Title Description

Given an integer sequence consisting of N elements, there are now two operations:

1 add a

At the end of the sequence, an integer a is added to form an integer sequence of length N + 1.

2 mid outputs the median of the current sequence

Median is the number that places a sequence in the middle after sorting it from small to large. (If the sequence length is even, it refers to the smaller of the two numbers in the middle)

Case 1:1 2 13 14 15 16 Median 13

Case 2:1 3 5 7 10 11 17 Median 7

Example 3:1 1 1 1 123 Median 1

Input and output format

Input format:

The initial sequence length of the first behavior is N. The second action is N integers, representing the sequence of integers, separated by spaces. The third action operand M is to perform M operations. Below are M lines, each of which is entered in the format described in the title.

Output format:

Output median for each mid operation

Input and Output Samples

6
1 2 13 14 15 16
5
add 5
add 3
mid
add 20
mid

5
13

Explain

For 30% of the data, 1 < N < 10,000, 0 < M < 1,000

For 100% data, 1 < N < 100,000, 0 < M < 10,000

The absolute value of an integer in a sequence does not exceed 1,000,000,000, and the number in the sequence may repeat.

Time limit of 1 second per test point

This question is not done casually.

Give me a mouthful of what I can think of.

1. $M < 10,000 dollars if vector s force insertion do not know whether A,

2. Use balance tree directly. I write splay. If I don't want to write code, I can use pb_ds to maintain the red-black tree in siz domain.

3. First offline, the weights are discretized, and then the weights are tree-checked.

4. Follow the practice of https://www.luogu.org/problemnew/show/P1801

#include<cstdio>
using namespace std;
const int MAXN = 1e6 + 10;
#define ls(x) ch[x][0]
#define rs(x) ch[x][1]
#define root  ch[0][1]
int fa[MAXN], val[MAXN], rev[MAXN], siz[MAXN], ch[MAXN][2], tot = 0;
bool ident(int x) {
    return ch[fa[x]][0] == x ? 0 : 1;
}
void connect(int x, int _fa, int opt) {
    fa[x] = _fa, ch[fa[x]][opt] = x;
}
void update(int x) {
    siz[x] = siz[ls(x)] + siz[rs(x)] + rev[x];
}
void rotate(int x) {
    int Y = fa[x], R = fa[Y];
    int Yson = ident(x), Rson = ident(Y);
    int B = ch[x][Yson ^ 1];
    connect(B, Y, Yson);
    connect(x, R, Rson);
    connect(Y, x, Yson ^ 1);
    //tag 
    update(Y); update(x);
}
void splay(int x, int to) {
    to = fa[to]; 
    while(fa[x] != to) {
        int y = fa[x];
        if(fa[y] == to) rotate(x); 
        else if(ident(x) == ident(y)) rotate(y), rotate(x);
        else rotate(x), rotate(x);
    }
}
int NewNode(int _fa, int _val) {
    val[++tot] = _val;
    fa[tot] = _fa;
    siz[tot] = rev[tot] = 1;
    return tot;
}
void insert(int x) {
    if(!root) {root = NewNode(0, x); return ;}
    int now = root;
    while(now) {
        siz[now]++;
        if(val[now] == x) {rev[now]++; return ;}
        int nxt = val[now] < x;
        if(!ch[now][nxt]) {ch[now][nxt] = NewNode(now, x); splay(ch[now][nxt], root); return ;}
        now = ch[now][nxt];
    }
}
int ARank(int x) {
    int now = root;
    while(now) {
        //if(siz[now] == x) return val[now];
        int used = siz[now] - siz[rs(now)];
        if(x > siz[ls(now)] && x <= used) return val[now];
        if(used < x) x = x - used, now = ch[now][1];
        else now = ch[now][0];
    }
}
char opt[5];
int main() {
#ifdef WIN32
    freopen("a.in", "r", stdin);
#endif
    int N;
    scanf("%d", &N);
    for(int i = 1; i <= N; i++) {
        int x; scanf("%d", &x);
        insert(x);
    }
    int Q;
    scanf("%d", &Q);
    while(Q--) {
        scanf("%s", opt + 1);
        if(opt[1] == 'a') {
            int x;
            scanf("%d", &x);
            insert(x); N++;    
        }
        else 
            printf("%d\n", ARank(N / 2 + (N & 1)));
    }
    return 0;
}