제출 #225260

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
225260		bukanYohandi	Solar Storm (NOI20_solarstorm)	C++14	100 / 100	1258 ms	126188 KiB

SolarStorm

#include <bits/stdc++.h>
/*
	Karya anak bangsa yang paling agung
	Author : Yohandi or... bukan..
*/

#define fs first
#define sc second
#define pb push_back
#define eb emplace_back

#define all(a) a.begin(),a.end()
#define lb(a,x) (lower_bound(all(a),x)-a.begin())
#define ub(a,x) (upper_bound(all(a),x)-a.begin())

#define rep(a,x,y) for(int a=(int)x;a<=(int)y;++a)
#define repd(a,x,y,d) for(int a=(int)x;a<=(int)y;a+=d)
#define res(a,x,y) for(int a=(int)x;a>=(int)y;--a)
#define resd(a,x,y,d) for(int a=(int)x;a>=(int)y;a-=d)

// Ordered Set, Ordered Multiset
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define o_set tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>
#define o_multiset tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update>
using namespace __gnu_pbds;
// .order_of_key(x) -> Number of elements less than x
// * .find_by_order(k) -> Kth smallest element (0-based)
// .erase(x) -> Remove all elements equal to x

#pragma GCC optimize("Ofast")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

using namespace std;
using lint=long long;

mt19937 rng(time(NULL));

lint Power(lint A,lint B,lint C){
	if(!B) return 1LL;
	lint tmp=Power(A,B>>1,C);
	return tmp*tmp%C*(B&1?A:1)%C;
}

int T,N,S,Le,Ri,Id;
int Par[1111111][21];
lint K,Ans;
lint V[1111111],D[1111111];
pair<int,int> P[1111111];

int Ancestor(int U,int Diff){
    for(int j=log2(N+1);j>=0;--j){
        if(Diff&(1<<j))
            U=Par[U][j];
    }
    return U;
}

int main(){
//	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	T=1;
//	cin>>T;
	rep(t,1,T){
        cin>>N>>S>>K;
        for(int i=1;i<=N-1;++i){
            cin>>D[i];
            D[i]+=D[i-1];
        }
        for(int i=1;i<=N;++i){
            cin>>V[i];
            V[i]+=V[i-1];
        }
        Le=1; Ri=1;
        for(int i=1;i<=N;++i){
            while(Le<=i&&D[i-1]-D[Le-1]>K) Le++;
            while(Ri<=N&&D[Ri-1]-D[i-1]<=K) Ri++;
            P[Le].fs=Ri;
            P[Le].sc=i;
        }
        for(int i=1;i<=N;++i){
            if(P[i].fs==0)
                P[i]=P[i-1];
        }
        P[N+1]=P[N];
        V[N+1]=V[N];
        memset(Par,-1,sizeof(Par));
        for(int i=1;i<=N+1;++i) Par[i][0]=P[i].fs;
        for(int j=1;j<=log2(N);++j)
            for(int i=1;i<=N+1;++i)
                if(Par[i][j-1]!=-1&&Par[Par[i][j-1]][j-1]!=-1)
                    Par[i][j]=Par[Par[i][j-1]][j-1];
        for(int i=1;i<=N;++i){
            if(Ans<V[Ancestor(i,S)-1]-V[i-1]){
                Ans=V[Ancestor(i,S)-1]-V[i-1];
                Id=i;
            }
        }
        cout<<S<<endl;
        for(int i=1;i<=S;++i){
            if(i!=1) cout<<" ";
            cout<<P[Id].sc;
            Id=P[Id].fs;
        }
        cout<<"\n";
	}
	return 0;
}

• Subtask 1: 10 / 10
• Subtask 2: 7 / 7
• Subtask 3: 11 / 11
• Subtask 4: 8 / 8
• Subtask 5: 18 / 18
• Subtask 6: 16 / 16
• Subtask 7: 30 / 30