Submission #677909

#	Time	Username	Problem	Language	Result	Execution time	Memory
677909		DwightKSchrute	Sprinkler (JOI22_sprinkler)	C++17	100 / 100	1160 ms	104780 KiB

sprinkler

/*
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
 */
#include<bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>

//using namespace __gnu_pbds;
using namespace std;

typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int>vi;
typedef vector<vector<int>>vvi;
typedef vector<ll>vl;
typedef vector<vl> vvl;
typedef pair<int,int>pi;
typedef pair<ll,ll> pl;
typedef vector<pl> vpl;
typedef vector<ld> vld;
typedef pair<ld,ld> pld;
typedef vector<pi> vpi;

//typedef tree<ll, null_type, less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
template<typename T> ostream& operator<<(ostream& os, vector<T>& a){os<<"[";for(int i=0; i<ll(a.size()); i++){os << a[i] << ((i!=ll(a.size()-1)?" ":""));}os << "]\n"; return os;}

#define all(x) x.begin(),x.end()
#define YES out("YES")
#define NO out("NO")
#define out(x){cout << x << "\n"; return;}
#define outfl(x){cout << x << endl;return;}
#define GLHF ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define print(x){for(auto ait:x) cout << ait << " "; cout << "\n";}
#define pb push_back
#define umap unordered_map


template<typename T>
void read(vector<T>& v){
    int n=v.size();
    for(int i=0; i<n; i++)
        cin >> v[i];
}
template<typename T>
vector<T>UNQ(vector<T>a){
    vector<T>ans;
    for(T t:a)
        if(ans.empty() || t!=ans.back())
            ans.push_back(t);
    return ans;
}



void solve();
int main(){
    GLHF;
    int t=1;
    //cin >> t;
    while(t--)
        solve();
}
ll n,m;
vvi g;
vl a;
vvl lazy;
vi up;
const int D = 40;
void dfs(int x,int p){
    for(int nbr:g[x])
        if(nbr!=p)
            up[nbr]=x,dfs(nbr,x);
}
ll qur(int x){
    ll ans=a[x];
    int d=0;
    for(; up[x]!=-1 && d<=D; d++){
        //I am on distance of i from x
        ans=(ans*lazy[x][d])%m;
        if(d+1<=D)
            ans=(ans*lazy[x][d+1])%m;
        x=up[x];
    }

    while(d<=D)
        ans=(ans*lazy[x][d])%m,d++;
    return ans;
}

void upd(ll x,ll d,ll w){
    for(; d>=0 && x!=-1; d--){
        lazy[x][d] = (lazy[x][d]*w)%m;
        x=up[x];
    }
}
void solve() {
    cin >> n >> m;
    g.resize(n+1);
    for(int i=0; i<n-1; i++){
        int u,v;
        cin >> u >> v;
        g[u].pb(v);
        g[v].pb(u);
    }
    a.resize(n+1);
    for(int i=1; i<=n; i++)
        cin >> a[i];

    up.resize(n+1,-1);
    lazy.resize(n+1,vl(D+1,1));
    dfs(1,-1);

    int q;
    cin >> q;
    while(q--){
        int op;
        cin >> op;
        if(op==2){
            int x;
            cin >> x;
            cout << qur(x) << "\n";
        }
        else{
            ll x,d,w;
            cin >> x >> d >> w;
            upd(x,d,w);
        }
    }
}

• Subtask 1: 3 / 3
• Subtask 2: 9 / 9
• Subtask 3: 29 / 29
• Subtask 4: 12 / 12
• Subtask 5: 30 / 30
• Subtask 6: 17 / 17