제출 #761058

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
761058		Red_Inside	선물상자 (IOI15_boxes)	C++17	0 / 100	1 ms	468 KiB

boxes

#include "boxes.h"
#include <bits/stdc++.h>

//#pragma GCC optimize("O1,O2,O3,Ofast,unroll-loops")

#define ll long long
#define f first
#define s second
#define pb push_back
#define mp make_pair
#define o cout<<"BUG"<<endl;
#define FOR(i, j, n) for(int j = i; j < n; ++j)
#define forn(i, j, n) for(int j = i; j <= n; ++j)
#define nfor(i, j, n) for(int j = n; j >= i; --j)
#define all(v) v.begin(), v.end()
#define ld long double
#define ull unsigned long long
 
using namespace std;
const int maxn=188500+10,LOG=17,mod=1e9+7;
int block = 200, timer = 0;
const double eps = 1e-9;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
 
#define bt(i) (1 << (i))
//#define int ll
const int inf=2e9;
#define y1 yy
#define pii pair <int, int>

vector <int> fi, se;

ll solve(int K, int L)
{
	int tek;
	tek = 0;
	ll S1 = 0;
	ll S2 = 0;
	ll ans = 0;
	for(auto i : fi) S1+=i;
	for(auto i : se) S2+=i;
	for(int i = (int)fi.size() - 1; i >= 0; --i)
	{
		if(tek < fi[i])
		{
			tek += K;
			ans += 2*i;
		}
		tek -= fi[i];
	}
	for(int i = (int)se.size() - 1; i >= 0; --i)
	{
		if(tek < se[i])
		{
			tek += K;
			ans += 2*i;
		}
		tek -= se[i];
	}
	ll ret = ans;
	ans = 0;
	for(int i = (int)fi.size() - 1; i >= 0; --i)
	{
		if(tek < fi[i])
		{
			fi[i] -= tek;
			S1 -= tek;
			tek = 0;
			if(S1 < K)
			{
				break;
			}
			tek += K;
			ans += 2*i;
		}
		tek -= fi[i];
		fi[i] = 0;
	}
	for(int i = (int)se.size() - 1; i >= 0; --i)
	{
		if(tek < se[i])
		{
			se[i] -= tek;
			S2 -= tek;
			tek = 0;
			if(S2 < K)
			{
				break;
			}
			tek += K;
			ans += 2*i;
		}
		tek -= se[i];
		se[i] = 0;
	}
	ret = min(ret, ans + L);
	return ret;
}

long long delivery(int N, int K, int L, int p[]) 
{
	ll ret = 0;
	N = 0;
	p[0] = 0;
	forn(1, i, L-1)
	{
		ret += p[i] / K * 2*min(i, L-i);
		p[i] %= K;
		N += p[i];
	}
	fi.clear();
	se.clear();
	if(L % 2 == 1)
	{
		int mid = L / 2;
		forn(0, i, mid)
		{
			fi.pb(p[i]);
		}
		se.pb(0);
		nfor(mid+1, i, L-1)
		{
			se.pb(p[i]);
		}
		return solve(K, L) + ret;
	}
	else
	{
		int mid = L / 2;
		forn(0, i, mid-1)
		{
			fi.pb(p[i]);
		}
		se.pb(0);
		nfor(mid + 1, i, L-1)
		{
			se.pb(p[i]);
		}
		ll ans = inf;
		forn(0, i, p[mid])
		{
			fi.pb(i);
			se.pb(p[mid]-i);
			ans = min(ans, solve(K, L));
			fi.pop_back();
			se.pop_back();
		}
		return ans + ret;
	}
}

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