UVA 10870 - Recurrences(矩陣快速冪)

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UVA 10870 - Recurrences

題目連結

題意：f(n) = a1 f(n - 1) + a2 f(n - 2) + a3 f(n - 3) + ... + ad f(n - d), for n > d.
已知前d項求第n項

思路：矩陣快速冪，對應矩陣為
|a1 a2 a3 ... ad|
|1 0 0 ... 0 0 0|
|0 1 0 ... 0 0 0|
|0 0 1 ... 0 0 0|
|0 0 0 ... 0 0 0|
|0 0 0 ... 1 0 0|
|0 0 0 ... 0 1 0|
|0 0 0 ... 0 0 1|

代碼：

#include <stdio.h>#include <string.h>const int N = 20;long long d, n, m, f[N];struct mat {long long n, v[N][N];mat(long long n = 0) {this->n = n;   memset(v, 0, sizeof(v));   }mat operator * (mat b) {mat ans = mat(n);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] * b.v[k][j] % m) % m + m) % m;    }   }  }  return ans; }};mat pow_mod(mat a, long long k) {mat ans(a.n);for (int i = 0; i < ans.n; i++)ans.v[i][i] = 1;while (k) {if (k&1) ans = ans * a;a = a * a;k >>= 1;}return ans;}int main() {while (~scanf("%lld%lld%lld", &d, &n, &m) && d) {mat a = mat(d);for (int i = 0; i < d; i++)scanf("%lld", &a.v[0][i]);for (int i = 1; i < d; i++)a.v[i][i - 1] = 1;for (int i = 0; i < d; i++)scanf("%lld", &f[i]);if (n <= d) printf("%lld\n", f[n - 1]);else {long long ans = 0;a = pow_mod(a, n - d); for (int i = 0; i < d; i++)ans = (ans + (f[d - i - 1] * a.v[0][i] % m + m)) % m;  printf("%lld\n", ans);  } }return 0;}