uva 10862 Connect the Cable Wires

原题：
Asif is a student of East West University and he is currently working for the EWUISP to meet his relatively high tuition fees. One day, as a part of his job, he was instructed to connect cable wires to N houses. All the houses lie in a straight line. He wants to use only the minimum number of cable wires required to complete his task such that all the houses receive the cable service. A house can either get the connection from the main transmission center or it can get it from a house to its immediate left or right provided the latter house is already getting the service. You are to write a program that determines the number of different combinations of the cable wires that is possible so that every house receives the service.
Example: If there are two houses then 3 combinations are possible as shown in the figure.

Figure: circles represent the transmission center and the small rectangles represent the houses.
Input
Each line of input contains a positive integer N (N ≤ 2000). The meaning of N is described in the above paragraph. A value of 0 for N indicates the end of input which should not be processed.
Output
For each line of input you have to output, on a single line, the number of possible arrangements. You can safely assume that this number will have less than 1000 digits.
Sample Input
1
2
3
0
Sample Output
1
3
8

中文：
给你一个圈和n个方块，现在要让你把这n个方块连接起来，可以方块和方块之间连接，也可以方块和圆连接，现在问你有多少种连接方法。

#include<bits/stdc++.h>using namespace std;const int maxn=1000;/*精度位数*//*(必选)类与基础功能定义，用法类似于unsigned(非负)*/class bign{    friend istream& operator>>(istream&,bign&);/*输入运算符友元*/    friend ostream& operator<<(ostream&,const bign&);/*输出运算符友元*/    friend bign operator+(const bign&,const bign&);/*加号运算符友元*/    friend bign operator*(const bign&,const bign&);/*乘号运算符友元*/    friend bign operator*(const bign&,int);/*高精度乘以低精度乘法友元*/    friend bign operator-(const bign&,const bign&);/*减号运算符友元*/    friend bign operator/(const bign&,const bign&);/*除法运算符友元*/    friend bign operator%(const bign&,const bign&);/*模运算符友元*/    friend bool operator<(const bign&,const bign&);/*逻辑小于符友元*/    friend bool operator>(const bign&,const bign&);/*逻辑大于符友元*/    friend bool operator<=(const bign&,const bign&);/*逻辑小于等于符友元*/    friend bool operator>=(const bign&,const bign&);/*逻辑大于等于符友元*/    friend bool operator==(const bign&,const bign&);/*逻辑等符友元*/    friend bool operator!=(const bign&,const bign&);/*逻辑不等符友元*/private:    int len,s[maxn];public:    bign(){memset(s,0,sizeof(s));len=1;}    bign operator=(const char* num)    {        int i=0,ol;        ol=len=strlen(num);        while(num[i++]=='0'&&len>1)        len--;        memset(s,0,sizeof(s));        for(i=0;i<len;i++)        s[i]=num[ol-i-1]-'0';        return *this;    }    bign operator=(int num)    {        char s[maxn];        sprintf(s,"%d",num);        *this=s;        return *this;    }    bign(int num){*this=num;}    bign(const char* num){*this=num;}    string str() const    {        int i;        string res="";        for(i=0;i<len;i++)res=char(s[i]+'0')+res;        if(res=="")res="0";        return res;    }};/*(可选)基本逻辑运算符重载*/bool operator<(const bign& a,const bign& b){    int i;    if(a.len!=b.len)return a.len<b.len;    for(i=a.len-1;i>=0;i--)        if(a.s[i]!=b.s[i])    return a.s[i]<b.s[i];    return false;}bool operator>(const bign& a,const bign& b){return b<a;}bool operator<=(const bign& a,const bign& b){return !(a>b);}bool operator>=(const bign& a,const bign& b){return !(a<b);}bool operator!=(const bign& a,const bign& b){return a<b||a>b;}bool operator==(const bign& a,const bign& b){return !(a<b||a>b);}/*(可选)加法运算符重载*/bign operator+(const bign& a,const bign& b){    int i,max=(a.len>b.len?a.len:b.len),t,c;    bign sum;    sum.len=0;    for(i=0,c=0;c||i<max;i++)    {        t=c;        if(i<a.len)t+=a.s[i];        if(i<b.len)t+=b.s[i];        sum.s[sum.len++]=t%10;        c=t/10;    }    return sum;}/*(可选)乘法运算符重载(高精度乘高精度)*/bign operator*(const bign& a,const bign& b){    int i,j;    bign res;    for(i=0;i<a.len;i++)    {        for(j=0;j<b.len;j++)        {            res.s[i+j]+=(a.s[i]*b.s[j]);            res.s[i+j+1]+=res.s[i+j]/10;            res.s[i+j]%=10;        }    }    res.len=a.len+b.len;    while(res.s[res.len-1]==0&&res.len>1)res.len--;    if(res.s[res.len])res.len++;    return res;}/*高精度乘以低精度(注意:必须是bign*int顺序不能颠倒,要么会与高精度乘高精度发生冲突*/bign operator*(const bign& a,int b){    int i,t,c=0;    bign res;    for(i=0;i<a.len;i++)    {        t=a.s[i]*b+c;        res.s[i]=t%10;        c=t/10;    }    res.len=a.len;    while(c!=0)    {        res.s[i++]=c%10;        c/=10;        res.len++;    }    return res;}/*(可选)减法运算符重载*/bign operator-(const bign& a,const bign& b){    bign res;    int i,len=(a.len>b.len)?a.len:b.len;    for(i=0;i<len;i++)    {        res.s[i]+=a.s[i]-b.s[i];        if(res.s[i]<0)        {            res.s[i]+=10;            res.s[i+1]--;        }    }    while(res.s[len-1]==0&&len>1)len--;    res.len=len;    return res;}/*(可选)除法运算符重载(注意:减法和乘法运算和>=运算符必选)*/bign operator/(const bign& a,const bign& b){    int i,len=a.len;    bign res,f;    for(i=len-1;i>=0;i--)    {        f=f*10;        f.s[0]=a.s[i];        while(f>=b)        {            f=f-b;            res.s[i]++;        }    }    while(res.s[len-1]==0&&len>1)len--;    res.len=len;    return res;}/*(可选)模运算符重载(注意:减法和乘法运算和>=运算符必选)*/bign operator%(const bign& a,const bign& b){    int i,len=a.len;    bign res,f;    for(i=len-1;i>=0;i--)    {        f=f*10;        f.s[0]=a.s[i];        while(f>=b)        {            f=f-b;            res.s[i]++;        }    }    return f;}/*(可选)X等运算符重载(注意:X法必选)*/bign& operator+=(bign& a,const bign& b){    a=a+b;    return a;}bign& operator-=(bign& a,const bign& b){    a=a-b;    return a;}bign& operator*=(bign& a,const bign& b){    a=a*b;    return a;}bign& operator/=(bign& a,const bign& b){    a=a/b;    return a;}/*可选前缀++/--与后缀++/--(注意:加法必选)*/bign& operator++(bign& a){    a=a+1;    return a;}bign& operator++(bign& a,int){    bign t=a;    a=a+1;    return t;}bign& operator--(bign& a){    a=a-1;    return a;}bign& operator--(bign& a,int){    bign t=a;    a=a-1;    return t;}istream& operator>>(istream &in,bign& x){    string s;    in>>s;    x=s.c_str();    return in;}ostream& operator<<(ostream &out,const bign& x){    out<<x.str();    return out;}bign f[2001];int main(){    ios::sync_with_stdio(false);    f[1]=1,f[2]=3;    for(int i=3;i<=2000;i++)        f[i]=3*f[i-1]-f[i-2];    int n;    while(cin>>n,n)        cout<<f[n]<<endl;    return 0;}

解答：

用递推的思想来分析，设i个方块有f(i)种连接方法，考虑第i个方块连接，那么，第i个方块可以连接到左侧的方块上，此时与前面连接好的i-1个方块组合有f(i-1)种可能，如果第i个方块连接到圆上，则左侧的i-1个方块可以有f(i-1)种连接方法，若是第i-1个方块连接到连接到第i个方块上，则左侧有f(i-2）种方法，同样第i-2个方块和第i-1可以连到第i个方块上以此类推
f(i)=f(i)*2+f(i-1)+f(i-2)+f(i-3)+…+f(1)
如果用上面的公式打表会超时
由于f(i-1)=2*f(i-2)+f(i-3)+…+f(1)
代入上面的式子，可以化简成f(i)=3*f(i-1)-f(i-2)
用大数打表过了