【UR #2】树上GCD

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 200003;
void read(int &k) {
	k = 0; int fh = 1; char c = getchar();
	for(; c < ‘0‘ || c > ‘9‘; c = getchar())
		if (c == ‘-‘) fh = -1;
	for(; c >= ‘0‘ && c <= ‘9‘; c = getchar())
		k = (k << 1) + (k << 3) + c - ‘0‘;
	k = k * fh;
}

bool vis[N];
struct node {
	int nxt, to;
} E[N];
int qu[N], ct[N], n, m, fa[N], de, deep[N], cnt = 0, point[N], root, sz[N], boo[N];
ll t[503][503], cn[N], sct[N], scn[N], ans2[N], S[N];

void ins(int x, int y) {E[++cnt].nxt = point[x]; E[cnt].to =  y; point[x] = cnt;}
void findrt(int x) {
	int p = 0, q = 0; qu[++q] = x;
	while (p != q) {
		int u = qu[++p];
		boo[u] = sz[u] = 1;
		for(int tmp = point[u]; tmp; tmp = E[tmp].nxt)
			if (!vis[E[tmp].to])
				qu[++q] = E[tmp].to;
	}
	for(int i = q; i; --i) {
		if (boo[qu[i]] && sz[qu[i]] * 2 >= q) {root = qu[i]; return;}
		sz[fa[qu[i]]] += sz[qu[i]];
		if (sz[qu[i]] * 2 >= q)
			boo[fa[qu[i]]] = 0;
	}
}//非递归找重心！！！国家队rank3的鏼爷Orz给SDOIers带来福利

void BFS(int x) {
	deep[x] = 1;
	int p = 0, q = 0; qu[++q] = x;
	while (p != q) {
		int u = qu[++p]; ++ct[deep[u]];
		for(int tmp = point[u]; tmp; tmp = E[tmp].nxt)
			if (!vis[E[tmp].to])
				deep[E[tmp].to] = deep[u] + 1, qu[++q] = E[tmp].to;
	}
	de = deep[qu[q]];
}

void Q(int x) {
	vis[x] = 1;
	int up = 0, upp = 0;
	for(int tmp = point[x]; tmp; tmp = E[tmp].nxt)
		if (!vis[E[tmp].to]) {
			BFS(E[tmp].to);
			up = max(up, de);
			for(int i = 1; i <= de; ++i)
				for(int j = i; j <= de; j += i)
					cn[i] += ct[j];
			for(int i = 1; i <= de; ++i) {
				ans2[i] += ct[i];
				sct[i] += ct[i];
				S[i] += scn[i] * cn[i];
				scn[i] += cn[i];
				ct[i] = cn[i] = 0;
			}
		}
	
	sct[0] = 1;
	int step = 0, son = x, line, to;
	for(int i = fa[x]; !vis[i] && i; son = i, i = fa[i]) {
		++step; to = 0;
		for(int tmp = point[i]; tmp; tmp = E[tmp].nxt)
			if (!vis[E[tmp].to] && E[tmp].to != son)
				BFS(E[tmp].to), to = max(to, de);
		de = to;
		upp = max(upp, de);
		for(int j = 1; j <= de; ++j)
			for(int k = j; k <= de; k += j)
				cn[j] += ct[k];
		line = min(de, m);
		for(int j = 1; j <= line; ++j) {
			to = step % j;
			if (t[j][to] == -1) {
				t[j][to] = 0;
				for(int k = (j - to) % j; k <= up; k += j)
					t[j][to] += sct[k];
			}
			S[j] += t[j][to] * cn[j];
		}
		for(int j = m + 1; j <= de; ++j)
			for(int k = (j - step % j) % j; k <= up; k += j)
				S[j] += sct[k] * cn[j];
		for(int j = 1; j <= de; ++j)
			ct[j] = cn[j] = 0;
		++ans2[step];
	}
	
	//下面是向ShallWe学的
	int L = 1, R = 0, tot = step + up;
	ll now = 0;
	for(int i = 2; i <= tot; ++i) {
		if (R + 1 < i) now += sct[++R];
		if (L + step < i) now -= sct[L++];
		ans2[i] += now;
	}
	//仰膜上方ShallWe的代码
	
	line = min(m, upp);
	for(int i = 1; i <= line; ++i)
		for(int j = 0; j < i; ++j)
			t[i][j] = -1;
	for(int i = 0; i <= up; ++i)
		sct[i] = scn[i] = 0;
	
	if (son != x) {
		findrt(son);
		Q(root);
	}
	for(int tmp = point[x]; tmp; tmp = E[tmp].nxt)
		if (!vis[E[tmp].to]) {
			findrt(E[tmp].to);
			Q(root);
		}
}

ll ans[N];
int main() {
	read(n); m = (sqrt(n));
	for(int i = 2; i <= n; ++i) {
		read(fa[i]);
		ins(fa[i], i);
	}
	
	memset(t, -1, sizeof(t));
	findrt(1);
	Q(root);
	for(int i = n - 1; i; --i)
		for(int j = i + i; j < n; j += i)
			S[i] -= S[j];
	for(int i = 1; i < n; ++i)
		printf("%lld\n", S[i] + ans2[i]);	
	
	return 0;
}