【BZOJ 1005】无根树的Prufer数列

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;

const int N = 10000 + 5, Prime_N = 1450 + 5;

int n, Prime[Prime_N], A[N], p[Prime_N], Ans[N], P = 0, Bg, l, K;

bool isPrime[N];

void Add(int num, int Flag)
{
    for (int i = 1; i <= P; i++)
    {
        if (Prime[i] > num) break;
        while (num % Prime[i] == 0)
        {
            p[i] += Flag;
            num /= Prime[i];
        }
    }
}

void Cheng(int num)
{
    for (int i = 1; i <= l; i++) Ans[i] *= num;
    for (int i = 1; i <= l; i++)
    {
        Ans[i + 1] += Ans[i] / 10;
        Ans[i] %= 10;
    }
    while (Ans[l + 1])
    {
        l++;
        Ans[l + 1] += Ans[l] / 10;
        Ans[l] %= 10;
    }
}

int main()
{
    scanf("%d", &n);    
    for (int i = 1; i <= n; i++) scanf("%d", A + i);
    for (int i = 1; i <= n; i++) isPrime[i] = true;
    isPrime[1] = false;
    for (int i = 2; i <= n; i++)
    {
        if (isPrime[i]) Prime[++P] = i;
        for (int j = 1; j <= P && i * Prime[j] <= n; j++)
            isPrime[i * Prime[j]] = false;
    }
    memset(p, 0, sizeof(p));
    Bg = n - 2;
    K = n;
    for (int i = 1; i <= n; i++)
    {
        if (A[i] != -1)
        {
            K--;
            for (int j = 1; j <= A[i] - 1; j++)
            {
                Add(Bg, 1);
                Bg--;
                Add(j, -1);
            }
        }
    }
    Ans[1] = 1;
    l = 1;
    for (int i = 1; i <= P; i++)
        for (int j = 1; j <= p[i]; j++)
            Cheng(Prime[i]);
    for (int i = 1; i <= Bg; i++) Cheng(K);
    for (int i = l; i >= 1; i--) printf("%d", Ans[i]);
    printf("\n");
    return 0;
}