首页 > 代码库 > hihocoder offer收割编程练习赛13 D 骑士游历
hihocoder offer收割编程练习赛13 D 骑士游历
思路:
矩阵快速幂。
实现:
1 #include <iostream> 2 #include <cstdio> 3 #include <vector> 4 using namespace std; 5 6 typedef long long ll; 7 typedef vector<ll> vec; 8 typedef vector<vec> mat; 9 10 const ll mod = 1e9 + 7; 11 12 ll n, x, y; 13 14 mat mul(mat & a, mat & b) 15 { 16 mat c(a.size(), vec(b[0].size())); 17 for (int i = 0; i < a.size(); i++) 18 { 19 for (int k = 0; k < b.size(); k++) 20 { 21 for (int j = 0; j < b[0].size(); j++) 22 { 23 c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % mod; 24 } 25 } 26 } 27 return c; 28 } 29 30 mat pow(mat a, ll n) 31 { 32 mat b(a.size(), vec(a.size())); 33 for (int i = 0; i < a.size(); i++) 34 { 35 b[i][i] = 1; 36 } 37 while (n > 0) 38 { 39 if (n & 1) 40 b = mul(b, a); 41 a = mul(a, a); 42 n >>= 1; 43 } 44 return b; 45 } 46 47 bool check(int x, int y) 48 { 49 int m = x / 8, n = x % 8, p = y / 8, q = y % 8; 50 if (abs(m - p) == 1 && abs(n - q) == 2) 51 return true; 52 if (abs(m - p) == 2 && abs(n - q) == 1) 53 return true; 54 return false; 55 } 56 57 void init(mat & x, mat & y, int p, int q) 58 { 59 for (int i = 0; i < 64; i++) 60 { 61 for (int j = 0; j < 64; j++) 62 { 63 if (check(i, j)) 64 x[i][j] = x[j][i] = 1; 65 else 66 x[i][j] = x[j][i] = 0; 67 } 68 } 69 for (int i = 0; i < 64; i++) 70 { 71 y[0][i] = 0; 72 } 73 y[0][p * 8 + q] = 1; 74 } 75 76 int main() 77 { 78 cin >> n >> x >> y; 79 x--, y--; 80 mat a(64, vec(64)); 81 mat m(1, vec(64)); 82 init(a, m, x, y); 83 a = pow(a, n); 84 m = mul(m, a); 85 ll cnt = 0; 86 for (int i = 0; i < 64; i++) 87 { 88 cnt += m[0][i]; 89 cnt %= mod; 90 } 91 cout << cnt << endl; 92 return 0; 93 }
hihocoder offer收割编程练习赛13 D 骑士游历
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