hdu2276---Kiki & Little Kiki 2 (Matrix)

Problem Description
There is n lights in a circle numbered from 1 to N. The left of light 1 are light n, and the Left of Light K (1< k<= N) are 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 isn't on, turn in it) at t+1 second (T >= 0), if the left Of light I am on!!! Given the initiation State, please find all lights ' state after M second. (2<= n <=, 1<= m<= 10^8)

Input
The input contains one or more data sets. The first line of each data set was an integer m indicate the time, the second line would be a string T, only contains ' 0 ' a nd ' 1 ', and its length n would 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 am on, and otherwise the light is off.

Output
For each data set, the output all lights ' state is at m seconds on one line. It is only contains character ' 0 ' and ' 1.

Sample Input

1 0101111 10 100000001

Sample Output

1111000 001000010

Source
HDU 8th Programming Contest Site (1)

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The state of each lamp is determined by the one on its left.
Constructs a n*n coefficient matrix, each row, the corresponding position of each column is 1,
For example, a coefficient matrix of 3 numbers
1 1 0
0 1 1
1 0 1
Then the initial state of 110, which is 101 after 1s
That is, [1 1 0] to multiply the new matrix obtained by the above matrix
Then we just need the matrix to speed up the power.

/************************************************************************* > File Name:hdu2276.cpp > Auth Or:alex > Mail: [email protected] > Created time:2015 March 12 Thursday 21:28 55 seconds ******************************** ****************************************/#include <map>#include <set>#include <queue>#include <stack>#include <vector>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace STD;Const DoublePI =ACOs(-1.0);Const intINF =0x3f3f3f3f;Const DoubleEPS =1e-15;typedef Long LongLL;typedefPair <int,int> PLL;intNclassmartix{ Public:intmat[ the][ the];        Martix (); Martixoperator* (ConstMartix &b)Const; martix&operator= (ConstMartix &b);}; Martix:: Martix () {memset(Mat,0,sizeof(MAT));} Martix Martix::operator* (ConstMartix &b)Const{Martix ret; for(inti =0; I < n; ++i) { for(intj =0; J < N; ++J) { for(intK =0; K < n; ++K) {Ret.mat[i][j] + = ( ThisMAT[I][K] * b.mat[k][j]); RET.MAT[I][J]%=2; }        }    }returnRET;} martix& Martix::operator= (ConstMartix &b) { for(inti =0; I < n; ++i) { for(intj =0; J < N; ++J) { ThisMAT[I][J] = B.mat[i][j]; }    }return* This;} Martix Fastpow (Martix A,intm) {//void Debug (Martix A);Martix ret; for(inti =0; I < n; ++i) {Ret.mat[i][i] =1; } while(m) {if(M &1) {ret = ret * A; } m >>=1;    A = a * A; }returnRET;}voidDebug (Martix A) { for(inti =0; I < n; ++i) { for(intj =0; J < N; ++J) {printf("%d", A.mat[i][j]); }printf("\ n"); }}Charstr[ the];intMain () {intM while(~scanf("%d", &m)) {scanf('%s ', str);        Martix F; n =strlen(str); Martix A; for(inti =0; I < n-1; ++i) {A.mat[i][i] = a.mat[i][i +1] =1; } A.mat[n-1][0] = A.mat[n-1][n-1] =1;//Debug (A);A = Fastpow (A, m); for(inti =0; I < n; ++i) {f.mat[0][i] = str[i]-' 0 '; } f = f * A; for(inti =0; I < n; ++i) {printf("%d", f.mat[0][i]); }printf("\ n"); }return 0;}

hdu2276---Kiki &amp; Little Kiki 2 (Matrix)