FZU Problem 1147 Tiling

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博客围绕Problem 1147 Tiling问题，即使用2x1或2x2瓷砖铺设2xn矩形的方式数量。输入为0到250的整数n，需输出对应铺设方式数量。因n范围大，long long存不下，解决方案是找规律，用大数相加，公式为F(n)=F(n - 2)*2+F(n - 1) 。

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Problem 1147 Tiling

Accept: 495 Submit: 1261 Time Limit: 1000 mSec Memory Limit : 32768 KB

Problem 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

Sample Output

3

171

2731

Source

Albert 2001

解决方案

找规律

由于n的范围太大了，long long都存不下，所以要用到大数相加

F(n)=F(n-2)*2+F(n-1)

F(0)=1 F(1)=1 F(2)=3

#include<iostream>
using namespace std;
int F[251][100];
void init() {
	memset(F, 0, sizeof(F));
	F[1][0] = 1;
	F[2][0] = 3;
	int i, c, j;
	for (i = 3; i <= 250; i++) {
		c = 0;
		for (j = 0; j <= 100; j++)
		{
			F[i][j] = F[i - 1][j] + 2 * F[i - 2][j] + c;
			c = F[i][j] / 10;
			F[i][j] %= 10;
		}
	}
}
int main() {
	int n, i, j;
	init();
	while (cin >> n) {
		if (n == 0)
			cout << 1 << endl;
		else
		{
			for (i = 99; i >= 0; i--)
				if (F[n][i])
					break;
			for (j = i; j >= 0; j--)
				cout << F[n][j];
			cout << endl;
		}
	}
}