uva 10862 Connect the Cable Wires

2019-11-08 01:14:08

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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. 这里写图片描述 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

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