Bzoj 2822: [AHOI2012] Tree house ladder

Description

The number of schemes to be spelled in a ladder-like form.

Sol

High precision +catalan number.

We can extend the last line wirelessly, and it's easy to see the number of Catalan.

\ (f_n=f_0f_{n-1}+f_1f_{n-2}+f_2f_{n-3}+...+f_{n-1}f_0\)

This is the number of Catalan, high-precision sticker board ...

Code

/************************************************************** problem:2822 User:beiyu language:c++ Result : Accepted time:20 ms memory:1308 kb****************************************************************/#include <cs tdio> #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&LT;LB) return 1;if (LA&GT;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&LT;A); }bool operator > (const big &a,const big &b) {return b<a;} BOOL operator >= (const big &a,const big &b) {return! A&LT;B); }bool operator = = (Const Big &a,const big &b) {return! A&GT;B) &&! (A&LT;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;}

　　

Bzoj 2822: [AHOI2012] Tree house ladder