【2017广西邀请赛】hdu 6185 Covering 矩阵快速幂

Problem Description

Bob's school has a big playground, boys and girls always play games here after school.

To protect boys and girls from getting hurt when playing happily on the playground, rich boy Bob decided to cover the playground using his carpets.

Meanwhile, Bob is a mean boy, so he acquired that his carpets can not overlap one cell twice or more.

He has infinite carpets with sizes of 1×2 and 2×1,
and the size of the playground is 4×n.

Can you tell Bob the total number of schemes where the carpets can cover the playground completely without overlapping?

 

Input

There are no more than 5000 test cases. 

Each test case only contains one positive integer n in a line.

1≤n≤1018

 

Output

For each test cases, output the answer mod 1000000007 in a line.

 

Sample Input

1
2

 

Sample Output

1
5

 

题意：

用1*2规格的地毯去铺4*n规格的地面，告诉你n，问有多少种不同的方案使得地面恰好被铺满且地毯不重叠。

思路：

骨牌覆盖的方法构造的矩阵太大，会超时。

手动推出递归式 f
=f[n-1]+5f[n-2]+f[n-3]-f[n-4]。然后再用矩阵快速幂即可。

推导过程：http://blog.csdn.net/elbadaernu/article/details/77825979

//
// main.cpp
// 1004
//
// Created by zc on 2017/9/6.
// Copyright © 2017年 zc. All rights reserved.
//
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<algorithm>
#define ll long long

using namespace std;

typedef vector<ll> vec;
typedef vector<vec> mat;
const ll MOD=1000000007;

mat mul(mat &A,mat &B)
{
mat C(A.size(),vec(B[0].size()));
for(int i=0;i<A.size();i++)
for(int k=0;k<B.size();k++)
for(int j=0;j<B[0].size();j++)
C[i][j]=(C[i][j]+A[i][k]*B[k][j])%MOD;
return C;
}

mat pow(mat A, ll n)
{
mat B(A.size(),vec(A.size()));
for(int i=0;i<A.size();i++) B[i][i]=1;
while(n>0)
{
if(n&1) B=mul(B,A);
A=mul(A,A);
n>>=1;
}
return B;
}

int ans[5]={0,1,5,11,36};

int main()
{
ll n;
mat A(4,vec(4));
for(int i=0;i<4;i++)
for(int j=0;j<4;j++)
A[i][j]=0;
A[0][0]=A[0][1]=A[1][2]=A[2][0]=A[2][3]=1;
A[1][0]=5;A[3][0]=-1;
while(~scanf("%lld",&n))
{
if(n<5)
{
printf("%d\n",ans
);
continue;
}
mat C=pow(A,n-4);
mat B(1,vec(4));
B[0][0]=36;B[0][1]=11;B[0][2]=5;B[0][3]=1;
B=mul(B,C);
printf("%lld\n",(B[0][0]+MOD)%MOD);
}
}