bzoj [1005] [HNOI2008]明明的烦恼

此题用到了prufer编码的性质，一个度为n的节点将会再其中出现n-1次，即可用C（n，m）进行求解。

并且加上高精度与素数分解。

#include <cstdio>
#define C (c=getchar())
using namespace std;
const int prime[170]={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};
int ans[170];
int big_num[10005];
int n,a,prufer,le,nos;
inline void read(int &a)
{
a=0;
static char c;int f=1;
while (C<'0'||c>'9') if (c=='-') f=-1;
while (c>='0'&&c<='9') a=a*10+c-'0',C;
a*=f;
return ;
}
inline void clac(int x)
{
register int i;
int p,j;
for (i=x+1;i<=prufer;i++)
{
p=i;j=1;
while (p>1)
{
while (p%prime[j]==0)
{
ans[j]++;
p/=prime[j];
}j++;
}
}
for (i=1;i<=prufer-x;i++)
{
p=i;j=1;
while (p>1)
{
while (p%prime[j]==0)
{
ans[j]--;
p/=prime[j];
}j++;
}
}
return ;
}
int main()
{
register int i,j;
read(n),prufer=n-2;
for (i=1;i<=n;i++)
{
read(a);
if (a==0||prufer-(a-1)<0) {printf("0");return 0;};
if (a==-1){nos++;continue;}a--;
clac(a);prufer-=a;
}
if (prufer)
{
int t=1;
while (nos>1)
{
while (nos%prime[t]==0)
{
ans[t]+=prufer;
nos/=prime[t];
}t++;
}
}
big_num[1]=1;le=1;
for (i=1;i<=168;i++)
{
while (ans[i])
{
ans[i]--;
for (j=1;j<=le;j++) big_num[j]*=prime[i];
for (j=1;j<=le;j++) if (big_num[j]>9) big_num[j+1]+=big_num[j]/10,big_num[j]%=10;
while (big_num[le+1]) le++,big_num[le+1]+=big_num[le]/10,big_num[le]%=10;
}
}
for (i=le;i>=1;i--) printf("%d",big_num[i]);
return 0;
}