HDU 2276 Kiki & Little Kiki 2 矩阵

构造出类似这样的矩阵

1 0 0 1

1 1 0 0

0 1 1 0

0 0 1 1

通过对2取模就可以模拟出开关灯的状态了。然后直接快速幂

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <climits>
#include <iostream>
#include <string>

using namespace std;
 
#define MP make_pair
#define PB push_back
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int INF = INT_MAX / 3;
const double eps = 1e-8;
const LL LINF = 1e17;
const double DINF = 1e60;
const int maxn = 105;

struct Matrix {
    int data[maxn][maxn], n, m;
    Matrix(int n = 0, int m = 0): n(n), m(m) {
        memset(data, 0, sizeof(data));
    }
};

Matrix operator * (Matrix a, Matrix b) {
    Matrix ret(a.n, b.m);
    for(int i = 1; i <= a.n; i++) {
        for(int j = 1; j <= b.m; j++) {
            for(int k = 1; k <= a.m; k++) {
                ret.data[i][j] += a.data[i][k] * b.data[k][j];
                ret.data[i][j] &= 1;
            }
        }
    }
    return ret;
}

Matrix pow(Matrix mat, int p) {
    Matrix ret(mat.n, mat.m);
    for(int i = 1; i <= mat.n; i++) ret.data[i][i] = 1;
    while(p) {
        if(p & 1) ret = ret * mat;
        mat = mat * mat;
        p >>= 1;
    }
    return ret;
}

char buf[128];
int n;

int main() {
    while(scanf("%d%s", &n, buf + 1) != EOF) {
        int len = strlen(buf + 1);
        Matrix A0(len, 1), A(len, len);
        for(int i = 1; i <= len; i++) A0.data[i][1] = buf[i] - '0';
        A.data[1][len] = A.data[1][1] = 1;
        for(int i = 2; i <= len; i++) A.data[i][i - 1] = A.data[i][i] = 1;
        A0 = pow(A, n) * A0;
        for(int i = 1; i <= len; i++) printf("%d", A0.data[i][1]);
        puts("");
    }
    return 0;
}