제출 #844646

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
844646		amine_aroua	Global Warming (CEOI18_glo)	C++17	62 / 100	48 ms	17780 KiB

glo

#include <bits/stdc++.h>
//#pragma GCC optimize("O3")
//#pragma GCC optimize("unroll-loops")
using namespace std;
#define int long long
#define vi vector<int>
#define vl vector<long long>
#define vii vector<pair<int,int>>
#define vll vector<pair<long long,long long>>
#define pb push_back
#define ll long long
#define ld long double
#define nl '\n'
#define boost ios::sync_with_stdio(false)
#define mp make_pair
#define se second
#define fi first
#define fore(i, y) for(int i = 0; i < y; i++)
#define forr(i,x,y) for(int i = x;i<=y;i++)
#define forn(i,y,x) for(int i = y; i >= x; i--)
#define all(v) v.begin(),v.end()
#define sz(v) (int)v.size()
#define clr(v,k) memset(v,k,sizeof(v))
#define rall(v) v.rbegin() , v.rend()
#define pii pair<int,int>
#define pll pair<ll , ll>

const ll MOD = 1e9 + 7;
const ll INF = 1e18 + 1;

ll gcd(ll a , ll b) {return b ? gcd(b , a % b) : a ;} // greatest common divisor (gcd)
ll lcm(ll a , ll b) {return a * (b / gcd(a , b));} // least common multiple (lcm)

// HERE IS THE SOLUTION
int n  , x;
pair<vi,vi> lis(vi v)
{
    vi dpA;
    vi pref(n);
    vi prefdpA(n);
    fore(i , n)
    {
        int cur = v[i];
        int idx = lower_bound(all(dpA) , cur) - dpA.begin();
        if(idx < sz(dpA))
        {
            dpA[idx] = cur;
        }
        else
        {
            dpA.pb(cur);
        }
        pref[i] = sz(dpA);
        prefdpA[i] = dpA.back();
    }
    return {pref , prefdpA};
}
signed main()
{
    boost;
    cin.tie(0);
    cout.tie(0);
//    freopen("io.txt" , "r" , stdin);
    cin>>n>>x;
    vi v(n);
    fore(i , n)
    {
        cin>>v[i];
    }
    vi pref , prefdpA;
    pair<vi , vi> ret = lis(v);
    pref = ret.fi;
    prefdpA = ret.se;
    vi suff , suffdpB;
    reverse(all(v));
    for(auto &X : v)X*=-1;
    ret = lis(v);
    suff = ret.fi;
    suffdpB = ret.se;
    reverse(all(suff));
    reverse(all(suffdpB));
    for(auto &X : suffdpB)X*=-1;
    int ans = pref[n - 1];
    fore(i , n - 1)
    {
        int idx = lower_bound(prefdpA.begin() , prefdpA.begin() + 1+ i , x + suffdpB[i + 1]) - prefdpA.begin() - 1;
        if(idx <= i && idx >= 0)
        {
            ans = max(ans , pref[idx] + suff[i +  1]);
        }
    }
    cout<<ans<<nl;
}

• Subtask 1: 5 / 5
• Subtask 2: 10 / 10
• Subtask 3: 0 / 13
• Subtask 4: 10 / 10
• Subtask 5: 20 / 20
• Subtask 6: 17 / 17
• Subtask 7: 0 / 25