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;
}