POJ 2506 Tiling (recursive + high precision)

" Topic link " click here~~

"The main topic "

In what many ways can you tile a 2xn rectangle by 2x1 or 2x2 tiles?
Here is a sample tiling of a 2x17 rectangle.

" problem-solving ideas ":

(1) A 2*2 lattice has three methods of filling:

Two of them sideways,

Two of them stood upright,

Put a 2*2.

(2) A recursive formula is obtained f[i]=f[i-1]+f[i-2]*2

And then it's a high-precision template.

Code

/*AUTHOR:HRW recursion + High precision! * * #include <stdio.h> #include <string.h> #include <iostream> #include <algorithm>using     namespace Std;const int Base=1e9;const int capacity=100;typedef long long huge;struct bigint{int Len;     int data[capacity];     BigInt (): Len (0) {} BigInt (const BigInt &v): Len (V.len) {memcpy (Data, V.data, len*sizeof*data);} BigInt (int V): Len (0) {for (; v>0;     v/=base) Data[len++]=v%base;}     BigInt &operator= (const BigInt &v) {len=v.len; memcpy (Data, V.data, Len*sizeof*data); return *this;}     int &operator[] (int Index) {return data[index];} int operator[] (int Index) const {return data[index];};     int compare (const BigInt &a, const BigInt &b) {if (A.len!=b.len) return a.len>b.len? 1:-1;     int i;     For (i=a.len-1;i>=0 && a[i]==b[i];i--);     if (i<0) return 0; Return a[i]>b[i]? 1:-1;}     BigInt operator+ (const BigInt &a,const BigInt &b) {int I,carry (0);     BigInt R; For(i=0;i<a.len| | i<b.len| |         carry>0;i++) {if (I<a.len) carry+=a[i];         if (I<b.len) carry+=b[i];         R[i]=carry%base;     Carry/=base;     } r.len=i; return R;}     BigInt operator-(const BigInt &a,const BigInt &b) {int I,carry (0);     BigInt R;     R.len=a.len;         for (i=0;i<r.len;i++) {r[i]=a[i]-carry;         if (I<b.len) r[i]-=b[i];         if (r[i]<0) carry=1,r[i]+=base;     else carry=0;     } while (r.len>0&&r[r.len-1]==0) r.len--; return R;}     BigInt operator* (const BigInt &a,const int &b) {int i;     Huge Carry (0);     BigInt R; for (i=0;i<a.len| |         carry>0;i++) {if (I<a.len) Carry+=huge (a[i]) *b;         R[i]=carry%base;     Carry/=base;     } r.len=i; return R;}     IStream &operator>> (IStream &in,bigint &v) {char Ch;         for (v=0;in>>ch;) {v=v*10+ (ch-' 0 ');     if (In.peek () <= ") break; } return in;} Ostream &operator<< (ostream &out,const BigInt &v) {int i;     out<< (v.len==0 0:v[v.len-1]);     for (i=v.len-2;i>=0;i--) for (int j=base/10;j>0;j/=10) out<<v[i]/j%10; return out;}    BigInt Fa[10000];int Main () {fa[0]=fa[1]=1,fa[2]=3;    for (int i=3;i<=250;i++) {fa[i]=fa[i-1]+fa[i-2]+fa[i-2];    } unsigned long long n;    while (cin>>n) {cout<<fa[n]<<endl; } return 0;}

POJ 2506 Tiling (recursive + high precision)