蓝桥杯 算法提高 递推求值

思路：

矩阵快速幂。

实现：

#include <iostream>
#include <cstdio>
#include <vector>
using namespace std;

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

const int mod = 99999999;

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.size(); k++)
        {
            for (int j = 0; j < b[0].size(); j++)
            {
                c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % mod;
            }
        }
    }
    return c;
}

mat pow(mat a, ll n)
{
    mat b(a.size(), vec(a.size()));
    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;
}

int main()
{
    mat F(7, vec(1));
    F[0][0] = 6;
    F[1][0] = 1;
    F[2][0] = 2;
    F[3][0] = 5;
    F[4][0] = 4;
    F[5][0] = 3;
    F[6][0] = 1;
    ll n;
    cin >> n;
    if (n <= 3)
    {
        printf("%d
", F[3-n][0]);
        printf("%d
", F[5-(n-1)][0]);
        return 0;
    }
    mat x(7, vec(7));
    for (int i = 0; i < 7; i++)
    {
        for (int j = 0; j < 7; j++)
        {
            x[i][j] = 0;
        }
    }
    x[0][2] = 2;
    x[0][3] = 1;
    x[0][6] = 5;
    x[1][0] = 1;
    x[2][1] = 1;
    x[3][0] = 1;
    x[3][2] = 3;
    x[3][5] = 2;
    x[3][6] = 3;
    x[4][3] = 1;
    x[5][4] = 1;
    x[6][6] = 1;
    mat res(7, vec(7));
    res = pow(x, n - 3);
    mat fuck(7, vec(1));
    fuck = mul(res, F);
    printf("%d
", fuck[0][0]);
    printf("%d
", fuck[3][0]);
    return 0;
}