Hdu2276-Kiki & little Kiki 2 (matrix power)

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Question:There are n lamps numbered from 1 to n. They are circled in a circle, that is to say, the left side of Lamp 1 is lamp n. If the light on the left of a lamp is on at T Second, the light state will be changed at t + 1 second. Enter m and initial light status. Output the status of all lights after M seconds.

Ideas:In fact, the sum of the States of each lamp is related to itself, so we can get a relational matrix. Assuming there are 6 lamps, the relational matrix can be obtained as follows:
(1, 0, 0, 0, 0, 1)
(1, 1, 0, 0, 0, 0)
(0, 1, 1, 0, 0, 0)
(0, 0, 1, 1, 0, 0)
(0, 0, 0, 1, 1, 0)
(0, 0, 0, 0, 1, 1)
In this way, we can obtain the relational matrix of N lamps, and then use the Matrix to quickly calculate the power.

PS: At the beginning, I used recursion for the fast power, but the stack was violent... Later it was changed to non-recursive.

Code:

# Include <iostream> # include <cstdio> # include <cstring> # include <cmath> # include <algorithm> using namespace STD; const int maxn = 105; struct mat {int s [maxn] [maxn]; int L; MAT (INT Len) {memset (S, 0, sizeof (s); L = Len ;} mat operator * (const mat & C) {mat ans (l); memset (ans. s, 0, sizeof (ans. s); For (INT I = 0; I <L; I ++) for (Int J = 0; j <L; j ++) {for (int K = 0; k <L; k ++) ans. s [I] [J] = (ANS. S [I] [J] + s [I] [k] * C. s [k] [J]); ans. s [I] [J] = ans. s [I] [J] % 2;} return ans ;}}; char STR [maxn]; int t; MAT pow_mod (MAT C, int K) {/* If (k = 1) return C; MAT a = pow_mod (C, K/2); MAT ans = A * A; If (K % 2) ans = ans * C; return ans; */MAT ans = C; k --; while (k) {If (K & 1) ans = ans * C; k> = 1; C = C * C;} return ans;} int main () {While (scanf ("% d", & T )! = EOF) {scanf ("% s", STR); int L = strlen (STR); mat c (l); For (INT I = 0; I <L; I ++) for (Int J = 0; j <L; j ++) {if (I = 0) C. s [I] [0] = C. s [I] [L-1] = 1; else C. s [I] [I-1] = C. s [I] [I] = 1;} mat TMP (l); For (INT I = 0; I <L; I ++) TMP. s [I] [0] = STR [I]-'0'; MAT ans = pow_mod (C, T); ans = ans * TMP; For (INT I = 0; I <L; I ++) printf ("% d", ans. s [I] [0]); printf ("\ n");} return 0 ;}

Hdu2276-Kiki & little Kiki 2 (matrix power)