Submission #1181849

#	Time	Username	Problem	Language	Result	Execution time	Memory
1181849		Zbyszek99	Modern Machine (JOI23_ho_t5)	C++20	37 / 100	487 ms	81904 KiB

Main

#include <bits/stdc++.h>
#define ll long long
#define ld long double
#define ull unsigned long long
#define ff first
#define ss second
#define pii pair<int,int>
#define pll pair<long long, long long>
#define vi vector<int>
#define vl vector<long long>
#define pb push_back
#define rep(i, b) for(int i = 0; i < (b); ++i)
#define rep2(i,a,b) for(int i = a; i <= (b); ++i)
#define rep3(i,a,b,c) for(int i = a; i <= (b); i+=c)
#define count_bits(x) __builtin_popcountll((x))
#define all(x) (x).begin(),(x).end()
#define siz(x) (int)(x).size()
#define forall(it,x) for(auto& it:(x))
using namespace std;
//mt19937 mt;void random_start(){mt.seed(chrono::time_point_cast<chrono::milliseconds>(chrono::high_resolution_clock::now()).time_since_epoch().count());}
//ll los(ll a, ll b) {return a + (mt() % (b-a+1));}
const int INF = 1e9+50;
const ll INF_L = 1e18+40;
const ll MOD = 1e9+7;

int n,m;
string s;

struct seg
{
    int l,r,k;
    bool operator<(const seg& other) const
    {
        if(l != other.l) return l < other.l;
        return r < other.r;
    }
};

struct node
{
    vector<seg> segs;
    node()
    {
        segs.pb({0,n,0});
    }
    void def()
    {
        segs = {{0,n,0}};
    }
    void default_a(int a)
    {
        segs = {};
        if(a*2 < n)
        {
            segs.pb({0,a-1,a+1});
        }
        else
        {
            if(n - (a+1) >= 0) segs.pb({0,n-(a+1),a+1});
            segs.pb({n-(a+1)+1,a-1,0});
        }
        int s = (a+a) % (n+1);
        if(s + (n - a) <= n)
        {
            segs.pb({a,n,s});
        }
        else
        {
            segs.pb({a,a+n-s,s});
            segs.pb({a+n-s+1,n,0});
        }
        sort(all(segs));
    }
    node operator+(const node& other)
    {
        node res;
        res.segs = {};
        forall(it,segs)
        {
            seg w = {it.k,(int)1e9,-1};
            auto iter = lower_bound(other.segs.begin(),other.segs.end(), w);
            iter--;
            int cur = it.k;
            int cur_poz = it.l;
            int r = it.k + (it.r - it.l);
            while(cur <= r)
            {
                if((*iter).r >= r)
                {
                    res.segs.pb({cur_poz,it.r,(*iter).k + cur - (*iter).l});
                    break;
                }
                else
                {
                    int d = (*iter).r - cur;
                    res.segs.pb({cur_poz,cur_poz + d,(*iter).k + cur - (*iter).l});
                    cur = (*iter).r+1;
                    cur_poz += d+1;
                    iter++;
                }
            }
        }
        sort(all(res.segs));
        return res;
    }
    int get_val(int k)
    {
        auto it = lower_bound(segs.begin(),segs.end(),(seg){k,(int)1e9,-1});
        it--;
        return (*it).k + (k - (*it).l);
    }
};

const int tree_siz = 1024*256-1;
node drzewo[tree_siz];

int get_pref(int akt, int p1, int p2, int s1, int s2, int cur)
{
    if(p2 < s1 || p1 > s2) return cur;
    if(p1 >= s1 && p2 <= s2)
    {
        return drzewo[akt-1].get_val(cur);
    }
    cur = get_pref(akt*2,p1,(p1+p2)/2,s1,s2,cur);
    cur = get_pref(akt*2+1,(p1+p2)/2+1,p2,s1,s2,cur);
    return cur;
}

void build_tree(vi& a)
{
    rep(i,m)
    {
        drzewo[tree_siz/2+i].default_a(a[i]);
    }
    rep2(i,tree_siz/2+m,tree_siz-1) drzewo[i].def();
    for(int i = tree_siz/2-1; i >= 0; i--)
    {
        drzewo[i] = drzewo[i*2+1] + drzewo[i*2+2];
    }
}

int main()
{
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    //random_start();
    cin >> n >> m >> s;
    int pref = 0;
    rep(i,n)
    {
        if(s[i] == 'R') pref++;
        else break;
    }
    vi A(m);
    rep(i,m) cin >> A[i];
    build_tree(A);
    int q;
    cin >> q;
    rep(i,q)
    {
        int l,r;
        cin >> l >> r;
        l--;
        r--;
        cout << get_pref(1,0,tree_siz/2,l,r,pref) << "\n";
    }
}

• Subtask 1: 0 / 3
• Subtask 2: 0 / 12
• Subtask 3: 0 / 10
• Subtask 4: 11 / 11
• Subtask 5: 26 / 26
• Subtask 6: 0 / 17
• Subtask 7: 0 / 21