Description：

Solution：

Code：

#include<algorithm>

#include<iostream>

#include<cstdlib>

#include<cstdio>

#include<cmath>

#include<map>

#include<cstring>

using namespace std;

inline int read()

{

	register int res = 0;

	register char c = getchar();

	while(!isdigit(c))c = getchar();

	while(isdigit(c))

	{

		res = (res << 1) + (res << 3) + c - '0';

		c = getchar();

	}

	return res;

}

int n;

#define MAXN 100010

int val[MAXN];

int b[MAXN];

map<int,int> fr;

int tot = 0;

struct edge

{

	int to,nxt;

}e[MAXN];

int edgenum = 0;

int lin[MAXN] = {0};

void add(int a,int b)

{

	++edgenum;e[edgenum].to = b;e[edgenum].nxt = lin[a];lin[a] = edgenum;

	return;

}

struct node

{

	int lc,rc;

	int sum;

}t[MAXN * 30];

int ptr = 0;

int newnode(){return ++ptr;}

int root[MAXN];

#define mid ((l + r) >> 1)

void insert(int &rt,int p,int l,int r)

{

	if(rt == 0)rt = newnode();

	++t[rt].sum;

	if(l == r)return;

	if(p <= mid)insert(t[rt].lc,p,l,mid);

	else insert(t[rt].rc,p,mid + 1,r);

	return;

}

int merge(int x,int y)

{

	if(!x)return y;if(!y)return x;

	t[x].lc = merge(t[x].lc,t[y].lc);

	t[x].rc = merge(t[x].rc,t[y].rc);

	t[x].sum = t[x].sum + t[y].sum;

	return x;

}

int query(int rt,int L,int R,int l,int r)

{

	if(L > R)return 0;

	if(L <= l && r <= R)return t[rt].sum;

	int res = 0;

	if(L <= mid)res += query(t[rt].lc,L,R,l,mid);

	if(R > mid)res += query(t[rt].rc,L,R,mid + 1,r);

	return res;

}

int ans[MAXN];

void dfs(int k)

{

	for(int i = lin[k];i != 0;i = e[i].nxt)

	{

		dfs(e[i].to);

		root[k] = merge(root[k],root[e[i].to]);

	}

	ans[k] = query(root[k],val[k] + 1,tot,1,tot);

	return;

}

int main()

{

	scanf("%d",&n);

	for(int i = 1;i <= n;++i)val[i] = read();

	for(int i = 1;i <= n;++i)b[i] = val[i];

	sort(b + 1,b + 1 + n);

	for(int i = 1;i <= n;++i)

		if(i == 1 || b[i] != b[i - 1])

			fr[b[i]] = ++tot;

	for(int i = 1;i <= n;++i)val[i] = fr[val[i]];

	for(int i = 2;i <= n;++i)add(read(),i);

	for(int i = 1;i <= n;++i)insert(root[i],val[i],1,tot);

	dfs(1);

	for(int i = 1;i <= n;++i)printf("%d\n",ans[i]);

	return 0;

}