제출 #1068519

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1068519		GrindMachine	게임 (IOI13_game)	C++17	63 / 100	846 ms	256000 KiB

game

#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;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) (int)a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x,y) ((x+y-1)/(y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i,n) for(int i = 0; i < n; ++i)
#define rep1(i,n) for(int i = 1; i <= n; ++i)
#define rev(i,s,e) for(int i = s; i >= e; --i)
#define trav(i,a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a,b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a,b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42
#endif

/*

already knew the solution idea (dynamic segtree with a treap/dynamic segtree in each node)

*/

const int MOD = 1e9 + 7;
const int N = 1e5 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

#include "game.h"

struct dynamic_segtree{
    struct node{
        node *l = nullptr, *r = nullptr;
        ll g = 0;
        node(){

        }
        node(ll x){
            g = x;
        }
    };

    node* root;
    int n;

    dynamic_segtree(){

    }

    dynamic_segtree(int n_){
        n = n_;
        root = new node();
    }

    void pupd(node* &u, int lx, int rx, int i, ll v){
        if(!u) u = new node();

        if(rx-lx == 1){
            u->g = v;
            return;
        }

        int mid = (lx+rx) >> 1;

        if(i < mid){
            pupd(u->l,lx,mid,i,v);
        }
        else{
            pupd(u->r,mid,rx,i,v);
        }

        u->g = 0;
        if(u->l) u->g = gcd(u->g,u->l->g);
        if(u->r) u->g = gcd(u->g,u->r->g);
    }

    void pupd(int i, ll v){
        pupd(root,0,n,i,v);
    }

    ll query(node* u, int lx, int rx, int l, int r){
        if(!u) return 0;
        if(lx >= r or rx <= l) return 0;
        if(lx >= l and rx <= r) return u->g;
        int mid = (lx+rx) >> 1;
        return gcd(query(u->l,lx,mid,l,r),query(u->r,mid,rx,l,r));
    }

    ll query(ll l, ll r){
        return query(root,0,n,l,r+1);
    }
};

struct seg2d{
    int n,m;

    struct node{
        node *l = nullptr, *r = nullptr;
        dynamic_segtree st;
        node(int m_){
            st = dynamic_segtree(m_);
        }
    };

    node* root;

    seg2d(){

    }

    seg2d(int n_, int m_){
        n = n_, m = m_;
        root = new node(m);
    }

    void pupd(node* &u, int lx, int rx, int i, int j, ll v){
        if(rx-lx == 1){
            u->st.pupd(j,v);
            return;
        }

        int mid = (lx+rx) >> 1;

        if(!u->l) u->l = new node(m);
        if(!u->r) u->r = new node(m);

        if(i < mid){
            pupd(u->l,lx,mid,i,j,v);
        }
        else{
            pupd(u->r,mid,rx,i,j,v);
        }

        ll g = gcd(u->l->st.query(j,j),u->r->st.query(j,j));
        u->st.pupd(j,g);
    }

    void pupd(int i, int j, ll v){
        pupd(root,0,n,i,j,v);
    }

    ll query(node* u, int lx, int rx, int l, int r, int x, int y){
        if(!u) return 0;
        if(lx >= r or rx <= l) return 0;
        if(lx >= l and rx <= r) return u->st.query(x,y);
        int mid = (lx+rx) >> 1;
        return gcd(query(u->l,lx,mid,l,r,x,y),query(u->r,mid,rx,l,r,x,y));
    }

    ll query(int l, int r, int x, int y){
        return query(root,0,n,l,r+1,x,y);
    }
};

seg2d st;

void init(int n, int m) {
    st = seg2d(n,m);
}

void update(int i, int j, long long v) {
    st.pupd(i,j,v);
}

long long calculate(int l1, int l2, int r1, int r2) {
    return st.query(l1,r1,l2,r2);
}

• Subtask 1: 10 / 10
• Subtask 2: 27 / 27
• Subtask 3: 26 / 26
• Subtask 4: 0 / 17
• Subtask 5: 0 / 20