Submission #310153

#	Time	Username	Problem	Language	Result	Execution time	Memory
310153		aZvezda	Boxes with souvenirs (IOI15_boxes)	C++14	20 / 100	1 ms	384 KiB

boxes

#include "boxes.h"

#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize ("O3")
//#pragma GCC target ("sse4")
#define endl "\n"
typedef long long ll;
template<class T, class T2> inline ostream &operator <<(ostream &out, const pair<T, T2> &x) { out << x.first << " " << x.second; return out;}
template<class T, class T2> inline istream &operator >>(istream &in, pair<T, T2> &x) { in >> x.first >> x.second; return in;}
template<class T, class T2> inline bool chkmax(T &x, const T2 &y) { return x < y ? x = y, 1 : 0; }
template<class T, class T2> inline bool chkmin(T &x, const T2 &y) { return x > y ? x = y, 1 : 0; }
const ll mod = 1e9 + 7;
#define out(x) "{" << (#x) << ": " << x << "} "

long long delivery(int N, int K, int L, int p[]) {
    ll l = 0, r = N - 1;
    long long ret = 0;
    ll k = K;
    while(true) {
        if(r - l + 1 <= k) {
            ll curr = min(L, min(2 * L - 2 * p[l], 2 * p[r]));
            for(ll i = l; i <= r - 1; i ++) {
                if(p[i] <= L / 2 && p[i + 1] >= (L + 1) / 2) {
                    chkmin(curr, p[i] * 2ll + 2ll * L - 2ll * p[i + 1]);
                }
            }
            ret += curr;
            break;
        }
        if(min(2ll * (L - p[r - k + 1]), (ll)L) < min(2ll * p[l + k - 1], (ll)L)) {
            ret += min(2ll * (L - p[r - k + 1]), (ll)L);
            r -= k;
        } else {
            ret += min(2ll * p[l + k - 1], (ll)L);
            l += k;
        }
    }
    return ret;
}

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