矩阵快速幂模板

其实跟普通的快速幂类似,只是普通乘法换成了矩阵乘法,所以时间复杂度为 $O(k^3logn)$($k$为矩阵大小)

#include<bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef vector<ll>  vec;
typedef vector<vec> mat;


const int mod = 1000000000 + 7;

void init(mat A)
{
    for(int i = 0;i < A.size();i++)
        A[i].clear();
}

//计算A*B%mod
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[0].size();k++)
            for(int j = 0;j < B[0].size();j++)
                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod;  //矩阵的各元素对mod取余
    return C;
}

//计算A^n
mat pow(mat A, ll n)
{
    mat B(A.size(), vec(A.size()));  //??  能自身与自身相乘,肯定是方阵
    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;
}

ll n;


int main()
{
    while(scanf("%lld", &n) == 1)
    {
        mat  A(2, vec(2)), B(2, vec(1));
        init(A); init(B);
        A[0][0] = 7; A[0][1] = 0;
        A[1][0] = 1; A[1][1] = 1;
        B[0][0] = 1;
        B[1][0] = 0;
        A = pow(A, n);
        A = mul(A, B);
        printf("%lld  %lld
", A[0][0], A[1][0]);
    }

    return 0;
}