hihocoder offer收割编程练习赛13 D 骑士游历

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 ll mod = 1e9 + 7;

ll n, x, y;

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;
}

bool check(int x, int y)
{
    int m = x / 8, n = x % 8, p = y / 8, q = y % 8;
    if (abs(m - p) == 1 && abs(n - q) == 2)
        return true;
    if (abs(m - p) == 2 && abs(n - q) == 1)
        return true;
    return false;
}

void init(mat & x, mat & y, int p, int q)
{
    for (int i = 0; i < 64; i++)
    {
        for (int j = 0; j < 64; j++)
        {
            if (check(i, j))
                x[i][j] = x[j][i] = 1;
            else
                x[i][j] = x[j][i] = 0;
        }
    }
    for (int i = 0; i < 64; i++)
    {
        y[0][i] = 0;
    }
    y[0][p * 8 + q] = 1;
}

int main()
{
    cin >> n >> x >> y;
    x--, y--;
    mat a(64, vec(64));
    mat m(1, vec(64));
    init(a, m, x, y);
    a = pow(a, n);
    m = mul(m, a);
    ll cnt = 0;
    for (int i = 0; i < 64; i++)
    {
        cnt += m[0][i];
        cnt %= mod;
    }
    cout << cnt << endl;
    return 0;
}