HDU Train Problem II （卡特兰数+大数）

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.
 
 


卡特兰数裸题，因为数据大，要用大数来做;
卡特兰数的知识链接—>http://www.cnblogs.com/kuangbin/archive/2012/03/21/2410516.html

下面附上代码
#include<iostream> 
#include<string> 
#include<iomanip> 
#include<cstring>
#include<algorithm> 
using namespace std; 

#define MAXN 9999
#define MAXSIZE 10
#define DLEN 4

class BigNum
{ 
private: 
	int a[500];    //可以控制大数的位数 
	int len;       //大数长度
public: 
	BigNum(){ len = 1;memset(a,0,sizeof(a)); }   //构造函数
	BigNum(const int);       //将一个int类型的变量转化为大数
	BigNum(const char*);     //将一个字符串类型的变量转化为大数
	BigNum(const BigNum &);  //拷贝构造函数
	BigNum &operator=(const BigNum &);   //重载赋值运算符，大数之间进行赋值运算

	friend istream& operator>>(istream&,  BigNum&);   //重载输入运算符
	friend ostream& operator<<(ostream&,  BigNum&);   //重载输出运算符

	BigNum operator+(const BigNum &) const;   //重载加法运算符，两个大数之间的相加运算 
	BigNum operator-(const BigNum &) const;   //重载减法运算符，两个大数之间的相减运算 
	BigNum operator*(const BigNum &) const;   //重载乘法运算符，两个大数之间的相乘运算 
	BigNum operator/(const int   &) const;    //重载除法运算符，大数对一个整数进行相除运算

	BigNum operator^(const int  &) const;    //大数的n次方运算
	int    operator%(const int  &) const;    //大数对一个int类型的变量进行取模运算    
	bool   operator>(const BigNum & T)const;   //大数和另一个大数的大小比较
	bool   operator>(const int & t)const;      //大数和一个int类型的变量的大小比较

	void print();       //输出大数
}; 
BigNum::BigNum(const int b)     //将一个int类型的变量转化为大数
{ 
	int c,d = b;
	len = 0;
	memset(a,0,sizeof(a));
	while(d > MAXN)
	{
		c = d - (d / (MAXN + 1)) * (MAXN + 1); 
		d = d / (MAXN + 1);
		a[len++] = c;
	}
	a[len++] = d;
}
BigNum::BigNum(const char*s)     //将一个字符串类型的变量转化为大数
{
	int t,k,index,l,i;
	memset(a,0,sizeof(a));
	l=strlen(s);   
	len=l/DLEN;
	if(l%DLEN)
		len++;
	index=0;
	for(i=l-1;i>=0;i-=DLEN)
	{
		t=0;
		k=i-DLEN+1;
		if(k<0)
			k=0;
		for(int j=k;j<=i;j++)
			t=t*10+s[j]-'0';
		a[index++]=t;
	}
}
BigNum::BigNum(const BigNum & T) : len(T.len)  //拷贝构造函数
{ 
	int i; 
	memset(a,0,sizeof(a)); 
	for(i = 0 ; i < len ; i++)
		a[i] = T.a[i]; 
} 
BigNum & BigNum::operator=(const BigNum & n)   //重载赋值运算符，大数之间进行赋值运算
{
	int i;
	len = n.len;
	memset(a,0,sizeof(a)); 
	for(i = 0 ; i < len ; i++) 
		a[i] = n.a[i]; 
	return *this; 
}
istream& operator>>(istream & in,  BigNum & b)   //重载输入运算符
{
	char ch[MAXSIZE*4];
	int i = -1;
	in>>ch;
	int l=strlen(ch);
	int count=0,sum=0;
	for(i=l-1;i>=0;)
	{
		sum = 0;
		int t=1;
		for(int j=0;j<4&&i>=0;j++,i--,t*=10)
		{
			sum+=(ch[i]-'0')*t;
		}
		b.a[count]=sum;
		count++;
	}
	b.len =count++;
	return in;

}
ostream& operator<<(ostream& out,  BigNum& b)   //重载输出运算符
{
	int i;  
	cout << b.a[b.len - 1]; 
	for(i = b.len - 2 ; i >= 0 ; i--)
	{ 
		cout.width(DLEN); 
		cout.fill('0'); 
		cout << b.a[i]; 
	} 
	return out;
}

BigNum BigNum::operator+(const BigNum & T) const   //两个大数之间的相加运算
{
	BigNum t(*this);
	int i,big;      //位数   
	big = T.len > len ? T.len : len; 
	for(i = 0 ; i < big ; i++) 
	{ 
		t.a[i] +=T.a[i]; 
		if(t.a[i] > MAXN) 
		{ 
			t.a[i + 1]++; 
			t.a[i] -=MAXN+1; 
		} 
	} 
	if(t.a[big] != 0)
		t.len = big + 1; 
	else
		t.len = big;   
	return t;
}
BigNum BigNum::operator-(const BigNum & T) const   //两个大数之间的相减运算 
{  
	int i,j,big;
	bool flag;
	BigNum t1,t2;
	if(*this>T)
	{
		t1=*this;
		t2=T;
		flag=0;
	}
	else
	{
		t1=T;
		t2=*this;
		flag=1;
	}
	big=t1.len;
	for(i = 0 ; i < big ; i++)
	{
		if(t1.a[i] < t2.a[i])
		{ 
			j = i + 1; 
			while(t1.a[j] == 0)
				j++; 
			t1.a[j--]--; 
			while(j > i)
				t1.a[j--] += MAXN;
			t1.a[i] += MAXN + 1 - t2.a[i]; 
		} 
		else
			t1.a[i] -= t2.a[i];
	}
	t1.len = big;
	while(t1.a[t1.len - 1] == 0 && t1.len > 1)
	{
		t1.len--; 
		big--;
	}
	if(flag)
		t1.a[big-1]=0-t1.a[big-1];
	return t1; 
} 

BigNum BigNum::operator*(const BigNum & T) const   //两个大数之间的相乘运算 
{ 
	BigNum ret; 
	int i,j,up; 
	int temp,temp1;   
	for(i = 0 ; i < len ; i++)
	{ 
		up = 0; 
		for(j = 0 ; j < T.len ; j++)
		{ 
			temp = a[i] * T.a[j] + ret.a[i + j] + up; 
			if(temp > MAXN)
			{ 
				temp1 = temp - temp / (MAXN + 1) * (MAXN + 1); 
				up = temp / (MAXN + 1); 
				ret.a[i + j] = temp1; 
			} 
			else
			{ 
				up = 0; 
				ret.a[i + j] = temp; 
			} 
		} 
		if(up != 0) 
			ret.a[i + j] = up; 
	} 
	ret.len = i + j; 
	while(ret.a[ret.len - 1] == 0 && ret.len > 1)
		ret.len--; 
	return ret; 
} 
BigNum BigNum::operator/(const int & b) const   //大数对一个整数进行相除运算
{ 
	BigNum ret; 
	int i,down = 0;   
	for(i = len - 1 ; i >= 0 ; i--)
	{ 
		ret.a[i] = (a[i] + down * (MAXN + 1)) / b; 
		down = a[i] + down * (MAXN + 1) - ret.a[i] * b; 
	} 
	ret.len = len; 
	while(ret.a[ret.len - 1] == 0 && ret.len > 1)
		ret.len--; 
	return ret; 
}
int BigNum::operator %(const int & b) const    //大数对一个int类型的变量进行取模运算    
{
	int i,d=0;
	for (i = len-1; i>=0; i--)
	{
		d = ((d * (MAXN+1))% b + a[i])% b;  
	}
	return d;
}
BigNum BigNum::operator^(const int & n) const    //大数的n次方运算
{
	BigNum t,ret(1);
	int i;
	if(n<0)
		exit(-1);
	if(n==0)
		return 1;
	if(n==1)
		return *this;
	int m=n;
	while(m>1)
	{
		t=*this;
		for( i=1;i<<1<=m;i<<=1)
		{
			t=t*t;
		}
		m-=i;
		ret=ret*t;
		if(m==1)
			ret=ret*(*this);
	}
	return ret;
}
bool BigNum::operator>(const BigNum & T) const   //大数和另一个大数的大小比较
{ 
	int ln; 
	if(len > T.len)
		return true; 
	else if(len == T.len)
	{ 
		ln = len - 1; 
		while(a[ln] == T.a[ln] && ln >= 0)
			ln--; 
		if(ln >= 0 && a[ln] > T.a[ln])
			return true; 
		else
			return false; 
	} 
	else
		return false; 
}
bool BigNum::operator >(const int & t) const    //大数和一个int类型的变量的大小比较
{
	BigNum b(t);
	return *this>b;
}

void BigNum::print()    //输出大数
{ 
	int i;   
	cout << a[len - 1]; 
	for(i = len - 2 ; i >= 0 ; i--)
	{ 
		cout.width(DLEN); 
		cout.fill('0'); 
		cout << a[i]; 
	} 
	cout << endl;
}
int main(void)    
{
	int i,j,n;
	BigNum f[1000];
	f[1]=1;
	for(i=2;i<=100;i++) {
		f[i]=f[i-1]*(4*i-2)/(i+1);    //核心代码就这么一句话，，写了辣么长的代码  也是醉了
	}
	while(scanf("%d",&n)!=EOF){
		cout<<f[n]<<endl;
	}
	return 0;
}