HDU-2276 Kiki & little Kiki 2

Description

There are n lights in a circle numbered from 1 to n. the left of light 1 is light N, and the left of light K (1 <k <= N) is the light k-1.At time of 0, some of them turn on, and others turn off.
Change the state of light I (if it's on, turn off it; if it is not on, turn on it) at t + 1 second (T> = 0 ), if the left of light I is on !!! Given the initiation state, please find all lights 'State after M second. (2 <=n <= 100, 1 <= m <= 10 ^ 8)

Input

The input contains one or more data sets. the first line of each data set is an integer m indicate the time, the second line will be a string T, only contains '0' and '1 ', and its length N will not exceed 100. it means all lights in the circle from 1 to n.
If the ith character of T is '1', it means the light I is on, otherwise the light is off.

Output

For each data set, output all lights 'State at M seconds in one line. It only contains character '0' and '1.

Sample Input

 
1010111110100000001
 

Sample output

 
1111000001000010
 

A: A light is arranged to give you an initial state, and then there will be such an operation every second: If the left side of the light is on, it will change the state, otherwise it will not change, leftmost reference rightmost

Train of Thought: A1 = (A1 + an) % 2, a2 = (A2 + A1) % 2 ......... an = (An + an-1) % 2

Then construct a matrix like: 1 0 0 1. Bit operations are much faster, because of the particularity of this question.

1 1 0 0

0 1 1 0

0 0 1 1

# Include <iostream> # include <cstring> # include <cstdio> # include <algorithm> # include <cmath> using namespace STD; const int maxn = 105; const int mod = 2; int CNT; struct matrix {int V [maxn] [maxn]; matrix () {} matrix (int x) {Init (); for (INT I = 0; I <maxn; I ++) V [I] [I] = x;} void Init () {memset (v, 0, sizeof (v);} Matrix Operator * (matrix const & B) const {matrix C; C. init (); For (INT I = 0; I <CNT; I ++) for (in T j = 0; j <CNT; j ++) for (int K = 0; k <CNT; k ++) C. V [I] [J] ^ = (V [I] [k] & B. V [k] [J]); Return C;} matrix operator ^ (int B) {matrix A = * This, res (1); while (B) {If (B & 1) RES = res * A; A = A * A; B >>= 1;} return res ;}} A, B, TMP; int main () {int t; char STR [maxn]; int num [maxn]; while (scanf ("% d", & T )! = EOF) {scanf ("% s", STR); CNT = strlen (STR); For (INT I = 0; I <CNT; I ++) num [I] = STR [I]-'0';. init ();. V [0] [cnt-1] =. V [0] [0] = 1; for (INT I = 1; I <CNT; I ++). V [I] [I] =. V [I] [I-1] = 1; TMP = a ^ t; int ans [maxn]; memset (ANS, 0, sizeof (ANS )); for (INT I = 0; I <CNT; I ++) if (Num [I]) for (Int J = 0; j <CNT; j ++) if (TMP. V [J] [I]) ans [J] = (ANS [J] + (TMP. V [J] [I] * num [I]) % mod) % MOD; For (INT I = 0; I <CNT; I ++) printf ("% d", ANS [I]); printf ("\ n");} return 0 ;}