Submission #284345

#	Time	Username	Problem	Language	Result	Execution time	Memory
284345		_7_7_	Boxes with souvenirs (IOI15_boxes)	C++14	0 / 100	1 ms	384 KiB

boxes

#include "boxes.h"
#include <bits/stdc++.h>                                           
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
 
using namespace std;
using namespace __gnu_pbds;
 
//#define int long long
//#pragma GCC optimize("Ofast")
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4")
 
 
#define file(s) freopen(s".in","r",stdin); freopen(s".out","w",stdout);
#define fastio ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define all(x) x.begin(), x.end()
#define sz(s) (int)s.size()
#define pb push_back
#define ppb pop_back
#define mp make_pair
#define s second
#define f first
 
 
 
 
typedef pair < long long, long long > pll;    
typedef pair < int, int > pii;  
typedef unsigned long long ull;         
typedef vector < pii > vpii;                                   	
typedef vector < int > vi;
typedef long double ldb;  
typedef long long ll;  
typedef double db;
 
typedef tree < int, null_type, less < int >, rb_tree_tag, tree_order_statistics_node_update > ordered_set;
 
const int inf = 1e9, maxn = 2e5 + 48, mod = 998244353, N = 1e7 + 12;
const int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1}, block = 300;
const pii base = mp(1171, 3307), Mod = mp(1e9 + 7, 1e9 + 9);
const db eps = 1e-12, pi = acos(-1);
const ll INF = 1e18;
                   
ll s1[N], s2[N];
int pos[N], pp[N];

ll delivery(int n, int k, int l, int p[]) {
	sort(p, p + n);
	int cur = 0, cnt = 0, j = 0;

	while (j < n && !p[j])
		++j;

	for (int i = j; i < n; ++i) {
		if (cnt == k) {
			cnt = 0;
			pos[cur++] = i - 1;
		}

		++cnt;
		if (!cur) 
			s1[i] = p[i] + min(p[i], l - p[i]);
		else 
			s1[i] = s1[pos[cur - 1]] + p[i] + min(p[i], l - p[i]);				
	}

	cur = 0, cnt = 0;
	for (int i = n - 1; i >= j; --i) {
	    pp[i] = l - p[i];

		if (cnt == k) {
			cnt = 0;
			pos[cur++] = i + 1;
		}
		
		++cnt;
		if (!cur)
			s2[i] = p[i] + min(p[i], l - p[i]);
		else
			s2[i] = s2[pos[cur - 1]] + p[i] + min(p[i], l - p[i]);	
	}

//	for (int i = 0; i < n; ++i)
//		cerr << s1[i] << ' ' << s2[i] << endl;

	ll ans = s1[n - 1];
	for (int i = 0; i < n - 1; ++i)
		ans = min(ans, s1[i] + s2[i + 1]);

	return ans;
}

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