, l*2ll);
    fo (i, A.size()) {
        ll j = A.size()+B.size() - i - 1 - k - 1;
        otans = min(otans, l + A[i] * 2ll + (j >= 0 ? (l - B[j]) * 2ll : 0ll));
    }
    ans += min(nans, otans);
    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