递推求值

最新推荐文章于 2019-06-13 23:56:59 发布

dongdongdong122

最新推荐文章于 2019-06-13 23:56:59 发布

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分类专栏：递推文章标签：递推

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Covering

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 481 Accepted Submission(s): 226

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

Source

2017ACM/ICPC广西邀请赛-重现赛（感谢广西大学）

根据转移的状态列出来式子，可以把无用的状态去掉，注意式子一定不要漏掉，漏掉的话就gg，比直接推数据不知道高到哪里去了呢。

代码：

#include<stdio.h>
#include<string.h>
#define mod 1000000007
#define N 6
typedef long long LL;
struct Matrix
{
    LL mat[N][N];
};

Matrix unit_matrix =
{
    1, 0, 0,0,0,0,
    0, 1, 0,0,0,0,
    0, 0, 1,0,0,0,
    0, 0, 0,1,0,0,
    0, 0, 0,0,1,0,
    0, 0, 0,0,0,1
}; //单位矩阵

Matrix mul(Matrix a, Matrix b) //矩阵相乘
{
    Matrix res;
    for(int i = 0; i < N; i++)
        for(int j = 0; j < N; j++)
        {
            res.mat[i][j] = 0;
            for(int k = 0; k < N; k++)
            {
                res.mat[i][j] += a.mat[i][k] * b.mat[k][j];
                res.mat[i][j] %= mod;
            }
        }
    return res;
}

Matrix pow_matrix(Matrix a, LL n)  //矩阵快速幂
{
    Matrix res = unit_matrix;
    while(n != 0)
    {
        if(n & 1)
            res = mul(res, a);
        a = mul(a, a);
        n >>= 1;
    }
    return res;
}

int main()
{
    LL n;
    Matrix tmp, arr;

    while(scanf("%I64d",&n)!=EOF)
    {
//        scanf("%lld%lld%lld%lld%lld%lld",&f1, &f2, &a, &b, &c, &n);
        memset(arr.mat, 0, sizeof(arr.mat));
        memset(tmp.mat, 0, sizeof(tmp.mat));
        arr.mat[0][0] = 1;
        arr.mat[1][0] = 1;
        arr.mat[2][0] = 0;
        arr.mat[3][0] = 1;
        arr.mat[4][0] = 1;
        arr.mat[5][0] = 1;
        tmp.mat[0][5] = 1,tmp.mat[0][0]=1,tmp.mat[0][4] = 1,tmp.mat[0][1]=1,tmp.mat[0][3]=1;
        tmp.mat[1][0] = 1,tmp.mat[1][4]=1;
        tmp.mat[2][3]=1;
        tmp.mat[3][0]=1,tmp.mat[3][2]=1;
        tmp.mat[4][0]=1,tmp.mat[4][1]=1;
        tmp.mat[5][0]=1;
        tmp.mat[1][1] = tmp.mat[1][2] = tmp.mat[2][0] = tmp.mat[2][1] = 0;
        Matrix p = pow_matrix(tmp, n-1);
        p = mul(p, arr);
        LL ans = (p.mat[0][0] ) % mod;
        printf("%I64d\n",ans);
    }
}