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.

 

 

 

 

 

 

 

 

 

 

 

 

 

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Solution:

             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;

    }

}

View Code