POJ 2506 Tilling

大意不再赘述。

思路：把前几个元素都推出来，然后高精度即可，可得到 ans[i] = ans[i-1] + ans[i-2]*2;

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
using namespace std;

const int MAXN = 1100;
int n;

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 = (const char *num)
	{
		len = strlen(num);
		for(int i = 0; i < len; i++) s[i] = num[len-i-1] - '0';
		return *this;
	}
	bign operator = (int num)
	{
		char s[MAXN];
		sprintf(s, "%d", num);
		*this = s;
		return *this;
	}
	bign operator +(const bign &b) const
	{
		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;
	}
	bign operator += (const bign &b)
	{
		*this = *this + b;
		return *this;
	}
	void clean()
	{
		while(len > 1 && !s[len-1]) len--;
	}
	bign operator * (const bign &b) const
	{
		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 < len; i++)
		{
			c.s[i+1] += c.s[i] / 10;
			c.s[i] %= 10;
		}
		c.clean();
		return c;
	}
	bign operator - (const bign &b) const
	{
		bign c;
		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;
	}
	bign operator / (const bign &b) const
	{
		bign c, f = 0;
		for(int i = 0; i < len; i++)
		{
			f = f*10;
			f.s[0] = s[i];
			while(f >= b)
			{
				f = f-b;
				c.s[i]++;
			}
		}
		c.len = len;
		c.clean();
		return c;
	}
	bool operator < (const bign &b) const
	{
		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
	{
		return !(*this < b);
	}
	void print()
	{
		for(int i = len-1; i >= 0; i--) printf("%d", s[i]);
		printf("\n");
	}
};

bign ans[300];

void init()
{
	ans[0] = 1;
	ans[1] = 1;
	for(int i = 2; i <= 250; i++)
	{
		ans[i] = ans[i-1] + ans[i-2] * 2;
	}
}

int main()
{
	init();
	while(~scanf("%d", &n))
	{
		ans[n].print();
	}
	return 0;
}