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;
}