蓝桥杯：递推求值（构造矩阵+矩阵快速幂）

http://lx.lanqiao.cn/problem.page?gpid=T396

题意：……

思路：自始至终不会构造矩阵，yr大佬手把手教我构造矩阵Orz。

先把有用的信息写成一个矩阵F，然后判断下一步要推出来的信息是怎么构成的，然后一行一行构造矩阵A。

最后将构造得到的矩阵A乘n次，再乘F，就可以得到答案了。

矩阵A的第一行就是F中的哪些信息可以构成F1(n)，就选择性用上，以此类推可以构造出来。

#include <bits/stdc++.h>
using namespace std;
#define MOD 99999999
typedef long long LL;

struct Matrix {
    LL mat[10][10]; int r, c;
    Matrix () { memset(mat, 0, sizeof(mat)); }
    Matrix (int a, int b) { r = a; c = b; memset(mat, 0, sizeof(mat)); }
    void init() { memset(mat, 0, sizeof(mat)); }
    void E() { memset(mat, 0, sizeof(mat)); for(int i = 1; i <= 9; i++) mat[i][i] = 1; }
};

Matrix operator * (Matrix a, Matrix b) {
    Matrix c = Matrix(a.r, b.c);
    for(int i = 1; i <= a.r; i++) {
        for(int k = 1; k <= a.c; k++) {
            for(int j = 1; j <= b.c; j++) {
                c.mat[i][j] = (c.mat[i][j] + a.mat[i][k] * b.mat[k][j] % MOD) % MOD;
            }
        }
    }
    return c;
}

Matrix operator ^ (Matrix a, LL k) {
    Matrix p = Matrix(7, 7), tmp = a;
    p.E();
    while(k) {
        if(k & 1) p = p * tmp;
        tmp = tmp * tmp;
        k >>= 1;
    }
    return p;
}

void print(Matrix now) {
    for(int i = 1; i <= now.r; i++) {
        for(int j = 1; j <= now.c; j++) {
            printf("%lld ", now.mat[i][j]);
        }
        puts("");
    }
}

int main() {
    Matrix now = Matrix(7, 7);
    now.mat[1][2] = 1; now.mat[1][5] = 2; now.mat[1][7] = 5;
    now.mat[2][1] = 1; now.mat[2][5] = 3; now.mat[2][6] = 2; now.mat[2][7] = 3;
    now.mat[3][1] = 1;
    now.mat[4][2] = 1;
    now.mat[5][3] = 1;
    now.mat[6][4] = 1;
    now.mat[7][7] = 1;
    Matrix f = Matrix(7, 1);
    f.mat[1][1] = 6; f.mat[2][1] = 5; f.mat[3][1] = 1;
    f.mat[4][1] = 4; f.mat[5][1] = 2; f.mat[6][1] = 3; f.mat[7][1] = 1;

    LL n; cin >> n;
    if(n == 2) { printf("1\n4\n"); return 0; }
    else if(n == 1) { printf("2\n3\n"); return 0; }
    n -= 3;
    Matrix ans = now ^ n;
    ans = ans * f;
    cout << ans.mat[1][1] << endl << ans.mat[2][1] << endl;
    return 0;
}