BZOJ 2822: [AHOI2012]树屋阶梯
Description
求拼成阶梯状的方案数.
Sol
高精度+Catalan数.
我们可以把最后一行无线延伸,所有就很容易看出Catalan数了.
\(f_n=f_0f_{n-1}+f_1f_{n-2}+f_2f_{n-3}+...+f_{n-1}f_0\)
这就是Catalan数了,高精贴板子...
Code
/************************************************************** Problem: 2822 User: BeiYu Language: C++ Result: Accepted Time:20 ms Memory:1308 kb ****************************************************************/ #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; } }; 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;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,Big &b){ Big c=1; for(;b!=0;b=b/2,a=a*a){ if(b.s[0] & 1) c=c*a; }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; } const int N = 1005; int cnt; int b[N],pr[N],minp[N],c[N]; void Pre(int t){ minp[1]=0; for(int i=2;i<=t;i++){ if(!b[i]) pr[++cnt]=i,minp[i]=cnt; for(int j=1;j<=cnt && i*pr[j]<=t;j++){ b[i*pr[j]]=1,minp[i*pr[j]]=j; if(i%pr[j]==0) break; } } } void Add(int x,int v){ while(x>1) c[minp[x]]+=v,x/=pr[minp[x]]; } int main(){ ios::sync_with_stdio(false); int n;Big ans=1,a,b; cin>>n; Pre(n*2); for(int i=n+1;i<=2*n;i++) Add(i,1); for(int i=1;i<n;i++) Add(i,-1); Add(n,-1),Add(n+1,-1); for(int i=1;i<=cnt;i++) a=pr[i],b=c[i],ans*=a^b; cout<<ans<<endl; return 0; }