nyoj 300 （矩陣快速冪）Kiki & Little Kiki 2

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爲1,
則調整當前燈的狀態,
0變爲1,1變爲0；否則不變。第1盞燈的前一盞燈是最後一盞燈。問最後每盞燈的狀態。
分析：通過模擬可以發現,
假設有n盞燈,第i盞燈的狀態爲f[i],則f[i] = (f[i] + f[i-1])%2;
又因爲這些燈形成了環,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
根據係數推出初始矩陣,然後構造出如下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<map>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<math.h>
#include<cstdio>
#include<sstream>
#include<numeric>//STL數值算法頭文件
#include<stdlib.h>
#include <ctype.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<functional>//模板類頭文件
using namespace std;

typedef long long ll;
const int maxn=1001;
const int INF=0x3f3f3f3f;


const int N = 102;

struct mat
{
    int r, c;
    int M[N][N];
    mat(int r, int c):r(r), c(c)
    {
        memset(M, 0, sizeof(M));
    }
};

mat mul(mat& A, mat& B)
{
    mat C(A.r, B.c);
    for(int i = 0; i < A.r; ++i)
        for(int j = 0; j < A.c; ++j)
            if(A.M[i][j])     //優化，只有state爲1的時候才需要改變
            {
                for(int k = 0; k < B.r; ++k)
                    if(B.M[j][k])
                        C.M[i][k] ^= A.M[i][j] & B.M[j][k];
            }
    return C;
}


mat pow(mat& A, int k)
{
    mat B(A.r, A.c);
    for(int i = 0; i < A.r; ++i) B.M[i][i] = 1;

    while(k)
    {
        if(k & 1) B = mul(B, A);
        A = mul(A, A);
        k >>= 1;
    }
    return B;
}

int main()
{
    int m;
    char t[105];

    while(scanf("%d %s", &m, t) != EOF)
    {
        int n = strlen(t);
        mat A(1, n);
        mat T(n, n);
        for(int i = 0; i < n; ++i)
        {
            A.M[0][i] = t[i] - '0';
            T.M[i][i] = T.M[i][(i + 1) % n] = 1;
        }

        T = pow(T, m);
        A = mul(A, T);

        for(int i = 0; i < n; ++i)
        {
            printf("%d", A.M[0][i]);
        }
        printf("\n");
    }

    return 0;
}