poj 2506

Tiling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8761		Accepted: 4209

Description

In how many ways can you tile a 2xn rectangle by 2x1 or 2x2 tiles? 
Here is a sample tiling of a 2x17 rectangle. 

Input

Input is a sequence of lines, each line containing an integer number 0 <= n <= 250.

Output

For each line of input, output one integer number in a separate line giving the number of possible tilings of a 2xn rectangle. 

Sample Input

2
8
12
100
200

Sample Output

3
171
2731
845100400152152934331135470251
1071292029505993517027974728227441735014801995855195223534251

分析:当铺垫的是ans[i]时，无非两种情形：ans[i-1]则只能在该基础上+2*1大小的砖块

      ans[i-2]则在该基础上+两块2*1的或一块2*2的

因而得到递推公式，然而从样例中可以看到。

这题还需要高精度计算，套刘爷的模板吧。

code:

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include <deque>
#include <vector>
using namespace std;

struct Bigint {
	static const int BASE = 100000000;
	static const int WIDTH = 8;
	vector<int> s;

	Bigint(long long num = 0) { *this = num; } //构造函数
	Bigint operator = (long long num) {
		s.clear();
		do {
			s.push_back(num%BASE);
			num /= BASE;
		} while (num > 0);
		return *this;
	}
	Bigint operator =(const string& str) {
		s.clear();
		int x, len = (str.length() - 1) / WIDTH + 1;
		for (int i = 0; i < len; i++) {
			int end = str.length() - i*WIDTH;
			int start = max(0, end - WIDTH);
			sscanf(str.substr(start, end - start).c_str(), "%d", &x);
			s.push_back(x);
		}
		return *this;
	}
	Bigint operator + (const Bigint& b) const {
		Bigint c;
		c.s.clear();
		for (int i = 0, g = 0;; i++) {
			if (g == 0 && i >= s.size() && i >= b.s.size()) break;
			int x = g;
			if (i < s.size()) x += s[i];
			if (i < b.s.size()) x += b.s[i];
			c.s.push_back(x%BASE);
			g = x / BASE;
		}
		return c;
	}
	Bigint operator * (const Bigint& b) const {
		Bigint c;
		c.s.clear();
		int x = 1;
		for (int i = 0, g = 0;; i++) {
			if (g == 0 && i >= s.size() && i >= b.s.size()) break;
			if (i < s.size()) x *= s[i];
			if(i < b.s.size()) x *= b.s[i];
			c.s.push_back(x%BASE);
			g = x / BASE;
		}
		return c;
	}
};

ostream& operator << (ostream &out, const Bigint& x)
{
	out << x.s.back();
	for (int i = x.s.size() - 2; i >= 0; i--)
	{
		char buf[20];
		sprintf(buf, "%08d", x.s[i]);
		for (int j = 0; j < strlen(buf); j++) out << buf[j];
	}
	return out;
}

istream& operator >> (istream &in, Bigint& x)
{
	string s;
	if (!(in >> s)) return in;
	x = s;
	return in;
}

int main(void)
{
	Bigint ans[255];
	ans[0] = 1;
	ans[1] = 1;
	//cout << ans[0] * ans[1] << endl;
	for (int i = 2; i <= 250; i++)
	{
		ans[i] = ans[i-2]+ans[i-2] + ans[i-1];
	}
	
	int n;
	while (cin >> n)
	{
		cout << ans[n] << endl;
	}
}