Matrix Power Series
Time limit:3000 Ms 

Memory limit:131072 K 
Total submissions:15417 

Accepted:6602 
Description
GivenN×NMatrixAAnd a positive integerK, Find the sumS=A+A2 +A3 +... +AK.
Input
The input contains exactly one test case. The first line of input contains three positive integersN(N≤ 30 ),K(K≤ 109) andM(M<104). Then followNLines each containingNNonnegative integers below 32,768, givingA'S elements in rowMajor Order.
Output
Output the elementsSModuloMIn the same wayAIs given.
Sample Input
2 2 40 11 1
Sample output
1 22 3
Source
Poj monthly  2007.06.03, Huang, Jinsong
Solution:
Given the matrix A (represented by Ori in the code below), and K, Mod, calculate a + A ^ 2 + A ^ 3 + ...... A ^ K and mod remainder.
Start with a loop of K times, recurrence, timeout...
View the question report, and use the binary summation.
The sum is represented by S (k.
When K is an even number:
For example, K = 6, then a + A ^ 2 + A ^ 3 + A ^ 4 + A ^ 5 + A ^ 6 = a + A ^ 2 + A ^ 3 + A ^ 3 * (a + A ^ 2 + A ^ 3)
S (K) = S (K/2) + A ^ (n/2) * s (K/2) that is S (k) = (E + A ^ (n/2) * S (n/2) (E is the unit matrix)
When K is an odd number:
S (K) = S (k1) * A ^ K, then the k1 is an even number, according to the above two points
PS: the code should be carefully written; otherwise, a small error will be checked for half a day ..... calculate the RET when two matrices are multiplied. arr [I] [J] + =. arr [I] [k] * B. arr [k] [J]; actually written as ret. arr [I] [J] + =. arr [I] [k] *. arr [k] [J]; t
Code:
# Include <iostream> # include <stdio. h> # include <string. h> using namespace STD; const int maxn = 31; int N, K, MOD; struct mat {int arr [maxn] [maxn]; MAT () {memset (ARR, 0, sizeof (ARR) ;}}; mat MUL (MAT a, mat B) {mat ret; For (INT I = 0; I <n; I ++) for (int K = 0; k <n; k ++) {if (. arr [I] [k]) for (Int J = 0; j <n; j ++) {ret. arr [I] [J] + =. arr [I] [k] * B. arr [k] [J]; If (Ret. arr [I] [J]> = mod) ret. arr [I] [J] % = MOD;} return ret;} mat add (MAT A, mat B) {mat an; For (INT I = 0; I <n; I ++) for (Int J = 0; j <n; j ++) {. arr [I] [J] =. arr [I] [J] + B. arr [I] [J]; If (. arr [I] [J]> = mod). arr [I] [J] % = MOD;} return an;} mat power (MAT P, int K) {If (k = 1) return P; MAT E; for (INT I = 0; I <n; I ++) E. arr [I] [I] = 1; if (k = 0) Return e; while (k) {If (K & 1) E = MUL (p, e ); P = MUL (p, p); k >>= 1;} return E;} void output (MAT ans) {for (INT I = 0; I <N; I ++) for (Int J = 0; j <n; j ++) {If (j = N1) c Out <ans. arr [I] [J] <Endl; else cout <ans. arr [I] [J] <";}} mat Cal (MAT Ori, int K) {If (k = 1) return Ori; If (K & 1) return add (CAL (ORI, k1), power (ORI, k); // when K is an odd number, minus 1 to an even number S (K) = S (K1) + Ori ^ K else return MUL (add (Power (ORI, 0), power (ORI, k> 1), Cal (ORI, k> 1 )); // when K is an even number, S (K) = (1 + Ori ^ (K/2) * s (K/2)} int main () {While (scanf ("% d", & N, & K, & mod )! = EOF) {mat Ori, ans; For (INT I = 0; I <n; I ++) for (Int J = 0; j <n; j ++) {scanf ("% d", & Ori. arr [I] [J]); If (Ori. arr [I] [J]> = mod) Ori. arr [I] [J] % = MOD;} ans = CAL (ORI, k); output (ANS);} return 0 ;}
[ACM] poj 3233 matrix power series (evaluate matrix A + A ^ 2 + A ^ 3... + A ^ K, bipartite summation)