BZOJ-1005 明明的烦恼

Prufer编码练习题，这个编码是跟树的生成计数有关系的。

推荐这篇博文：http://www.cnblogs.com/zhj5chengfeng/archive/2013/08/23/3278557.html 介绍地挺全面+生动形象

会了Prufer之后这道题还要用上组合数学来高精度计算。

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

#define rep(i, l, r) for(int i=l; i<=r; i++)
#define down(i, l, r) for(int i=l; i>=r; i--)
#define N 12345
#define MAX 1<<30
#define Q 1000000

using namespace std;
int read()
{
	int x=0, f=1; char ch=getchar();
	while (ch<'0' || ch>'9') { if (ch=='-') f=-1; ch=getchar(); }
	while (ch>='0' && ch<='9') { x=x*10+ch-'0' ; ch=getchar(); }
	return x*f;
}

int n, d[N], k, s, m[N], l;

void Mult(int x)
{
	rep(i, 1, l) m[i]*=x;
	rep(i, 1, l) m[i+1]+=m[i]/Q, m[i]%=Q;
	while (m[l+1]) l++, m[l+1]+=m[l]/Q, m[l]%=Q;
}

void Div(int x)
{
	int a=0;
	down(i, l, 1) a=a*Q+m[i], m[i]=a/x, a%=x;
	while (m[l]==0 && l>1) l--;
}

int main()
{
	n=read(); rep(i, 1, n) d[i]=read();
	rep(i, 1, n) if (d[i]!=-1) k++, s+=d[i]-1;
	if (s>n-2) printf("0"); else
	{
		m[l=1] = 1;
		rep(i, 1, n-2) Mult(i);
		rep(i, 1, n-2-s) Div(i);
		rep(i, 1, n-2-s) Mult(n-k);
		rep(i, 1, n) if (d[i]!=-1) rep(j, 1, d[i]-1) Div(j);
		printf("%d", m[l]);
		down(i, l-1, 1) 
			if (m[i]>=100000) printf("%d", m[i]);
			else if (m[i]>=10000) printf("0%d", m[i]);
			else if (m[i]>=1000) printf("00%d", m[i]);
			else if (m[i]>=100) printf("000%d", m[i]);
			else if (m[i]>=10) printf("0000%d", m[i]);
			else printf("00000%d", m[i]);
	}
	return 0;
}