矩阵快速幂习题练习——Jzzhu and Sequences

题目

Jzzhu has invented a kind of sequences, they meet the following property:

You are given x and y, please calculate f n modulo 1000000007 (109 + 7).

输入

The first line contains two integers x and y (|x|, |y| ≤ 109). The second line contains a single integer n (1 ≤ n ≤ 2·109).

输出

Output a single integer representing f n modulo 1000000007 (109 + 7).

示例

输入

2 3
3

输出

1

输入

0 -1
2

输出

1000000006

Note

In the first sample, f 2 = f 1 + f 3, 3 = 2 + f 3, f 3 = 1.

In the second sample, f 2 =  - 1;  - 1 modulo (109 + 7) equals (109 + 6).

解题

方法一：找规律

#include 

using namespace std;

const int MOD = 1000000007;
const int N = 6;

int ans[N];

int main()
{
    int x, y, n;

    cin >> x >> y >> n;

    // Init ans
    ans[0] = x - y;
    ans[1] = x;
    ans[2] = y;
    ans[3] = y - x;
    ans[4] = -x;
    ans[5] = -y;

    int result = ans[n % N];
    if (result >= 0)
        cout << result % MOD << endl;
    else
        cout << (result % MOD + MOD) % MOD << endl;

    return 0;
}

方法二：矩阵快速幂

#include 
#include 

using namespace std;

const int MOD = 1000000007;
const int N = 2;

class Matrix
{
public:
    int n;
    long long int m[N][N];

    Matrix()
    {
        memset(m, 0, sizeof(m));
    }

    Matrix operator*(const Matrix &y)
    {
        Matrix z;

        for (int i = 0; i < N; i++)
            for (int j = 0; j < N; j++)
                for (int k = 0; k < N; k++)
                {
                    z.m[i][j] += m[i][k]%MOD * y.m[k][j]%MOD;
                    z.m[i][j] %= MOD;
                }

        return z;
    }
};

Matrix Matrix_Powmul(int x, int y, Matrix t, int n)
{
    Matrix z;
    z.m[0][0] = y;
    z.m[0][1] = x;
    z.m[1][0] = 0;
    z.m[1][1] = 0;

    while (n)
    {
        if (n & 1)
        {
            z = z * t;
        }
        n >>= 1;
        t = t*t;
    }

    return z;
}

int main()
{
    int x, y, n;
    int result;

    cin >> x >> y >> n;

    if (n==1)
        result = x;
    else if (n==2)
        result = y;
    else
    {
        Matrix t;
        t.m[0][0] = 1;
        t.m[0][1] = 1;
        t.m[1][0] = -1;
        t.m[1][1] = 0;

        Matrix f = Matrix_Powmul(x, y, t, n-2);

        result = f.m[0][0]%MOD;
    }

    if (result>=0)
        result%=MOD;
    else
    {
        result = result + MOD; // 语句：result = f.m[0][0]%MOD; 已经取模。
    }

    cout<<result<<endl;
}