BZOJ 1005([HNOI2008]明明的烦恼-Prufer数列-树与数组的一一对应) (二)

2014-11-24 02:32:54 · 作者: · 浏览: 5
nt n,degree[MAXN],T=0,blank_pos;
struct mul_arr
{
int a[MAXN];
int& operator[](int i){return a[i];}
mul_arr(){memset(a,0,sizeof(a));}
mul_arr(int m,int k) //pow(m,k)
{
memset(a,0,sizeof(a));
for(int i=2;i*i<=m;i++)
while (!(m%i))
{
a[i]+=k;m/=i;
}
if (m>1) a[m]+=k;
}
friend mul_arr operator-(mul_arr a,mul_arr b){For(i,n) a[i]-=b[i]; return a;}
friend mul_arr operator+(mul_arr a,mul_arr b){For(i,n) a[i]+=b[i]; return a;}
void print(){For(i,n) cout< }_ans,jc;
struct Highn
{
int a[MAXN],size;
int& operator[](int i){return a[i];}
Highn():size(0){memset(a,0,sizeof(a));}
Highn(int x)
{
size=0;memset(a,0,sizeof(a));
while (x)
{
a[++size]+=x%F;
x/=F;
}
}
friend Highn operator*(Highn a,Highn b)
{
Highn c;
For(i,a.size)
For(j,a.size)
{
c[i+j-1]+=a[i]*b[j];
c[i+j]+=c[i+j-1]/F;
c[i+j-1]%=F;
}
c.size=a.size+b.size;
while (c.size&&!c[c.size]) c.size--;
return c;
}
friend Highn operator*(Highn a,int b)
{
For(i,a.size) a[i]*=b;
For(i,a.size) a[i+1]+=a[i]/F,a[i]%=F;
a.size++;while (a.size&&!a[a.size]) a.size--;
return a;
}

void print()
{
printf("%d",a[size]);
ForD(i,size-1)
{
printf("%04d",a[i]);
}puts("");
}
}ans;
int v[MAXN]={0};
int main()
{
freopen("bzoj1005.in","r",stdin);
scanf("%d",&n);blank_pos=n-2;
int un_blank_pos=0;
For(i,n)
{
scanf("%d",°ree[i]);
if (degree[i]^-1)
{
if (!degree[i]&&n>1){puts("1");return 0;}
else if (!degree[i]) {puts("0");return 0;}
blank_pos-=(--degree[i]);v[degree[i]]--;un_blank_pos+=degree[i]++;
}else T++;
}
v[n-2]++;v[blank_pos]--;
// For(i,n) cout< _ans=mul_arr(T,blank_pos);
if (un_blank_pos+blank_pos!=n-2)
{
puts("0");return 0;
}
// cout< // _ans.print();
For(i,n)
{
jc=jc+mul_arr(i,1);
// cout<<' ';jc.print();
if (v[i]) For(j,n) _ans[j]+=v[i]*jc[j];
// _ans.print();
}
// For(i,n) cout<<_ans[i]<<' ';cout< ans=1;
For(i,n)
For(j,_ans[i])
ans=ans*i;//,ans.print();
ans.print();
return 0;
}