#include<bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = (a); i < (b); i++)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define ST first
#define ND second
#define PB push_back
using ll = long long;
using ld = long double;
using pi = pair<int,int>;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
#ifdef LOCAL
#define DTP(x, y) \
auto operator<<(auto &o, auto a)->decltype(y, o) \
{ \
o << "("; \
x; \
return o << ")"; \
}
DTP(o << a.first << ", " << a.second, a.second);
DTP(for (auto i : a) o << i << ", ", all(a));
void dump(auto... x) { ((cerr << x << ", "), ...) << '\n'; }
#define debug(x...) cerr << setw(4) << __LINE__ << ":[" #x "]: ", dump(x)
#else
#define debug(...) 0
#endif
void Solve()
{
int n;
cin >> n;
vi A(2*n);
rep(i, 0, n)
cin >> A[i];
rep(i, n, 2*n-1)
A[i]=A[i-n];
A[2*n-1]=2000'000;
int k=log2(n)+1;
vvi P(2*n, vi(k+1));
stack<pi> S;
S.push(pi(A[2*n-1], 2*n-1));
P[2*n-1][0]=2*n-1;
for(int i=2*n-2; i>=0; --i){
while(S.top().ST<=A[i])
S.pop();
P[i][0]=S.top().ND;
S.push(pi(A[i], i));
}
rep(i, 1, k+1)
rep(j, 0, 2*n)
P[j][i]=P[P[j][i-1]][i-1];
int res=1;
rep(i, 0, n){
int curr=1;
int ind=i;
for(int j=k; j>=0; --j){
if(P[ind][j]<i+n){
curr+=(1 << j);
ind=P[ind][j];
}
if(curr+(1 << j)-1<res)
break;
}
if(curr>res)
res=curr;
}
cout << res << "\n";
return;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
int t=1;
// cin >> t;
while(t--)
{
Solve();
}
return 0;
}
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 | #include<bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i = (a); i < (b); i++) #define all(x) begin(x), end(x) #define sz(x) (int)(x).size() #define ST first #define ND second #define PB push_back using ll = long long; using ld = long double; using pi = pair<int,int>; using vi = vector<int>; using vvi = vector<vi>; using vll = vector<ll>; #ifdef LOCAL #define DTP(x, y) \ auto operator<<(auto &o, auto a)->decltype(y, o) \ { \ o << "("; \ x; \ return o << ")"; \ } DTP(o << a.first << ", " << a.second, a.second); DTP(for (auto i : a) o << i << ", ", all(a)); void dump(auto... x) { ((cerr << x << ", "), ...) << '\n'; } #define debug(x...) cerr << setw(4) << __LINE__ << ":[" #x "]: ", dump(x) #else #define debug(...) 0 #endif void Solve() { int n; cin >> n; vi A(2*n); rep(i, 0, n) cin >> A[i]; rep(i, n, 2*n-1) A[i]=A[i-n]; A[2*n-1]=2000'000; int k=log2(n)+1; vvi P(2*n, vi(k+1)); stack<pi> S; S.push(pi(A[2*n-1], 2*n-1)); P[2*n-1][0]=2*n-1; for(int i=2*n-2; i>=0; --i){ while(S.top().ST<=A[i]) S.pop(); P[i][0]=S.top().ND; S.push(pi(A[i], i)); } rep(i, 1, k+1) rep(j, 0, 2*n) P[j][i]=P[P[j][i-1]][i-1]; int res=1; rep(i, 0, n){ int curr=1; int ind=i; for(int j=k; j>=0; --j){ if(P[ind][j]<i+n){ curr+=(1 << j); ind=P[ind][j]; } if(curr+(1 << j)-1<res) break; } if(curr>res) res=curr; } cout << res << "\n"; return; } int main() { ios_base::sync_with_stdio(false); cin.tie(0); int t=1; // cin >> t; while(t--) { Solve(); } return 0; } |
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