[HDU 1757] A Simple Math Problem (matrix fast power)

[HDU 1757] A Simple Math Problem (matrix fast power)

Question:

If x <10 f (x) = x.
If x> = 10 f (x) = a0 * f (x-1) + a1 * f (X-2) + a2 * f (X-3) + ...... + A9 * f (X-10 );
And ai (0 <= I <= 9) can only be 0 or 1.
For k and m, evaluate f (k) % m.

Solution:

F (x) = a0 * f (x-1) + a1 * f (X-2) + a2 * f (X-3) + ...... + A9 * f (X-10)

K is 10 ^ 9 orders of magnitude, so it cannot be done in recursive mode. This type of question can be done by constructing a matrix and using a matrix's rapid power.

The constructed matrix is:

| 010 ...... 0 | | F0 | | F1 |
| 0010 ...... 0 | | F1 | | F2 |
| ...... 1 |* |... | =|... |
| A9A8... A0 | | F9 | | F10 |

Code:

/* ID: wuqi9395@126.comPROG: LANG: C ++ */# include# Include
  
   
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              # Include # define INF (1 <30) # define PI acos (-1.0) # define mem (a, B) memset (a, B, sizeof ()) # define For (I, n) for (int I = 0; I <n; I ++) typedef long ll; using namespace std; const int maxn = 20; const int maxm = 20; ll mod, k; struct Matrix {int n, m; int a [maxn] [maxm]; void clear () {n = m = 0; memset (a, 0, sizeof (a);} Matrix operator * (const Matrix & B) const {// implement Matrix multiplication Matrix tmp; tmp. clear (); tmp. n = n; tmp. m = B. m; for (int I = 0; I <n; I ++) for (int j = 0; j <B. m; j ++) for (int k = 0; k <m; k ++) {tmp. a [I] [j] + = a [I] [k] * B. a [k] [j]; tmp. a [I] [j] % = mod;} return tmp ;}}; Matrix A, B; void init () {. clear (); // matrix A is the constructed matrix. n =. m = 10; for (int I = 0; I <9; I ++). a [I] [I + 1] = 1; B. clear (); B. n = 10; B. m = 1; // matrix B is the first 10 count of f (x) for (int I = 0; I <10; I ++) B. a [I] [0] = I;} Matrix Matrix_pow (Matrix A, ll k, ll mod) {// Matrix fast power Matrix res; res. clear (); res. n = res. m = 10; for (int I = 0; I <10; I ++) res. a [I] [I] = 1; while (k) {if (k & 1) res = A * res; k >>=1; A = A * ;} return res;} int main () {init (); while (cin> k> mod) {int x; for (int I = 0; I <10; I ++) {scanf ("% d", & x);. a [9] [9-I] = x;} if (k <10) {printf ("% lld \ n", k % mod );} else {Matrix res = Matrix_pow (A, k-9, mod); res = res * B; cout <