Description

We are given a rooted tree of n vertices. The vertices are to be labeled with numbers1, 2,..., n so that each label is unique and the heap condition holds, i.e. the label of any vertex is less than the label of its parent. How many such labellings exist? Since this number may be quite large, calculate only its remainder modulo m .

Input 

The input contains several tree descriptions. The first line contains the number of input treest(t250) . Each tree description begins with a line containing the size of the tree n(1n500000) and an integer m(2m10⁹) . n - 1 lines follow, i -th of which containsp(i + 1) , the number of the parent of thei + 1 -th vertex (1p(i + 1)i) . Vertex number 1 will be the root in each tree, so its parent will not be given. Total size of the input will not exceed 50MB.

Output 

For each tree output the number of its valid labellings modulo given m .

Explanation for sample: The 8 possible labellings from the last example test case are as follows:

Sample Input 

4 
3 1000000
1 
1 
4 1000000
1 
1 
1 
5 1000000
1 
2 
3 
4 
5 1000000
1 
1 
3 
3

Sample Output 

2 
6 
1 
8题意：给你一棵n个结点的有根树，要求给结点标号1～n，使得不同结点标号不同，且每个非根结点结点的标号比父结点小，求方案数。思路：这题和UVA - 11174 Stand in a Line 是一样的，不过因为m不是素数，所以所以我们要把分子分母分解质因子，然后快速幂取模。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>
typedef long long ll;
using namespace std;
const int maxn = 510005;

int pre[maxn], cnt[maxn];
int isleaf[maxn];
int prime[maxn], num_prime, cprime[maxn], isprime[maxn];

queue<int> q;
ll n, mod;

void init() {
	memset(isprime, 1, sizeof(isprime));
	num_prime = 0;
	int cnt = 0;
	isprime[0] = isprime[1] = 0;
	for (int i = 2; i < maxn; i++)
		if (isprime[i]) {
			isprime[i] = num_prime;
			prime[num_prime++] = i;
			for (int j = 2; j * i < maxn; j++)
				isprime[i * j] = 0;
		}
}

void bfs() {
	for (int i = 1; i <= n; i++)
		cnt[i] = 1;
	while (!q.empty()) {
		int u = q.front();
		q.pop();
		if (u == 1)
			continue;
		cnt[pre[u]] += cnt[u];
		isleaf[pre[u]]--;
		if (!isleaf[pre[u]])
			q.push(pre[u]);
	}
}

void fac(int a, int v) {
	for (int i = 0; a != 1; i++) {
		while (a % prime[i] == 0) {
			cprime[i] += v; 
			a /= prime[i];
		}
		if (isprime[a]) {
			cprime[isprime[a]] += v;
			break;
		}
	}
}

ll pow_mod(int a, int b) {
	ll x = a, tmp = 1;
	while (b) {
		if (b & 1)
			tmp = (tmp * x) % mod;
		b >>= 1;
		x = (x * x) % mod;
	}
	return tmp;
}

ll solve() {
	memset(cprime, 0, sizeof(cprime));
	for (int i = 2; i < n; i++)
		fac(i, 1);
	for (int i = 2; i <= n; i++)
		fac(cnt[i], -1);
	ll res = 1;
	for (int i = 0; i <= n && res != 0; i++)
		if (cprime[i]) 
			res = (res * pow_mod(prime[i], cprime[i])) % mod;
	return res;
}

int main() {
	init();
	int t;
	scanf("%d", &t);
	while (t--) {
		scanf("%lld%lld", &n, &mod);
		pre[1] = -1;
		memset(isleaf, 0, sizeof(isleaf));
		for (int i = 2; i <= n; i++) {
			scanf("%d", &pre[i]);
			isleaf[pre[i]]++;
		}
		for (int i = 1; i <= n; i++)
			if (!isleaf[i])
				q.push(i);
		bfs();	
		printf("%lld\n", solve());
	}
	return 0;
}


 

UVA - 1436 Counting heaps