矩阵快速幂HDU6030

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long
using namespace std;

typedef __int64 int64;
const int N = 3;

__int64 k , MOD = 1e9+7;
int arr[50] ;
struct Matrix{
    int64 mat[N][N];
    Matrix operator*(const Matrix& m)const{
        Matrix tmp;
        for(int i = 0 ; i < N ; i++){
            for(int j = 0 ; j < N ; j++){
                tmp.mat[i][j] = 0;
                for(int k = 0 ; k < N ; k++){
                    tmp.mat[i][j] += mat[i][k]*m.mat[k][j]%MOD;
                    tmp.mat[i][j] %= MOD;
                }
            }
        }
        return tmp;
    }
};

int64 Pow(Matrix &m , __int64 k){
    Matrix ans;
    memset(ans.mat , 0 , sizeof(ans.mat));
    for(int i = 0 ; i < N ; i++)
        ans.mat[i][i] = 1;
    while(k){
        if(k&1)
            ans = ans*m;
        k >>= 1;
        m = m*m;
    }
    int64 sum = 0;
    for(int i = 0 ; i < N ; i++){
        for(int j = 0; j < N; j++){
            sum += ans.mat[i][j]*arr[j+1]%MOD;
            sum %= MOD;
        }
    }
    return sum;
}

void init(Matrix &m){
    memset(m.mat , 0 , sizeof(m.mat));
    m.mat[0][0] = 1;
    m.mat[0][2] = 1;
    m.mat[1][0] = 1;
    m.mat[2][1] = 1;
}

int main(){
    Matrix m;
    arr[1] = 1;
    arr[2] = 1;
    arr[3] = 1;
    int t;
    scanf("%d",&t);
    while(t--){
            scanf("%I64d",&k);
            init(m);
            if(k == 2)
                printf("2
");
            else
                printf("%I64d
" , Pow(m , k-2));//输入n时计算矩阵的m次方，则Pow(n-(n-m))
    }
    return 0;
}