BZOJ 1005 [HNOI2008] 明明的烦恼

Description

　　自从明明学了树的结构,就对奇怪的树产生了兴趣......给出标号为1到N的点,以及某些点最终的度数,允许在

任意两点间连线,可产生多少棵度数满足要求的树?

Input

　　第一行为N(0 < N < = 1000),

接下来N行,第i+1行给出第i个节点的度数Di,如果对度数不要求,则输入-1

Output

　　一个整数,表示不同的满足要求的树的个数,无解输出0

Sample Input

3

1

-1

-1

Sample Output

2

HINT

　　两棵树分别为1-2-3;1-3-2

Source

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

purfer序列+高精度+质因数分解~

purfer序列真神奇~推荐看一下http://www.cnblogs.com/zhj5chengfeng/archive/2013/08/23/3278557.html~

注意像π一样的符号是从1乘到a[i]-1，不是直接乘a[i]-1，改过来以后代码就A了~

#include<cstdio>

int n,x,cnt,tot,tot1,a[1001],num[169],ans[10001],len,c[169]={0,2,3,5,7,11,13,17,19,23,29,31,37,41,
43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,
157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,
269,271,277,281,283,293,307,311,313,317,331,337,347,349,353,359,367,373,379,383,
389,397,401,409,419,421,431,433,439,443,449,457,461,463,467,479,487,491,499,503,
509,521,523,541,547,557,563,569,571,577,587,593,599,601,607,613,617,619,631,641,
643,647,653,659,661,673,677,683,691,701,709,719,727,733,739,743,751,757,761,769,
773,787,797,809,811,821,823,827,829,839,853,857,859,863,877,881,883,887,907,911,
919,929,937,941,947,953,967,971,977,983,991,997};

void del(int u,int v)
{
for(int i=1;u>1 && i<=168;i++) while(!(u%c[i])) num[i]+=v,u/=c[i];
}

void cheng(int v)
{
for(int i=1;i<=len;i++) ans[i]*=v;
for(int i=1;i<=len;i++)
if(ans[i]>=10)
{
ans[i+1]+=ans[i]/10;ans[i]%=10;
if(i==len) len++;
}
}

int main()
{
scanf("%d",&n);
if(n==1)
{
scanf("%d",&x);
if(!x) printf("1");
else printf("0");
return 0;
}
for(int i=1;i<=n;i++)
{
scanf("%d",&x);
if(!x)
{
printf("0");return 0;
}
if(x!=-1) a[++cnt]=x,tot+=x-1;
else tot1++;
}
if(tot>n-2)
{
printf("0");return 0;
}
for(int i=n-tot-1;i<=n-2;i++) del(i,1);
for(int i=1;i<=cnt;i++)
for(int j=1;j<a[i];j++) del(j,-1);
del(n-cnt,n-2-tot);
ans[1]=len=1;
for(int i=1;i<=168;i++) while(num[i]--) cheng(c[i]);
for(int i=len;i;i--) printf("%d",ans[i]);
return 0;
}