HDU - 5950(矩阵快速幂)

矩阵快速幂要推出递推式，左矩阵右矩阵，带入模板
分享一篇博客：https://blog.csdn.net/u012061345/article/details/52224623

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <stdlib.h>
#include <string.h>
using namespace std;
#define ll long long
#define debug(x, y) cout <<x<<" "<<y<<"\n"
#define lson l , m , rt << 1
#define rson m + 1 , r , rt << 1 | 1
#define endl "\n"
#define pii pair<int, int>
#define inf 100000000
#define gg cout<<"_________"<<endl
const int maxn = 222222;
const int N = 20;
const ll mod = 2147493647;
struct matrix
{
    int row, col;
    ll mat[N][N];
    matrix(int _row=0, int _col=0)
    {
        init(_row, _col);
    }
    void init(int _row, int _col)
    {
        row = _row, col = _col;
        memset(mat, 0, sizeof mat);
    }
    matrix operator* (matrix b)
    {
        matrix c(row, b.col);
        for(int i = 1; i <= row; i++)
            for(int j = 1; j <= b.col; j++)
                for(int k = 1; k <= col; k++)
                    c.mat[i][j] = (c.mat[i][j] + mat[i][k] * b.mat[k][j] % mod) % mod;
        return c;
    }
};

matrix mod_pow(matrix a, ll b, ll p)
{
    matrix ans(7, 7);
    for(int i = 1; i <= 7; ++i) ans.mat[i][i] = 1;
    while(b)
    {
        if(b&1) ans = ans*a;
        a = a*a;
        b >>= 1;
    }
    return ans;
}

matrix standard(7, 7);
ll mp[8][8] = {
        0,0,0,0,0,0,0,0,
        0,1,2,1,4,6,4,1,
        0,1,0,0,0,0,0,0,
        0,0,0,1,4,6,4,1,
        0,0,0,0,1,3,3,1,
        0,0,0,0,0,1,2,1,
        0,0,0,0,0,0,1,1,
        0,0,0,0,0,0,0,1
};
void init()
{
    for(int i = 1; i <= 7; ++i)
        for(int j = 1; j <= 7; ++j)
            standard.mat[i][j] = mp[i][j];
}
int main()
{
    int t;
    init();
    scanf("%d", &t);
    while(t--)
    {
        ll n, a, b;
        scanf("%lld%lld%lld", &n, &a, &b);
        if(n == 1) printf("%lld\n", a);
        else if(n == 2) printf("%lld\n", b);
        else {
            matrix f(7, 7);
            f.mat[1][1] = (ll) b;  f.mat[2][1] = (ll) a;
            f.mat[3][1] = (ll) 16; f.mat[4][1] = (ll) 8;
            f.mat[5][1] = (ll) 4;  f.mat[6][1] = (ll) 2;
            f.mat[7][1] = (ll) 1;
            f = mod_pow(standard, n-2, mod)*f;
            printf("%lld\n", f.mat[1][1]);
        }
    }
    return 0;
}