結果

問題 No.318 学学学学学
ユーザー ojisan_IT
提出日時 2018-04-21 10:44:36
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 906 ms / 2,000 ms
コード長 3,416 bytes
コンパイル時間 1,669 ms
コンパイル使用メモリ 181,828 KB
実行使用メモリ 15,744 KB
最終ジャッジ日時 2024-06-22 18:32:42
合計ジャッジ時間 10,207 ms
ジャッジサーバーID
(参考情報)
judge2 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 26
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
template<class T> using vt = vector<T>;
template<class T> using vvt = vector<vt<T>>;
template<class T> using ttt = tuple<T,T>;
using tii = tuple<int,int>;
using vi = vector<int>;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define pb push_back
#define mt make_tuple
#define ALL(a) (a).begin(),(a).end()
#define FST first
#define SEC second
#define DEB cerr<<"!"<<endl
#define SHOW(a,b) cerr<<(a)<<" "<<(b)<<endl
#define DIV int(1e9+7)
const int INF = (INT_MAX/2);
const ll LLINF = (LLONG_MAX/2);
const double eps = 1e-8;
//const double PI = M_PI;
inline ll pow(ll x,ll n,ll m){ll r=1;while(n>0){if((n&1)==1)r=r*x%m;x=x*x%m;n>>=1;}return r%m;}
inline ll lcm(ll d1, ll d2){return d1 / __gcd(d1, d2) * d2;}
/*
西 1118-1190
--This is old Japanese poem expressing the desire to die under tree on the full moon night.(written by Saigyo)
In my research, there is possibility that the tree may be seg-tree.
*/
/*Coding Space*/
// This Segtree written by qcw_pro(my team mate).
template <class M, class O> // M: monoid, O: operator monoid
struct SegTree {
using MM = function <M(M, M)>;
using MO = function <M(M, O)>;
using OO = function <O(O, O)>;
using OI = function <O(O, int)>;
const int n, h; // n: array size, h: height of tree
const M m_id; // identity element of monoid
const O o_id; // identity element of operator monoid
MM mm; MO mo; OO oo; OI oi;
vector<M> m; // monoid array
vector<O> o; // operator monoid array
SegTree(int n, M m_id, O o_id, MM mm, MO mo, OO oo,
OI oi = [](const O& a, int b) { return a; })
:n(n), h(sizeof(int)*CHAR_BIT - __builtin_clz(n)),
m_id(m_id), o_id(o_id), mm(mm), mo(mo), oo(oo), oi(oi),
m(n * 2, m_id), o(n * 2, o_id) {}
inline void apply(int i, const O& v, int k) {
m[i] = mo(m[i], oi(v, k));
if(i < n) o[i] = oo(o[i], v);
}
inline void build(int i) {
while(i >>= 1)
if(o[i] == o_id)
m[i] = mm(m[i * 2], m[i * 2 + 1]);
}
inline void push(int i) {
int s{ i >> h ? h : h - 1 };
for(int k{ 1 << (s - 1) }; s > 0; --s, k >>= 1) {
int p{ i >> s };
if(o[p] == o_id) continue;
apply(p * 2, o[p], k);
apply(p * 2 + 1, o[p], k);
o[p] = o_id;
}
}
void modify(int l, int r, const O& v) {
l += n, r += n;
push(l), push(r - 1);
int ll{ l }, rr{ r };
for(int k{ 1 }; l < r; l >>= 1, r >>= 1, k <<= 1) {
if(l & 1) apply(l++, v, k);
if(r & 1) apply(r - 1, v, k);
}
build(ll), build(rr - 1);
}
M query(int l, int r) { // [l,r)
l += n, r += n;
push(l), push(r - 1);
M a{ m_id }, b{ m_id };
for(; l < r; l >>= 1, r >>= 1) {
if(l & 1) a = mm(a, m[l++]);
if(r & 1) b = mm(m[r - 1], b);
}
return mm(a, b);
}
};
//SegTree<input type, operator type> (size, M_I, O_I, MM, MO, OO)
int F (int a, int b){return max(a, b);};
SegTree<int,int> st(100005, 0, 0, F, F,F);
int main(){
int n; cin >> n;
map<int,vector<int>> in;
rep(i,n){
int ii; cin >> ii;
in[ii].pb(i);
}
for(auto&& i:in){
int a; int b;
if(i.SEC.size() == 0) continue;
a = i.SEC[0];
b = *(--(i.SEC.end()));
st.modify(a,b+1,i.FST);
cerr << a <<" "<< b <<" "<< i.FST << endl;
}
rep(i,n)
cout << st.query(i,i+1) << (i == n-1?"\n":" ");
}
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