POJ 2084

第一题组合数学题。可以使用递推，设1与其他各数分别连边，假设N=3;若1-4，则圆分成两部分计数，此时可以利用乘法原理。（高精度）

#include <cstdio>
#include <cstring>
#include <iostream>
#include <string>

using namespace std;

const int maxn = 200;
struct bign {
	int len, s[maxn];

	bign() {
		memset(s, 0, sizeof(s));
		len = 1;
	}

	bign(int num) {
		*this = num;
	}

	bign(const char* num) {
		*this = num;
	}

	bign operator =(int num) { //直接以整数赋值
		char s[maxn];
		sprintf(s, "%d", num);
		*this = s;
		return *this;
	}

	bign operator =(const char* num) { //以字符串赋值
		len = strlen(num);
		for(int i = 0; i < len; i++)
			s[i] = num[len - i - 1] - '0';
		return *this;
	}

	string str() const { //将bign转化成字符串
		string res = "";
		for(int i = 0; i < len; i++)
			res = (char) (s[i] + '0') + res;
		if(res == "")
			res = "0";
		return res;
	}

	bign operator +(const bign& b) const { //重载+号运算
		bign c;
		c.len = 0;
		for(int i = 0, g = 0; g || i < max(len, b.len); i++) {
			int x = g;
			if(i < len) x += s[i];
			if(i < b.len) x += b.s[i];
			c.s[c.len++] = x % 10;
			g = x / 10;
		}
		return c;
	}

	void clean() { //去掉前到0
		while(len > 1 && !s[len - 1])
			len--;
	}

	bign operator *(const bign& b) { //重载*号运算
		bign c;
		c.len = len + b.len;
		for(int i = 0; i < len; i++)
			for(int j = 0; j < b.len; j++)
				c.s[i + j] += s[i] * b.s[j];
		for(int i = 0; i < c.len - 1; i++) {
			c.s[i + 1] += c.s[i] / 10;
			c.s[i] %= 10;
		}
		c.clean();
		return c;
	}

	bign operator -(const bign& b) { //重载-号运算
		bign c;
		c.len = 0;
		for(int i = 0, g = 0; i < len; i++) {
			int x = s[i] - g;
			if(i < b.len)
				x -= b.s[i];
			if(x >= 0)
				g = 0;
			else {
				g = 1;
				x += 10;
			}
			c.s[c.len++] = x;
		}
		c.clean();
		return c;
	}

	bool operator <(const bign& b) const { //重载<号运算
		if(len != b.len)
			return len < b.len;
		for(int i = len - 1; i >= 0; i--)
			if(s[i] != b.s[i])
				return s[i] < b.s[i];
		return false;
	}

	bool operator >(const bign& b) const { //重载>号运算
		return b < *this;
	}

	bool operator <=(const bign& b) { //重载<=号运算
		return !(b > *this);
	}

	bool operator ==(const bign& b) { //重载>=号运算
		return !(b < *this) && !(*this < b);
	}

	bign operator +=(const bign& b) { //重载+=号运算
		*this = *this + b;
		return *this;
	}
};

istream& operator >>(istream &in, bign& x) { //重载输入运算符
	string s;
	in >> s;
	x = s.c_str();
	return in;
}

ostream& operator <<(ostream &out, const bign& x) { //重载输出运算符
	out << x.str();
	return out;
}



bign ans[210];

void initial(){
	ans[2]=1;
	ans[4]=2;
	for(int i=5;i<=200;i++){
		ans[i]=ans[i-2]+ans[i-2];
		for(int j=3;j<i;j++){
			ans[i]=ans[i]+ans[j-2]*ans[i-j];
		}
	}
}

int main(){
	initial();
	int n;
	while(scanf("%d",&n),n!=-1){
		cout<<ans[2*n]<<endl;
	}
	return 0;
}