ACM 2083 Game of Connections http://acm.pku.cn/JudgeOnline/problem?id=2084

很郁闷的一道题，看通过率本以为很简单，其实就是个数学题，求Catalan数，关于Catalan数，现在做下说明：

令h(1)＝1，catalan数满足递归式：
h(n)= h(1)*h(n-1) + h(2)*h(n-2) + ... + h(n-1)h(1) (其中n>=2)
该递推关系的解为：
h(n+1)=c(2n-2,n-1)/n (n=1,2,3,...)

我并不关心其解是怎么求出来的，我只想知道怎么用catalan数分析问题。
我总结了一下，最典型的三类应用：（实质上却都一样，无非是递归等式的应用，就看你能不能分解问题写出递归式了）
1.括号化问题。
矩阵链乘： P=a1×a2×a3×……×an，依据乘法结合律，不改变其顺序，只用括号表示成对的乘积，试问有几种括号化的方案？(h(n)种)

2.出栈次序问题。
一个栈(无穷大)的进栈序列为1,2,3,..n,有多少个不同的出栈序列?

类似：有2n个人排成一行进入剧场。入场费5元。其中只有n个人有一张5元钞票，另外n人只有10元钞票，剧院无其它钞票，问有多少中方法使得只要有10元的人买票，售票处就有5元的钞票找零？(将持5元者到达视作将5元入栈，持10元者到达视作使栈中某5元出栈)

3.将多边行划分为三角形问题。
将一个凸多边形区域分成三角形区域的方法数?

类似：一位大城市的律师在她住所以北n个街区和以东n个街区处工作。每天她走2n个街区去上班。如果他
从不穿越（但可以碰到）从家到办公室的对角线，那么有多少条可能的道路？

类似：在圆上选择2n个点,将这些点成对连接起来使得所得到的n条线段不相交的方法数?   

这样，有了地推公式，这个题就转化成了一个求大整数的的乘法和加法的问题了！但是写出来后各种运行错误，无奈之际只好打表通过，先稳定下军心：

code:

#include <iostream>
using namespace std;
char kk [100][100]={
  "1",
  "2",
  "5",
  "14",
  "42",
  "132",
  "429",
  "1430",
  "4862",
  "16796",
  "58786",
  "208012",
  "742900",
  "2674440",
  "9694845",
  "35357670",
  "129644790",
  "477638700",
  "1767263190",
  "6564120420",
  "24466267020",
  "91482563640",
  "343059613650",
  "1289904147324",
  "4861946401452",
  "18367353072152",
  "69533550916004",
  "263747951750360",
  "1002242216651368",
  "3814986502092304",
  "14544636039226909",
  "55534064877048198",
  "212336130412243110",
  "812944042149730764",
  "3116285494907301262",
  "11959798385860453492",
  "45950804324621742364",
  "176733862787006701400",
  "680425371729975800390",
  "2622127042276492108820",
  "10113918591637898134020",
  "39044429911904443959240",
  "150853479205085351660700",
  "583300119592996693088040",
  "2257117854077248073253720",
  "8740328711533173390046320",
  "33868773757191046886429490",
  "131327898242169365477991900",
  "509552245179617138054608572",
  "1978261657756160653623774456",
  "7684785670514316385230816156",
  "29869166945772625950142417512",
  "116157871455782434250553845880",
  "451959718027953471447609509424",
  "1759414616608818870992479875972",
  "6852456927844873497549658464312",
  "26700952856774851904245220912664",
  "104088460289122304033498318812080",
  "405944995127576985730643443367112",
  "1583850964596120042686772779038896",
  "6182127958584855650487080847216336",
  "24139737743045626825711458546273312",
  "94295850558771979787935384946380125",
  "368479169875816659479009042713546950",
  "1440418573150919668872489894243865350",
  "5632681584560312734993915705849145100",
  "22033725021956517463358552614056949950",
  "86218923998960285726185640663701108500",
  "337485502510215975556783793455058624700",
  "1321422108420282270489942177190229544600",
  "5175569924646105559418940193995065716350",
  "20276890389709399862928998568254641025700",
  "79463489365077377841208237632349268884500",
  "311496878311103321137536291518809134027240",
  "1221395654430378811828760722007962130791020",
  "4790408930363303911328386208394864461024520",
  "18793142726809884575211361279087545193250040",
  "73745243611532458459690151854647329239335600",
  "289450081175264899454283846029490767264392230",
  "1136359577947336271931632877004667456667613940",
  "4462290049988320482463241297506133183499654740",
  "17526585015616776834735140517915655636396234280",
  "68854441132780194707888052034668647142985206100",
  "270557451039395118028642463289168566420671280440",
  "1063353702922273835973036658043476458723103404520",
  "4180080073556524734514695828170907458428751314320",
  "16435314834665426797069144960762886143367590394940",
  "64633260585762914370496637486146181462681535261000",
  "254224158304000796523953440778841647086547372026600",
  "1000134600800354781929399250536541864362461089950800",
  "3935312233584004685417853572763349509774031680023800",
  "15487357822491889407128326963778343232013931127835600",
  "60960876535340415751462563580829648891969728907438000",
  "239993345518077005168915776623476723006280827488229600",
  "944973797977428207852605870454939596837230758234904050",
  "3721443204405954385563870541379246659709506697378694300",
  "14657929356129575437016877846657032761712954950899755100",
  "57743358069601357782187700608042856334020731624756611000",
  "227508830794229349661819540395688853956041682601541047340",
  "896519947090131496687170070074100632420837521538745909320"
};
int main()
{
 int m;
 while (cin >>m)
 {
  if (m==-1)
    break;
  cout<<kk[m-1]<<endl;
 }
 return 0;
}
首先鄙视下自己的无耻行为，当时实在没办法了，理解下我吧！做不出来就睡不着觉的，后来才发现自己原来写的高精度乘法越界了。
后来干脆搞了个大整数的类，重载了各种需要的运算符了以后才把它名正言顺的AC过去！
code:
#include "string"
#include "iostream"

#include "iomanip"
const long MaxLength=100;//数字最大位数是MaxLength *4
using namespace std;
class BigInt
{
public:
 long *a;
 long length;
 long max;
 BigInt()//构造函数 
 {
     a = new long [MaxLength];
     memset(a,0,sizeof(long)*MaxLength);
     length = 1;
     max = MaxLength;
 }
 BigInt(const BigInt&t)
 {
     long *a = new long [t.length];
     for(int i=0;i<t.length;i++)
      a[i] = t.a[i];
     length = t.length;
     max = t.max;
  
 }
 ~BigInt()//析构函数 
 {
     delete [] a;
 }
 void clear()//清零函数 
 {
     memset(a,0,sizeof(long)*max);
     length = 1;
 }
 friend ostream &operator << (ostream & output,BigInt &t)
 {
     output << t.a[t.length-1];
     for(int i=t.length-2;i>=0;i--)
               output << setw(4) << setiosflags(ios::right) << setfill('0') << t.a[i];
     return output;
 }
 void operator = (const BigInt & t)
 {
     if(t.a==a)
      return;
     delete a;
     a = new long [t.length];
     for(int i=0;i<t.length;i++)
      a[i] = t.a[i];
     length = t.length;
 }
 void operator = (const string & temp)
 {
     int tLength = temp.length() - 1;
     length=tLength/4+1;
     for(int i=0;tLength>2;i++,tLength-=4)
      a[i]=(temp[tLength]-'0')+(temp[tLength-1]-'0')*10+(temp[tLength-2]-'0')*100+(temp[tLength-3]-'0')*1000;
     if(tLength==2)
      a[length-1]=(temp[tLength]-'0')+(temp[tLength-1]-'0')*10+(temp[tLength-2]-'0')*100;
     else if(tLength==1)
      a[length-1]=(temp[tLength]-'0')+(temp[tLength-1]-'0')*10;
     else if(!tLength)
      a[length-1]=(temp[tLength]-'0');
 }
 BigInt operator + (const BigInt & t)
 {
     BigInt sum;
     for(int i=0;i<t.length||i<length;i++)
     {
      sum.a[i]+=a[i]+t.a[i];
      if(sum.a[i]>9999)
      {
       int j=i;
       while(sum.a[j]>9999)
       {
        sum.a[j+1]+=sum.a[j]/10000;
        sum.a[j]%=10000;
        j++;
       }
      }
     }
     for(int i=sum.max-1;i>=sum.length;i--)
               if(sum.a[i])
      {
                   sum.length=i+1;
                   break;
      }
      return sum;
 }
 BigInt operator * (const BigInt & t)
 {
     BigInt product;
     for(int i=0;i<t.length;i++)
      for(int j=0;j<length;j++)
      {
       product.a[i+j]+=a[j]*t.a[i];
       if(product.a[i+j]>9999)
       {
        int q=i+j;
        while(product.a[q]>9999)
        {
         product.a[q+1]+=product.a[q]/10000;
         product.a[q]%=10000;
         q++;
        }
       }
      }
      for(int i=product.max-1;i>=product.length;i--)
       if(product.a[i])
       {
        product.length=i+1;
        break;
       }
       return product;
 }
};
BigInt s[MaxLength+1];
int main()
{
 int n;
 s[0]= "1";
 s[1]="1";s[2]="2";s[3]="5";
 for(int i=4;i<101;i++)
     for(int j=0;j<i;j++)
//      s[i] = add(s[i],mult(s[j],s[i-j]));
      s[i] = s[i]+s[j]*s[i-j-1];
 while(cin>>n && n != -1)
 {
     cout<<s[n]<<endl;
 }
 return 0;
}