BZOJ 1089: [SCOI2003]严格n元树

Decription

询问深度为 \(d\) 的 \(n\) 元树个数, \(n\) 元树所有节点的孩子都有 \(n\) 个.

Sol

递推+高精度.

\(f[i]\) 表示深度为 \(i\) 的 \(n\) 元树个数,我这里深度是从 \([1,k+1]\) 的...

转移就是从上方添加一个节点,子节点任选然后再减去不合法的方案.

\(f[i]=(\sum ^{i-1} _{j=1} f[j])^n-(\sum ^{i-2} _{j=1} f[j])^n\)

Code

#include<cstdio>
#include<cmath>
#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;
 
typedef long long LL;
const int B = 10;
const int W = 1;
 
struct Big{
    vector<int> s;
    void clear(){ s.clear(); }
     
    Big(LL num=0){ *this=num; }
    Big operator = (LL x){
        clear();
        do{ s.push_back(x%B),x/=B; }while(x);
        return *this;
    }
    Big operator = (const string &str){
        clear();
        int x,len=(str.length()-1)/W+1,l=str.length();
        for(int i=0;i<len;i++){
            int tt=l-i*W,st=max(0,tt-W);
            sscanf(str.substr(st,tt-st).c_str(),"%d",&x);
            s.push_back(x);
        }return *this;
//      clear();reverse(str.begin(),str.end());
//      int x,len=(str.length()-1)/W+1,l=str.length();
//      for(int i=0;i<len;i++){
//          int st=i,tt=min(i+W,l);
//          sscanf(str.substr(st,tt-st).c_str(),"%d",&x);
//          s.push_back(x);
//      }return *this;
    }
//  Big operator = (char *str){
//      clear();reverse(str.begin(),str.end());
//      int x,len=(str.length()-1)/W+1,l=str.length();
//      for(int i=0;i<len;i+=W){
//          int s=i,t=min(i+W,l);
//          sscanf(str(s,t-s),"%d",&x);
//          s.push_back(x);
//      }return *this;
//  }
};
 
istream& operator >> (istream & in,Big &a){
    string s;
    if(!(in>>s)) return in;
    a=s;return in;
}
 
ostream& operator << (ostream &out,const Big &a){
    cout<<a.s.back();
    for(int i=a.s.size()-2;i>=0;i--){
        cout.width(W),cout.fill('0'),cout<<a.s[i];
    }return out;
}
 
bool operator < (const Big &a,const Big &b){
    int la=a.s.size(),lb=b.s.size();
    if(la<lb) return 1;if(la>lb) return 0;
    for(int i=la-1;~i;i--){
        if(a.s[i]<b.s[i]) return 1;
        if(a.s[i]>b.s[i]) return 0;
    }return 0;
}
bool operator <= (const Big &a,const Big &b){ return !(b<a); }
bool operator > (const Big &a,const Big &b){ return b<a; }
bool operator >= (const Big &a,const Big &b){ return !(a<b); }
bool operator == (const Big &a,const Big &b){ return !(a>b) && !(a<b); }
bool operator != (const Big &a,const Big &b){ return a>b || a<b ; }
 
 
Big operator + (const Big &a,const Big &b){
    Big c;c.clear();
    int lim=max(a.s.size(),b.s.size()),la=a.s.size(),lb=b.s.size(),i,g,x;
    for(i=0,g=0;;i++){
        if(g==0 && i>=lim) break;
        x=g;if(i<la) x+=a.s[i];if(i<lb) x+=b.s[i];
        c.s.push_back(x%B),g=x/B;
    }i=c.s.size()-1;
    while(c.s[i]==0 && i) c.s.pop_back(),i--;
    return c;
}
Big operator - (const Big &a,const Big &b){
    Big c;c.clear();
    int i,g,x,la=a.s.size(),lb=b.s.size();
    for(i=0,g=0;i<la;i++){
        x=a.s[i]-g;
        if(i<lb) x-=b.s[i];
        if(x>=0) g=0;else g=1,x+=B;
        c.s.push_back(x);
    }i=c.s.size()-1;
    while(c.s[i]==0 && i) c.s.pop_back(),i--;
    return c;
}
Big operator * (const Big &a,const Big &b){
    Big c;
    int i,j,la=a.s.size(),lb=b.s.size(),lc=la+lb;
    c.s.resize(lc,0);
    for(i=0;i<la;i++) for(j=0;j<lb;j++) c.s[i+j]+=a.s[i]*b.s[j];
    for(i=0;i<lc;i++) c.s[i+1]+=c.s[i]/B,c.s[i]%=B;
    i=lc-1;while(c.s[i]==0 && i) c.s.pop_back(),i--;
    return c;
}
Big operator / (const Big &a,const Big &b){
    Big c,f=0;
    int la=a.s.size(),i;
    c.s.resize(la,0);
    for(i=la-1;~i;i--){
        f=f*B,f.s[0]=a.s[i];
        while(f>=b) f=f-b,c.s[i]++;
    }i=la-1;while(c.s[i]==0 && i) c.s.pop_back(),i--;
    return c;
}
Big operator % (const Big &a,const Big &b){
    Big c=a-(a/b)*b;
    return c;
}
Big operator += (Big &a,const Big &b){ return a=a+b; }
Big operator -= (Big &a,const Big &b){ return a=a-b; }
Big operator *= (Big &a,const Big &b){ return a=a*b; }
Big operator /= (Big &a,const Big &b){ return a=a/b; }
Big operator %= (Big &a,const Big &b){ return a=a%b; }
Big operator ^ (Big a,int b){
	Big res=1LL;
	for(;b;b>>=1,a=a*a) if(b&1) res=res*a;
	return res;
}

const int N = 35;

Big f[N],tmp,lst;int n,k;
int main(){
	cin>>n>>k;
	f[1]=1,f[2]=1,tmp=2,lst=1;
	for(int i=3;i<=k+1;i++) f[i]=(tmp^n)-(lst^n),tmp+=f[i],lst+=f[i-1];
	cout<<f[k+1]<<endl;
    return 0;
}