SGU 197.Nice Patterns Strike Back

时间限制：0.5s

空间限制：6M

题意：

给出长n（n<=10^100）和宽m（m<=5）的地面,铺上黑色和白色的地板,使得没有任意一个2*2大小的地面铺同种颜色的方案数是多少.

状态压缩,一个数字的对应为01分别代表铺白色和黑色地板,对于每一列有(1<<m)种状态.

我们可以构造一个矩阵,mat[i][j]代表,第一列是状态是i,第二列是j的方案数,显然mat[i][j]不是0就是1,而且很容易判断构造.

然后我们只要对这个矩阵进行快速幂运算,幂为(n-1),当然不要忘记取模,最后把mat所有元素加起来就是我们想要的答案了.

由于n比较大,所以要做大数的减一,和除以二的运算.

code

#include <iostream>
#include <cstring>
#include <string>
using namespace std;
struct Mat {
    int mat[100][100];
} mx;
int pow[109];
int n, m, mod, len;
Mat operator * (Mat a, Mat b) {
    Mat c;
    memset (c.mat, 0, sizeof c.mat);
    for (int k = 0; k <  (1 << m); k++)
        for (int i = 0; i <  (1 << m); i++)
            for (int j = 0; j <  (1 << m); j++)
                (c.mat[i][j] += (a.mat[i][k] * b.mat[k][j]) % mod) %= mod;
    return c;

}
inline int div2() {
    int ans[103] = {0};
    int i, res = 0;
    for(i = 0; i < len; ++i) {
        ans[i] = (pow[i]+res*10)/2;
        res = (pow[i]+res*10)%2;
    }
    if(ans[0] == 0) len--;
    for(i = 0+(ans[0] == 0); i < len+(ans[0] == 0); i++)
        pow[i-(ans[0] == 0)] = ans[i];
    return res;
}
Mat operator ^ (Mat a, int pow[]) {
    Mat c;
    for (int i = 0; i <  (1 << m); i++)
        for (int j = 0; j <  (1 << m); j++)
            c.mat[i][j] = (i == j);
    while (len) {
        if (div2() )
                     c = c * a;
        a = a * a;
    }
    return c;
}
string s;
int main() {
    ios::sync_with_stdio (0);
    while(cin >> s >> m >> mod){
    for (int i = 0; i < s.size(); ++i)
        pow[i] = s[i] - ‘0‘;
    len = s.size();
    for (int i = len - 1; i >= 0; i--) {
        if (pow[i]) {
            --pow[i];
            break;
        }
        else
            pow[i] = 9;
    }
    if (pow[0] == 0) {
        for (int i = 0; i < len - 1; i++) pow[i] = 9;
        pow[--len]=0;
    }
    for (int i = 0; i < (1 << m); i++)
        for (int j = 0; j < (1 << m); j++) {
            mx.mat[i][j] = 1;
            for (int k = 0, tem = i ^ j; k < m - 1; k++)
                if (  (tem & 1 << k) == 0  && (tem & 1 << k + 1) == 0
&&( ( (i & 1 << k) > 0 && (i & 1 << k + 1) > 0) || ( ( (i & 1 << k) == 0 && (i & 1 << k + 1) == 0) ) ) ) {
                    mx.mat[i][j] = 0;
                    break;
                }
        }
    mx = mx ^ pow;
    int ans = 0;
    for (int i = 0; i < (1 << m); i++)
        for (int j = 0; j < (1 << m); j++) {
            ans += mx.mat[i][j];
            while (ans >= mod) ans -= mod;
        }
    cout << ans << endl;
    }
}