HDU 6050 Funny Function (矩阵快速幂)

Description

函数\(F_{x,y}\)满足：

\(F_{1,1}=F_{1,2}=1\)
\(F_{1,j}=F_{1,j-1}+2F_{1,j-2}\ (j\geqslant 3)\)
\(F_{i,j}=\sum_{k=j}^{j+N-1}{F_{i-1,k}}\ (i \geqslant 2, j \geqslant 1)\)

给出整数\(N\)和\(m\)，计算\(F_{m,1}\ mod\ 1e9+7\)。

Input

第一行一个整数\(T\)表示用例组数，接下来\(T\)行，每行两个整数\(N\)和\(m\)。\(1 \leqslant N,m \leqslant 2^{63}\)。

Output

对于每组用例输出\(F_{m,1}\ mod\ 1e9+7\)。

Sample Input

2
2 2
3 3

Sample Output

2
33

Solution

\(F_{1,j}=F_{i,j-1}+2F_{i,j-2}\ (i \geqslant 1, j \geqslant 3)\\\)

\(\sum_{k=1}^j{F_{i,k}}=F_{i,j+1}-F_{i,1}\ (j为偶数)\\ \sum_{k=1}^j{F_{i,k}}=F_{i,j+1}+F_{i,1}-F_{i,2}\ (j为奇数)\)

\((F_{i,1},F_{i,2})=(F_{i-1,n+1}-F_{i-1,1},F_{i-1,n+2}-F_{i-1,2})\)
\(=(F_{i-1,n+1},F_{i-1,n+2})-(F_{i-1,1},F_{i-1,2})\)
\(=(F_{i-1,1},F_{i-1,2})A^n-(F_{i-1,1},F_{i-1,2})\)
\(=(F_{i-1,1},F_{i-1,2})(A^n-B_0)\ (n为偶数)\)

\((F_{m,1},F_{m,2})=(F_{1,1},F_{1,2})(A^n-B_0)^{m-1}\ (n为偶数)\)

\((F_{m,1},F_{m,2})=(F_{1,1},F_{1,2})(A^n-B_1)^{m-1}\ (n为奇数)\)

\(其中A，B_0，B_1均为二阶矩阵\)

Code

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const ll mod = 1e9 + 7;
const int N = 5;

const int nn = 2;
struct Matrix
{
    ll a[5][5], n;
    Matrix(ll x = 0, ll b = 0, ll c = 0, ll d = 0)
    {
        n = 2;
        a[1][1] = x; a[1][2] = b;
        a[2][1] = c; a[2][2] = d;
    }
    ll* operator[](int i)
    {
        return a[i];
    }
};

Matrix operator-(Matrix A, Matrix B)
{
    ll n = A.n;
    Matrix C;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            C[i][j] = ((A[i][j] - B[i][j]) % mod + mod) % mod;
    return C;
}

Matrix operator*(Matrix A, Matrix B)
{
    int n = A.n;
    Matrix C;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
            for (int k = 1; k <= n; k++)
                C[i][j] = (C[i][j] + A[i][k] * B[k][j] % mod) % mod;
    return C;
}

Matrix power(Matrix A, ll n, ll mod)
{
    Matrix C(1, 0, 0, 1);
    while (n)
    {
        if (n & 1) C = C * A;
        A = A * A;
        n >>= 1;
    }
    return C;
}

int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        ll n, m;
        scanf("%lld%lld", &n, &m);
        Matrix A(0, 2, 1, 1);
        Matrix B = n & 1 ? Matrix(-1, 2, 1, 0) : Matrix(1, 0, 0, 1);
        A = power(A, n, mod) - B;
        A = power(A, m - 1, mod);
        ll ans = (A[1][1] + A[2][1]) % mod;
        printf("%lld\n", ans);
    }
    return 0;
}

Link

http://acm.hdu.edu.cn/showproblem.php?pid=6050