订单编号(stl)

// Problem: 字典序最大的子序列
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/20960/1036
// Memory Limit: 65536 MB
// Time Limit: 2000 ms
// 2022-03-03 20:19:50
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
int n;
set<int> s;
map<int, int> st;
int a[500005];
void solve() {
    cin >> n;
    int maxv = 0;
    int minv = 0x3f3f3f3f;
    for (int i = 1; i<= n; i ++) {
        cin >> a[i];
        minv = min (minv, a[i]);
    }
    for (int i = minv; i <= minv + n - 1; i ++)
        s.insert(i);
    for (int i = 1; i<= n; i ++) {
        if (s.count(a[i])) {
            cout << a[i] << " ";
            s.erase(a[i]);
        }
        else {
            auto it= s.lower_bound(a[i]);
            cout << *it << " ";
            s.erase(it);
        }
    }
}
int main () {
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

用set维护[minv, minv + n - 1]的数。可以利用lower_bound返回第一个大于等于x的数