HDU 6185Covering（矩阵快速幂）

Covering

Time Limit : 5000/2500ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 2   Accepted Submission(s) : 2

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

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

12

Sample Output

15

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

想法：

1. 推导公式:F[i]   = F[i-1] + 5*F[i-2] + F[i-3] - F[i-4] 

代码;

#include<bits/stdc++.h>using namespace std;typedef long long ll;const int maxn = 4;const ll mod= 1e9+7;struct Matrix{    ll tmp[maxn][maxn];}a;void init(){    for(int i=0;i<maxn;i++)    {        for(int j=0;j<maxn;j++)        {            a.tmp[i][j]=0;        }    }    a.tmp[0][0]=a.tmp[0][2]=1;    a.tmp[0][1]=5;    a.tmp[0][3]=-1;    a.tmp[1][0]=a.tmp[2][1]=a.tmp[3][2]=1;    return;}Matrix mul(Matrix a,Matrix b){    Matrix ans;    for(int i=0;i<maxn;i++)        for(int j=0;j<maxn;j++)        {            ans.tmp[i][j]=0;            for(int k=0;k<maxn;k++)            {                ans.tmp[i][j]+=(a.tmp[i][k]*b.tmp[k][j]+mod)%mod;                ans.tmp[i][j]%=mod;            }        }    return ans;}void fun(Matrix ans,ll k){    for(int i=0;i<maxn;i++)    {        for(int j=0;j<maxn;j++)        {            a.tmp[i][j]=(i==j);        }    }    while(k)    {        if(k%2)  a=mul(a,ans);        ans=mul(ans,ans);        k/=2;    }    return ;}int main(){    Matrix t;    ll n;    for(int i=0;i<maxn;i++)    {        for(int j=0;j<maxn;j++)        {            t.tmp[i][j]=0;        }    }    t.tmp[0][0]=36;    t.tmp[1][0]=11;    t.tmp[2][0]=5;    t.tmp[3][0]=1;    while(~scanf("%lld",&n))    {        init();        if(n<=4)        {            printf("%lld\n",t.tmp[4-n][0]);            continue;        }        fun(a,n-4);        a=mul(a,t);        printf("%lld\n",a.tmp[0][0]%mod);    }    return 0;}