이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#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,Log;
int Par[1111111][21];
lint K,Ans,Sum;
lint V[1111111],D[1111111];
pair<int,int> P[1111111];
int Ancestor(int U,int Diff){
for(int j=Log;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;
Log=log2(N+1);
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;
}
P[N+1].fs=N+1;
P[N+1].sc=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<=Log;++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){
Sum=V[Ancestor(i,S)-1]-V[i-1];
if(Ans<Sum){
Ans=Sum;
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;
}
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