【卡特兰数+高精度】HDU1023Train Problem II

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1023

Problem Description

As we all know the Train Problem I, the boss of the Ignatius Train Station want to know if all the trains come in strict-increasing order, how many orders that all the trains can get out of the railway.

 

Input

The input contains several test cases. Each test cases consists of a number N(1<=N<=100). The input is terminated by the end of file.

 

Output

For each test case, you should output how many ways that all the trains can get out of the railway.

 

Sample Input

   
   
   
   

    
    
    
    1
2
3
10
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
2
5
16796


    
    
    
    
 
     
 
      Hint 
     
 The result will be very large, so you may not process it by 32-bit integers. 
    
   
   
   
   

代码：

#include<iostream>
#include<string>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
string s[110];
//  高精度加法；
string add(string aa,string bb)
{
    string ans="";
    int a[10100]={0},b[10100]={0};
    int aLen=aa.size();
    int bLen=bb.size();
    int MaxLen=max(aLen,bLen);
    for(int i=aLen-1;i>=0;i--)
        a[aLen-i-1]=aa[i]-'0';
    for(int i=bLen-1;i>=0;i--)
        b[bLen-i-1]=bb[i]-'0';
    for(int i=0;i<MaxLen;i++){
        a[i]+=b[i];
        if(a[i]>=10){
            a[i+1]++;
            a[i]%=10;
        }
    }
    if(a[MaxLen]) ans=a[MaxLen]+'0';
    for(int i=MaxLen-1;i>=0;i--)
        ans+=a[i]+'0';
    return ans;
}
//  高精度乘法；
string mul(string aa,string bb)
{
    string ans="";
    int a[10100]={0},b[10100]={0},c[10100]={0};
    int aLen=aa.size();
    int bLen=bb.size();
    for(int i=aLen-1;i>=0;i--)
        a[aLen-i-1]=aa[i]-'0';
    for(int i=bLen-1;i>=0;i--)
        b[bLen-i-1]=bb[i]-'0';
    for(int i=0;i<aLen;i++)
        for(int j=0;j<bLen;j++)
            c[i+j]+=a[i]*b[j];
    for(int i=0;i<=aLen+bLen;i++){
        c[i+1]+=c[i]/10;
        c[i]%=10;
    }
    int MaxLen=aLen+bLen;
    while(c[MaxLen]==0) MaxLen--;
    for(int i=MaxLen;i>=0;i--)
        ans+=c[i]+'0';
    return ans;
}
int main()
{
    s[0]="1";s[1]="1";
    for(int i=2;i<101;i++){
        s[i]="0";
        for(int j=0;j<i;j++){
            s[i]=add(s[i],mul(s[j],s[i-j-1]));
        }
    }
    int n;
    while(cin>>n){
        cout<<s[n]<<endl;
    }
    return 0;
}