HDU 1134 卡特兰数 大数乘法除法

Problem Description
This is a small but ancient game. You are supposed to write down the numbers 1, 2, 3, ... , 2n - 1, 2n consecutively in clockwise order on the ground to form a circle, and then, to draw some straight line segments to connect them into number pairs. Every number must be connected to exactly one another. And, no two segments are allowed to intersect.

It's still a simple game, isn't it But after you've written down the 2n numbers, can you tell me in how many different ways can you connect the numbers into pairs Life is harder, right

Input
Each line of the input file will be a single positive number n, except the last line, which is a number -1. You may assume that 1 <= n <= 100.

Output
For each n, print in a single line the number of ways to connect the 2n numbers into pairs.

Sample Input
2
3
-1

Sample Output
2
5

  #include   
#include   
#include   
using namespace std; 
int n; 
#define BASE 10000  
#define UNIT 4  
#define FORMAT "%04d"  
 
class BigNum{ 
public: 
    int a[20]; 
    int length; 
    BigNum(const int k){ //用小于BASE的k初始化大数  
        memset(a, 0, sizeof(a)); 
        a[0] = k; 
        length = 1; 
    } 
    BigNum(){ 
        memset(a, 0, sizeof(a)); 
        length = 0; 
    } 
    BigNum operator * (const BigNum & B){ 
        BigNum ans; 
        int i,j,up=0,num; 
        for(i=0; i 0) 
                ans.a[i+j] = up; 
        } 
        ans.length = i+j; 
        while(ans.a[ans.length -1] == 0 && ans.length > 1) 
            ans.length--; 
        return ans; 
    } 
    BigNum operator /(const int & k) const{  // k < BASE, 对此题适用  
        BigNum ans; 
        int down=0,i,num; 
        for(i=length-1; i>=0; i--){ 
            num = ( (down * BASE) + a[i] ) / k; 
            down =  ( (down * BASE) + a[i] ) % k; 
            ans.a[i] = num; 
        } 
        ans.length = length; 
        while(ans.a[ans.length-1] == 0 && ans.length > 1) 
            ans.length -- ; 
        return ans; 
    } 
    void print(){ 
        printf("%d", a[length-1]); 
        for(int i=length-2; i>=0; i--) 
            printf(FORMAT,a[i]); 
    } 
}; 
 
//f(n) = C(2n,n)/(n+1)  
int main(){ 
    BigNum nums[101]; 
    nums[1] = BigNum(1); 
    nums[2] = BigNum(2); 
    nums[3] = BigNum(5); 
    for(int i=4; i<=100; i++){ 
        nums[i] = nums[i-1] * (4*i-2)/(i+1); 
    } 
    int n; 
    while(scanf("%d", &n), n>0){ 
        nums[n].print(); 
        printf("\n"); 
    } 
    return 0; 
} 

#include 
#include 
#include 
using namespace std;
int n;
#define BASE 10000
#define UNIT 4
#define FORMAT "%04d"

class BigNum{
public:
 int a[20];
 int length;
 BigNum(const int k){ //用小于BASE的k初始化大数
  memset(a, 0, sizeof(a));
  a[0] = k;
  length = 1;
 }
 BigNum(){
  memset(a, 0, sizeof(a));
  length = 0;
 }
 BigNum operator * (const BigNum & B){
  BigNum ans;
  int i,j,up=0,num;
  for(i=0; i 0)
    ans.a[i+j] = up;
  }
  ans.length = i+j;
  while(ans.a[ans.length -1] == 0 && ans.length > 1)
   ans.length--;
  return ans;
 }
 BigNum operator /(const int & k) const{  // k < BASE, 对此题适用
   BigNum ans;
  int down=0,i,num;
  for(i=length-1; i>=0; i--){
   num = ( (down * BASE) + a[i] ) / k;
   down =  ( (down * BASE) + a[i] ) % k;
   ans.a[i] = num;
  }
  ans.length = length;
  while(ans.a[ans.length-1] == 0 && ans.length > 1)
   ans.length -- ;
  return ans;
 }
 void print(){
  printf("%d", a[length-1]);
  for(int i=length-2; i>=0; i--)
   printf(FORMAT,a[i]);
 }
};

//f(n) = C(2n,n)/(n+1)
int main(){
 BigNum nums[101];
 nums[1] = BigNum(1);
 nums[2] = BigNum(2);
 nums[3] = BigNum(5);
 for(int i=4; i<=100; i++){
  nums[i] = nums[i-1] * (4*i-2)/(i+1);
 }
 int n;
 while(scanf("%d", &n), n>0){
  nums[n].print();
  printf("\n");
 }
 return 0;
}