POJ2663（矩阵+递推）

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最新推荐文章于 2021-01-18 10:51:21 发布

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Tri Tiling

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9786		Accepted: 5023

Description

In how many ways can you tile a 3xn rectangle with 2x1 dominoes?
Here is a sample tiling of a 3x12 rectangle.

Input

Input consists of several test cases followed by a line containing -1. Each test case is a line containing an integer 0 <= n <= 30.

Output

For each test case, output one integer number giving the number of possible tilings.

Sample Input

Sample Output

3
153
2131

Source

Waterloo local 2005.09.24

n为奇数肯定为0，n为偶数，每次都是加两列，我们把两列看为一列，如果这一列与前面分开就只有三种方法即3*a[n-2],如果这一列不与前面的分开，那么不可分解矩形都只有两种情况所以为2*（a[n-4]+a[n-6]+……a[0])

化简即为a[n]=4*a[n-2]-a[n-4]

#include<iostream>
#include<cstdlib>
#include<stdio.h>
#define ll long long
using namespace std;
ll a[31];
int main()
{
    ll n;
    a[0]=1;a[2]=3;
    for(int i=4;i<=30;i+=2)
    a[i]=4*a[i-2]-a[i-4];
    while(scanf("%lld",&n))
    {
        if(n==-1) break;
        printf("%lld\n",a[n]);
    }
}