# BZOJ 1089: [SCOI2003]严格n元树

## Sol

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

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

## 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;
}


posted @ 2016-10-23 21:35  北北北北屿  阅读(173)  评论(0编辑  收藏  举报