Submission #1212350

#	Time	Username	Problem	Language	Result	Execution time	Memory
1212350		Lemser	Boxes with souvenirs (IOI15_boxes)	C++20	100 / 100	375 ms	199864 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;
    vll dpa(A.size()), dpb(B.size());
    fo (i, A.size()) {
        if (i <= k-1) dpa[i] = A[i]*2ll;
        else dpa[i] = dpa[i-k] + A[i]*2ll;
    }
    fo (i, B.size()) {
        if (i <= k-1) dpb[i] = (l - B[i]) * 2ll;
        else dpb[i] = dpb[i-k] + (l - B[i]) * 2ll;
    }
    ans = (A.size() > 0ll ? dpa[A.size()-1]: 0ll) + (B.size() > 0ll ? dpb[B.size()-1] : 0ll);
    for (ll i = 0; i <= A.size(); i++) {
        ll ri = i-1;
        if (A.size()-ri-1 >= k) continue;
        ll j = A.size()+B.size() - ri - 1 - k - 1;
        // cout << ri << ' ' << j << ' ' << l << '\n';
        ans = min(ans, (ri >= 0 ? dpa[ri] : 0ll) + l + (j >= 0 ? dpb[j] : 0ll));
    }
    return ans;
}

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