Original address

Idea: Change deletion operation to add operation, and use the method of time reversal to deal with it. So imagine the number of reverse pairs that can be increased by inserting a number xxx into position iii for a sequence?

The contribution of the insertion point is calculated in two parts, one is that the larger number before position i can form an inverse pair, the other is that the smaller number after position i can form an inverse pair.

Therefore, the problem can be regarded as a three-dimensional partial ordering problem (adding time, adding location, adding value size).
So we can solve this problem perfectly by sorting one-dimensional time, dividing and conquering cdqcdq, and then dealing with one-dimensional tree array.

#include <bits/stdc++.h>

#define eps 1e-8
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define CLR(x, y) memset((x),y,sizeof(x))
#define fuck(x) cerr << #x << "=" << x << endl;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int seed = 131;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const int N = 1e5 + 5;
int n, m;
int a[maxn], b[maxn];
int bit[maxn];
int vis[maxn];
int pos[maxn];

int lowbit(int x) {
    return x & -x;
}

void add(int i, int x) {
    while (i < N) {
        bit[i] += x;
        i += lowbit(i);
    }
}

int query(int i) {
    int sum = 0;
    while (i) {
        sum += bit[i];
        i -= lowbit(i);
    }
    return sum;
}

struct node {
    int t, p, val;
    ll ans;

    bool operator<(const node &a) const {
        return t < a.t;
    }
} e[maxn], t[maxn];

ll ans[maxn];

void cdq(int l, int r) {
    if (l == r) return;
    int mid = (l + r) / 2;
    cdq(l, mid);
    cdq(mid + 1, r);
    int i = l, j = mid + 1;
    int cnt = l;
    while (i <= mid || j <= r) {
        if ((j > r) || (i <= mid && e[i].p < e[j].p)) {
            add(e[i].val, 1);
            t[cnt++] = e[i++];

        } else {
            ll sum = (i - l) - query(e[j].val);
            e[j].ans += sum;
            t[cnt++] = e[j++];
        }
    }
    for (int i = l; i <= mid; i++) {
        add(e[i].val, -1);
    }
    for (int i = l; i <= r; i++) {
        e[i] = t[i];
    }
}

void cdq2(int l, int r) {
    if (l == r) return;
    int mid = (l + r) / 2;
    cdq2(l, mid);
    cdq2(mid + 1, r);
    int i = l, j = mid + 1;
    int cnt = l;
    while (i <= mid || j <= r) {
        if ((j > r) || (i <= mid && e[i].p > e[j].p)) {
            add(e[i].val, 1);
            t[cnt++] = e[i++];
        } else {
            ll sum =  query(e[j].val);
            e[j].ans += sum;
            t[cnt++] = e[j++];
        }
    }
    for (int i = l; i <= mid; i++) {
        add(e[i].val, -1);
    }
    for (int i = l; i <= r; i++) {
        e[i] = t[i];
    }
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
#endif
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        pos[a[i]] = i;
    }
    for (int i = 1; i <= m; i++) {
        scanf("%d", &b[i]);
        vis[b[i]] = 1;
    }
    int tmp = m;
    for (int i = 1; i <= n; i++) {
        if (!vis[i]) b[++tmp] = i;
    }
    for (int i = n; i >= 1; i--) {
        e[i].t = n - i + 1;
        e[i].p = pos[b[i]];
        e[i].val = b[i];
    }
    sort(e + 1, e + 1 + n);
    cdq(1, n);
    CLR(bit, 0);
    CLR(t, 0);
    sort(e + 1, e + 1 + n);
    cdq2(1, n);
    sort(e + 1, e + 1 + n);


    for (int i = 1; i <= n; i++) {
        ans[i] = ans[i - 1] + e[i].ans;
    }
    for (int i = n; i > n - m; i--) printf("%lld\n", ans[i]);
    return 0;
}