Hdu2276Kiki & Little Kiki 2矩阵

ai = (ai-1  + ai) %2 然后构造个 n*n的矩阵A,  时间K之后的状态就是  A^k  *  B(给出的字符串)

 

#include <cstdio>

#include <algorithm>

#include <iostream>

#include <string.h>

typedef long long LL;

using namespace std;

int n;

struct Matrix

{

    int m[104][105];

};



Matrix Mul(Matrix a, Matrix b)

{

    Matrix  ans;

    for (int i = 0; i < n; i++){

        for (int j = 0; j < n; j++){

            ans.m[i][j] = 0;

            for (int k = 0; k < n; k++){

                ans.m[i][j] += a.m[i][k] * b.m[k][j];

                ans.m[i][j] %= 2;

            }

        }

    }

    return ans;

}



Matrix quick(Matrix a, int b)

{

    Matrix ans;

    for (int i = 0; i < n; i++)

    for (int j = 0; j < n; j++)

        ans.m[i][j] = (i == j);

    while (b){

        if (b & 1){

            ans = Mul(ans, a);

        }

        a = Mul(a, a);

        b >>= 1;

    }

    return ans;

}



Matrix init()

{

    Matrix ans;

    for (int i = 0; i < n; i++)

    for (int j = 0; j < n; j++)

        ans.m[i][j] = 0;

    ans.m[n - 1][0] = 1;

    for (int i = 0; i<n; i++)

        ans.m[i][i] = 1;

    for (int i = 0; i<n - 1; i++)

        ans.m[i][i + 1] = 1;

    return ans;

}



int main()

{

    int k;

    char str[1000];

    int a[1000];

    while (cin >> k){

        scanf("%s", str);

        n = strlen(str);

        Matrix ans = init();

        Matrix  t = quick(ans, k);

        for (int i = 0; i < n; i++)

            a[i] = str[i] - '0';

        for (int j = 0; j<n; j++){

            int sum = 0;

            for (int k = 0; k<n; k++){

                sum += a[k] * t.m[k][j];

            }

            sum %= 2;

            cout << sum;

        }

        cout << endl;

    }

    return 0;

}