Poj3233 (second division and second division of matrix)

The question is very simple:

Description

GivenN×NMatrixAAnd a positive integerK, Find the sumS=A+A2 +A3 +... +AK.

Output

S mod m

Range: N(N≤ 30 ),K(K≤ 109) andM(M<104 ).

Obviously, violence cannot solve the problem, which is very interesting and difficult to think about.

What I take is recursive binary solution:

If K is an even number, it can be converted to (a + A ^ 2 +... + A ^ (K/2) + A ^ (K/2) * (a + A ^ 2 +... + A ^ (K/2 ));

For example, K = 6 can be converted to (a + A ^ 2 + A ^ 3) + (a ^ 3) * (a + A ^ 2 + A ^ 3 ), the problem is converted to a + A ^ 2 + A ^ 3, and then a ^ 3 is quickly multiplied.

When K is an odd number: can be converted into a K-1 solution + (a ^ K)

This topic is very interesting, that is, a second division is used for sum, and a second division is used for power.

I wrote a little slowly. I reloaded a lot of symbols and had obsessive-compulsive disorder. In order to be nice ....

13577997

Dengyaolong

3233

Accepted

896 K

1672 Ms

C ++

2867b

17:10:35

/*************************************** * *******************> OS: linux 3.small-24-generic (MINT-17)> author: yaolong> mail: [email protected]> time: october 16: 05 minutes 45 seconds ******************************** * *************************/# include <iostream> # include <cstdio> # include <string >#include <cstring> using namespace STD; struct matrix {int A [31] [31]; int size; int MOD; matrix (int s, int m): size (s), MOD (m) {for (INT I = 1; I <= s; I ++) {A [I] [I] = 1; // matrix of units} Matrix Operator * (const Matrix & RHs) const // simple multiplication {matrix res (size, MoD); For (INT I = 1; I <= size; I ++) {for (Int J = 1; j <= size; j ++) {res. A [I] [J] = 0; For (int K = 1; k <= size; k ++) {res. A [I] [J] = (res. A [I] [J] + (A [I] [k] * RHS. A [k] [J]) % mod ;}}return res ;}matrix operator + (const Matrix & RHs) const // simpler addition {matrix res (size, MoD); For (INT I = 1; I <= size; I ++) {for (Int J = 1; j <= size; j ++) {res. A [I] [J] = (a [I] [J] + RHS. A [I] [J]) % mod ;}return res;} matrix operator ^ (INT p) // fast power {matrix R (size, MoD ); matrix base (* This); While (p) {If (P & 1) {r = r * base;} base = base * base; P >>= 1 ;} return R;} void print () // output {for (INT I = 1; I <= size; I ++) {for (Int J = 1; j <= size; j ++) {If (j> 1) {cout <"" ;}cout <A [I] [J];} cout <Endl ;}} void input () // input {for (INT I = 1; I <= size; I ++) {for (Int J = 1; j <= size; j ++) {CIN> A [I] [J]; A [I] [J] = A [I] [J] % mod ;}}}; matrix powerplus (matrix m, int K) {If (k = 1) // returns {return m;} If (K & 1) {Matrix T = powerplus (M, k> 1); Return (M ^ K) only once) + T + (M ^ (k> 1) * t;} else {Matrix T = powerplus (M, k> 1 ); return T + (M ^ (k> 1) * t ;}} int main () {int size, K, MOD; cin> size> K> MOD; matrix MTR (size, MoD); MTR. input (); (powerplus (MTR, k )). print (); Return 0 ;}

Poj3233 (second division and second division of matrix)