Submission #711069

#	Time	Username	Problem	Language	Result	Execution time	Memory
711069		zeta7532	Crossing (JOI21_crossing)	C++17	0 / 100	437 ms	7272 KiB

Main

#include <bits/stdc++.h>
#pragma GCC target ("avx2")
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
using namespace std;
using ll = int64_t;
const ll mod = 1e9+7;
#define fi first
#define se second
#define all(x) x.begin(),x.end()
#define rep(i,x) for(ll i=0;i<x;i++)
#define faster ios::sync_with_stdio(false);cin.tie(nullptr)
 
template <typename T>
struct RMQ{
    const T INF = 0;
    ll N;
    vector<T> dat,lazy;
    RMQ(ll N_,vector<ll> &rui):N(),dat(N_*4,INF),lazy(N_*4,INF){
        ll x=1;
        while(N_>x) x*=2;
        N=x;
    }
    void eval(ll k,ll l,ll r,vector<ll> &rui){
        if(lazy[k]==-1) return;
        if(k<N-1){
            lazy[k*2+1]=lazy[k];
            lazy[k*2+2]=lazy[k];
        }
        dat[k]=lazy[k]*((rui[r-l]-1+mod*((rui[r-l]+1)%2))/2);
        dat[k]%=mod;
        lazy[k]=-1;
    }
    void update(ll a,ll b,T x,ll k,ll l,ll r,vector<ll> &rui){
        eval(k,l,r,rui);
        if(a<=l&&r<=b){
            lazy[k]=x;
            eval(k,l,r,rui);
        }else if(a<r&&l<b){
            update(a,b,x,k*2+1,l,(l+r)/2,rui);
            update(a,b,x,k*2+2,(l+r)/2,r,rui);
            dat[k]=dat[k*2+1]+dat[k*2+2]*rui[(l+r)/2-l];
            dat[k]%=mod;
        }
    }
    void update(ll a,ll b,T x,vector<ll> &rui){update(a,b,x,0,0,N,rui);}
    T query_sub(ll a,ll b,ll k,ll l,ll r,vector<ll> &rui){
        eval(k,l,r,rui);
        if(r<=a||b<=l){
            return INF;
        }else if(a<=l&&r<=b){
            return dat[k];
        }else{
            T vl=query_sub(a,b,k*2+1,l,(l+r)/2,rui);
            T vr=query_sub(a,b,k*2+2,(l+r)/2,r,rui);
            return (vl+vr*rui[(l+r)/2-l])%mod;
        }
    }
    T query(ll a,ll b,vector<ll> &rui){return query_sub(a,b,0,0,N,rui);}
};
 
int main(){
    ll N;
    cin >> N;
    vector<ll> rui(800000);
    rui[0]=1;
    for(ll i=0;i<800000-1;i++) rui[i+1]=(rui[i]*3)%mod;
    vector<vector<ll>> S(N,vector<ll>(3));
    rep(i,3){
        rep(j,N){
            char c;
            cin >> c;
            if(c=='J') S[j][i]=0;
            if(c=='O') S[j][i]=1;
            if(c=='I') S[j][i]=2;
        }
    }
    set<ll> ans;
    rep(i,27){
        if(i==0) continue;
        vector<ll> cnt(N,0);
        ll i0=i%3;
        ll i1=(i%9)/3;
        ll i2=i/9;
        if((i0+i1+i2)%3!=1) continue;
        rep(j,N) cnt[j]+=S[j][0]*i0;
        rep(j,N) cnt[j]+=S[j][1]*i1;
        rep(j,N) cnt[j]+=S[j][2]*i2;
        rep(j,N) cnt[j]%=3;
        ll ANS=0;
        rep(j,N) ANS=(ANS+cnt[j]*rui[j])%mod;
        ans.insert(ANS);
    }
    ll Q;
    cin >> Q;
    RMQ<ll> T(N,rui);
    rep(i,N){
        char c;
        cin >> c;
        if(c=='J') T.update(i,i+1,0,rui);
        if(c=='O') T.update(i,i+1,1,rui);
        if(c=='I') T.update(i,i+1,2,rui);
    }
    rep(i,Q+1){
        ll x=T.query(0,N,rui);
        if(ans.find(x)!=ans.end()){
            cout << "Yes" << '\n';
        }    
        else cout << "No" << '\n';
        if(i==Q) break;
        ll l,r;
        char c;
        cin >> l >> r >> c;
        l--;
        ll C=-1;
        if(c=='J') C=0;
        if(c=='O') C=1;
        if(c=='I') C=2;
        T.update(l,r,C,rui);
    }
}

• Subtask 1: 0 / 3
• Subtask 2: 0 / 23
• Subtask 3: 0 / 23
• Subtask 4: 0 / 51