矩阵快速幂模板

struct mat{
    LL a,b;
    mat(LL A=0,LL B=0):a(A),b(B){}

    bool isIdentity()const{ return a==1 && b==0; }

    mat operator+(const mat &rhs)const{
        return mat( (a+rhs.a)%MOD, (b+rhs.b)%MOD );
    }

    mat operator*(const mat &rhs)const{
        return mat(
                (a*rhs.a+b*rhs.b)%MOD,
                (a*rhs.b+b*(rhs.a+rhs.b))%MOD);
    }

    mat power(LL b)const{
        mat a=*this,ans(1,0);
        while(b){
            if(b&1)ans=ans*a;
            b>>=1; a=a*a;
        }
        return ans;
    }
};

来源题目：http://codeforces.com/contest/719/problem/E

代码来源：http://codeforces.com/contest/719/standings 的 sk_aswd

 

或者用下面这个模板，不过要一直用struct而已

来自http://www.cnblogs.com/heimao5027/p/5923401.html

struct Node{
    LL mat[2][2];
    void reset(){memset(mat, 0, sizeof(mat));}
    void getone(){
        reset();
        mat[0][0] = mat[1][1] = 1;
    }
};

inline Node mul(Node A, Node B){
    Node ans; ans.reset();
    for (int i = 0; i < 2; i++)
        for (int j = 0; j < 2; j++)
            for (int k = 0; k < 2; k++)
                ans.mat[i][j] = (ans.mat[i][j] + A.mat[i][k] * B.mat[k][j]) % mod;
    return ans;
}

inline Node add(Node a, Node b){
    Node ans; ans.reset();
    for (int i = 0; i < 2; i++)
        for (int j = 0; j < 2; j++)
            ans.mat[i][j] = (a.mat[i][j] + b.mat[i][j]) % mod;
    return ans;
}

inline Node kpow(LL k){
    Node ans; ans.reset();
    ans.mat[0][0] = ans.mat[1][1] = 1;
    Node A;
    A.mat[0][0] = A.mat[1][0] = A.mat[0][1] = 1;
    A.mat[1][1] = 0;
    while (k){
        if (k & 1) ans = mul(ans, A);
        A = mul(A, A);
        k >>= 1;
    }
    return ans;
}