Kiki & Little Kiki 2

時間限制：5000?ms ?|? 內存限制：65535?KB 難度：4 描述

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) 輸入

The input contains no more than 1000 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.

輸出

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

樣例輸入

1 0101111 10 100000001

樣例輸出

1111000 001000010

題意：給出一些燈的初始狀態（用0、1表示），對這些燈進行m次變換；若當前燈的前一盞燈的狀態為1，則調整當前燈的狀態，0變為1,1變為0；否則不變。第1盞燈的前一盞燈是最后一盞燈。問最后每盞燈的狀態。

分析：通過模擬可以發現，假設有n盞燈，第i盞燈的狀態為f[i],則f[i] = (f[i] + f[i-1])%2;又因為這些燈形成了環，則f[i] = (f[i] + f[(n+i-2)%n+1])%2. 這樣初始狀態形成一個1*n的矩陣，然后構造出如下n*n的矩陣：

1 ?1 ?0…… 0 ? 0

0 ?1 ?1…… 0 ? 0

……………………

1 ?0 ?0…… 0 ? 1

每次乘以這個矩陣得出的結果就是下一個狀態。

#include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N = 105; char state[N]; struct Matrix {Matrix() {}void Init(int n) {row = n; col = n;memset(mat, 0, sizeof(mat));for(int i = 0; i < n; i++)mat[i][i] = 1;}void Init(int m, int n) {row = m; col = n;memset(mat, 0, sizeof(mat));}int row, col;int mat[N][N];const Matrix &Pow(int n); };const Matrix &operator *(const Matrix &A, const Matrix &B) {Matrix res;res.Init(A.row, B.col);for(int i = 0; i < A.row; i++) {for(int j = 0; j < A.col; j++) {if(A.mat[i][j]) {for(int k = 0; k < B.col; k++) {if(B.mat[j][k])res.mat[i][k] = res.mat[i][k] ^ (A.mat[i][j] & B.mat[j][k]);}}}}return res; }const Matrix &Matrix::Pow(int n) {Matrix tmp, pre;tmp = *this;pre.Init(this->row);while(n) {if(n&1) pre = tmp * pre;tmp = tmp * tmp;n >>= 1;}return pre; }int main() {int m;Matrix A, B, ans;while(~scanf("%d", &m)) {scanf("%s",state);int len = strlen(state);A.Init(1, len);for(int i = 0; i < len; i++)A.mat[0][i] = state[i] - '0';B.Init(len);for(int i = 0; i < len; i++) {B.mat[i][i] = 1;B.mat[i][(i+1)%len] = 1;}ans = A * B.Pow(m);for(int i = 0; i < len; i++)printf("%d", ans.mat[0][i]);printf("\n");}return 0; }

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