HDU 5015 233 Matrix

题目链接

思路：矩阵快速幂，观察没一列，第一个和为左边加最上面，第二个可以拆为左边2个加最上面，第三个可以拆为为左边3个加最上面，这样其实只要把每一列和每一列右边那列的233构造出一个矩阵，进行矩阵快速幂即可

代码：

#include 
  
   
#include 
   
     typedef long long ll; const int N = 15; const int MOD = 10000007; int n, m; struct mat { ll v[N][N]; mat() {memset(v, 0, sizeof(v));} mat operator * (mat c) { mat ans; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) for (int k = 0; k < n; k++) ans.v[i][j] = (ans.v[i][j] + v[i][k] * c.v[k][j] % MOD) % MOD; return ans; } }; mat pow_mod(mat x, int k) { mat ans; for (int i = 0; i < n; i++) ans.v[i][i] = 1; while (k) { if (k&1) ans = ans * x; x = x * x; k >>= 1; } return ans; } int main() { while (~scanf("%d%d", &n, &m)) { mat A; A.v[0][0] = 1; A.v[1][0] = 1; A.v[1][1] = 10; n += 2; for (int i = 2; i < n; i++) { for (int j = 1; j <= i; j++) A.v[i][j] = 1; } A = pow_mod(A, m); ll ans = 0; ll tmp[N]; tmp[0] = 3; tmp[1] = 233; for (int i = 2; i < n; i++) scanf("%I64d", &tmp[i]); for (int i = 0; i < n; i++) ans = (ans + tmp[i] * A.v[n - 1][i] % MOD) % MOD; printf("%I64d\n", ans); } return 0; }