Bzoj 1089 scoi2003 strict n element tree dynamic planning + High Precision

Define a strict n-element tree with a depth of D as the root depth of 0, the deepest node depth is d, and each non-leaf node has a tree with EXACTLY n subnodes

Given N and D, find the number of strict n-element trees with a depth of D.

The recursive part of this question is not difficult. First, we set the depth to I. The strict n-element tree has a total of F [I] prefixes and

It is not difficult to find a strict n-element tree below I consists of two parts: a root node, n sub-trees, where the depth of each sub-tree does not exceed the I-1

Each subtree has s [I-1] to plant a total of N subtree so s [I] = s [I-1] ^ n

Hmm? Is there something missing? Yes, there is another situation where the strict n-element tree itself is a root node.

So s [I] = s [I-1] ^ n + 1

Then let's take a look at the example... Mom... High precision...

In fact, high precision is not difficult to photograph... My high precision is a one-pass operator overload... Even cout is overloaded .... By the way, I learned a lot.

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<iomanip>using namespace std;struct long_int{int num[300],cnt;void operator = (int y){num[1]=y;cnt=1;}int& operator [] (int x){return num[x];}}S[20];void operator *= (long_int &x,long_int &y){long_int z=S[19];int i,j;for(i=1;i<=x.cnt;i++)for(j=1;j<=y.cnt;j++){z[i+j-1]+=x[i]*y[j];z[i+j]+=z[i+j-1]/10000;z[i+j-1]%=10000;}z.cnt=x.cnt+y.cnt;if(!z[z.cnt])--z.cnt;x=z;}void operator ++ (long_int &x){int i=1;x[1]++;while(x[i]==10000)x[i]=0,x[++i]++;}long_int operator - (long_int &x,long_int &y){long_int z=S[19];int i;for(i=1;i<=x.cnt;i++){z[i]+=x[i]-y[i];if(z[i]<0)z[i]+=10000,z[i+1]--;if(z[i])z.cnt=i;}return z;}long_int operator ^ (long_int x,int y){long_int z=S[19];z=1;while(y){if(y&1) z*=x;x*=x;y>>=1;}return z;}ostream& operator << (ostream& os,long_int x){int i;os<<x[x.cnt];for(i=x.cnt-1;i;i--)os<<setfill('0')<<setw(4)<<x[i];return os;}int n,d;int main(){int i;cin>>n>>d;if(!d){puts("1");return 0;}S[0]=1;for(i=1;i<=d;i++)S[i]=S[i-1]^n,++S[i];cout<<S[d]-S[d-1]<<endl;}

Bzoj 1089 scoi2003 strict n element tree dynamic planning + High Precision