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 + ;

 ll n, x, y;

 mat mul(mat & a, mat & b)
 {
     mat c(a.size(), vec(b[].size()));
     for (int i = ; i < a.size(); i++)
     {
         for (int k = ; k < b.size(); k++)
         {
             for (int j = ; j < b[].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 = ; i < a.size(); i++)
     {
         b[i][i] = ;
     }
     while (n > )
     {
         if (n & )
             b = mul(b, a);
         a = mul(a, a);
         n >>= ;
     }
     return b;
 }

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

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

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