题意:
题解:
AC代码
/* Author : zzugzx Lang : C++ Blog : blog.csdn.net/qq_43756519 */ #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pb push_back #define mp make_pair #define all(x) (x).begin(), (x).end() #define endl '\n' #define SZ(x) (int)x.size() typedef long long ll; typedef unsigned long long ull; typedef pair<int, int> pii; typedef pair<ll, ll> pll; const int mod = 1e9 + 9; const int MOD = 998244353; const double eps = 1e-6; const double PI = acos(-1.0); const int maxn = 1e6 + 10; const int N = 1000; const ll inf = 0x3f3f3f3f; const int dir[][2]={{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}}; int main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); // freopen("in.txt", "r", stdin); // freopen("out.txt", "w", stdout); int n; cin >> n; set<int> s; map<int, int> d; s.insert(inf), s.insert(-inf); d[inf] = d[-inf] = -1; ll ans = 0; while (n--) { int x; cin >> x; auto p = s.lower_bound(x); d[x] = max(d[*prev(p)], d[*p]) + 1; s.insert(x); ans += d[x]; cout << ans << endl; } return 0; }