HDU 2276

http://acm.hdu.edu.cn/showproblem.php?pid=2276

矩阵乘法可以解决的一类灯泡开关问题

/*
转移关系为
now left now*
1    0    1
1    1    0
0    1    1
0    0    0

可知F[i] = (f[i] + f[(n+i-2)%n+1]) % 2 

得到n*n的关系矩阵是
1 1 0 0 ... 0
0 1 1 0 ... 0
0 0 1 1 ... 0
. . . . ... 0
. . . . ... 0
. . . . ... 0
0 0 0 0 ... 1
1 0 0 0 ... 1
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>

using namespace std;

#define MOD 2

#define Mat 105 //矩阵大小

struct mat{//矩阵结构体，a表示内容，r行c列 矩阵从1开始
    int a[Mat][Mat];
    int r, c;
    mat() {
        r = c = ;
        memset(a, , sizeof(a));
    }
};  

void print(mat m) {
    //printf("%d\n", m.size);
    for(int i = ; i < m.r; i++) {
        for(int j = ; j < m.c; j++) printf("%d ", m.a[i][j]);
        putchar('\n');
    }
}  

mat mul(mat m1, mat m2, int mod) {
    mat ans  = mat();
    ans.r = m1.r, ans.c = m2.c;
    for(int i = ; i <= m1.r; i++)
        for(int j = ; j <= m2.r; j++)
            if(m1.a[i][j])
                for(int k = ; k <= m2.c; k++)
                    ans.a[i][k] = (ans.a[i][k] + m1.a[i][j] * m2.a[j][k]) % mod;
    return ans;
} 

mat quickmul(mat m, int n, int mod) {
    mat ans = mat();
    for(int i = ; i <= m.r; i++) ans.a[i][i] = ;
    ans.r = m.r, ans.c = m.c;
    while(n) {
        if(n & ) ans = mul(m, ans, mod);
        m = mul(m, m, mod);
        n >>= ;
    }
    return ans;
}  

/*
初始化ans矩阵
mat ans = mat();
ans.r = R, ans.c = C;
ans = quickmul(ans, n, mod);
*/

char a[];

int main() {
    int m;
    while(~scanf("%d", &m)) {
        scanf("%s", a);
        int n = strlen(a);
        mat A = mat();
        A.r = , A.c = n;
        for(int i = ; i <= n; i++)
            A.a[][i] = a[i-] - '';
        mat G = mat();
        G.r = G.c = n;
        for(int i = ; i < n; i++) {
            G.a[i][i] = G.a[i][i+] = ;
        }
        G.a[n][] = G.a[n][n] = ;
        mat ans = quickmul(G, m, MOD);
        ans = mul(A, ans, MOD);
        for(int i = ; i <= n; i++)
            printf("%d", ans.a[][i]);
        putchar('\n');
    }
    return ;
}