NYOJ 300 && hdu 2276 Kiki & Little Kiki 2 （矩阵高速功率）

pid=300">Kiki & Little Kiki 2

时间限制：5000 ms  |  内存限制：65535 KB

难度：4

描写叙述

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)

输入

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.

输出

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

例子输入

1
0101111
10
100000001

例子输出

1111000
001000010

题意：给出一些灯的初始状态（用0、1表示）。对这些灯进行m次变换。若当前灯的前一盏灯的状态为1，则调整当前灯的状态。0变为1,1变为0；否则不变。第1盏灯的前一盏灯是最后一盏灯。

问最后每盏灯的状态。

分析：通过模拟能够发现，如果有n盏灯。第i盏灯的状态为f[i],则f[i] = (f[i] + f[i-1])%2;又由于这些灯形成了环，则f[i] = (f[i] + f[(n+i-2)%n+1])%2. 这样初始状态形成一个1*n的矩阵。然后构造出例如以下n*n的矩阵：

1  1  0…… 0   0

0  1  1…… 0   0

……………………

1  0  0…… 0   1

每次乘以这个矩阵得出的结果就是下一个状态。

#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;
}

版权声明：本文博主原创文章，博客，未经同意不得转载。