题目链接

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;
}