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>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
#define FOR(i, a, b) for (int i=a; i<(b); i++)
#define F0R(i, a) for (int i=0; i<(a); i++)
#define FORd(i,a,b) for (int i = (b)-1; i >= a; i--)
#define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--)
#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
const int MOD = 1000000007;
const ll INF = 1e18;
const int MX = 1000001;
int N, ad[1000001];
vi h;
void compress() {
map<int,int> m;
F0R(i,N) m[h[i]] = 0;
int co = 0;
for (auto& a: m) a.s = co++;
F0R(i,N) h[i] = m[h[i]];
}
int main() {
ios_base::sync_with_stdio(0); cin.tie(0);
cin >> N;
h.resize(N+2);
FOR(i,1,N+1) cin >> h[i];
compress();
F0R(i,N+1) {
ad[min(h[i],h[i+1])] ++;
ad[max(h[i],h[i+1])] --;
}
int ans = 0;
F0R(i,1000001) {
if (i) ad[i] += ad[i-1];
ans = max(ans,ad[i]);
}
cout << ans/2;
}
/* Look for:
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* special cases (n=1?)
* overflow (ll vs int?)
* array bounds
*/
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