快速矩阵幂+DFS构造矩阵+大数 ACdream1214 Nice Patterns Strike Back

传送门：点击打开链接

题意：告诉你矩阵大小是n*m，要求矩阵中不能有2*2的白色子矩阵或者黑色子矩阵，最后种类数模P

思路：如果不是大数，这道题还是非常有意思的。。对于专门卡C++的题目也是醉了...因为n太大了，而m最大也只有5，很明显是大数上的快速矩阵幂。

问题是如何构造出矩阵出来，之前做过骨牌的题目，就是利用DFS来构造的，感觉这道题在思路上是一样的，同样也是利用DFS先构造出矩阵

然后直接大数+快速矩阵幂撸一发就行了

#include<map>
#include<set>
#include<cmath>
#include<stack>
#include<queue>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define FIN freopen("input.txt","r",stdin)
using namespace std;

const int matMX = 50;
const int DLEN = 4;
const int MAXN = 9999;

class BN {
public:
    int a[500];
    int len;

    BN(const int b = 0) {
        int c, d = b;
        len = 0;
        memset(a, 0, sizeof(a));
        while(d > MAXN) {
            c = d - (d / (MAXN + 1)) * (MAXN + 1);
            d = d / (MAXN + 1);
            a[len++] = c;
        }
        a[len++] = d;
    }
    BN(const char *s) {
        int t, k, index, L, i;
        memset(a, 0, sizeof(a));
        L = strlen(s);
        len = L / DLEN;
        if(L % DLEN) len++;
        index = 0;
        for(i = L - 1; i >= 0; i -= DLEN) {
            t = 0;
            k = i - DLEN + 1;
            if(k < 0) k = 0;
            for(int j = k; j <= i; j++) {
                t = t * 10 + s[j] - '0';
            }
            a[index++] = t;
        }
    }
    BN operator/(const int &b)const {
        BN ret;
        int i, down = 0;
        for(int i = len - 1; i >= 0; i--) {
            ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
            down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
        }
        ret.len = len;
        while(ret.a[ret.len - 1] == 0 && ret.len > 1) ret.len--;
        return ret;
    }
    bool operator>(const BN &T)const {
        int ln;
        if(len > T.len) return true;
        else if(len == T.len) {
            ln = len - 1;
            while(a[ln] == T.a[ln] && ln >= 0) ln--;
            if(ln >= 0 && a[ln] > T.a[ln]) return true;
            else return false;
        } else return false;
    }
    BN operator-(const BN &T)const {
        int i, j, big;
        bool flag;
        BN t1, t2;
        if(*this > T) {
            t1 = *this;
            t2 = T;
            flag = 0;
        } else {
            t1 = T;
            t2 = *this;
            flag = 1;
        }
        big = t1.len;
        for(i = 0; i < big; i++) {
            if(t1.a[i] < t2.a[i]) {
                j = i + 1;
                while(t1.a[j] == 0) j++;
                t1.a[j--]--;
                while(j > i) t1.a[j--] += MAXN;
                t1.a[i] += MAXN + 1 - t2.a[i];
            } else t1.a[i] -= t2.a[i];
        }
        t1.len = big;
        while(t1.a[t1.len - 1] == 0 && t1.len > 1) {
            t1.len--;
            big--;
        }
        if(flag) t1.a[big - 1] = 0 - t1.a[big - 1];
        return t1;
    }
    int operator%(const int &b)const {
        int i, d = 0;
        for(int i = len - 1; i >= 0; i--) {
            d = ((d * (MAXN + 1)) % b + a[i]) % b;
        }
        return d;
    }
};

LL mod;
struct Mat {
    int m, n;
    LL S[matMX][matMX];
    Mat(int a, int b) {
        m = a; n = b;
        memset(S, 0, sizeof(S));
    }
    Mat(int a, int b, LL w[][matMX]) {
        m = a; n = b;
        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                S[i][j] = w[i][j];
            }
        }
    }
};

Mat mat_mul(Mat A, Mat B) {
    Mat C(A.m, B.n);
    for(int i = 0; i < A.m; i++) {
        for(int j = 0; j < B.n; j++) {
            for(int k = 0; k < A.n; k++) {
                C.S[i][j] = (C.S[i][j] + A.S[i][k] * B.S[k][j]) % mod;
            }
        }
    }
    return C;
}

Mat Blank(int m, int n) {
    Mat ret(m, n);
    for(int i = 0; i < m; i++) {
        ret.S[i][i] = 1;
    }
    return ret;
}

Mat mat_pow(Mat A, BN b) {
    Mat ret = Blank(A.m, A.n);
    while(b > 0) {
        if(b % 2) ret = mat_mul(ret, A);
        A = mat_mul(A, A);
        b = b / 2;
    }
    return ret;
}

int m;
LL TB[matMX][matMX];
void DFS(int x, int y, int p) {
    if(p == m) {
        TB[x][y] = 1;
        return;
    }
    DFS(x << 1 | 1, y << 1, p + 1);
    DFS(x << 1, y << 1 | 1, p + 1);
    if(!p || (!((x & 1) && (y & 1)))) DFS(x << 1 | 1, y << 1 | 1, p + 1);
    if(!p || (!(!(x & 1) && !(y & 1)))) DFS(x << 1, y << 1, p + 1);
}

int main() {
    //FIN;
    char word[500];
    while(~scanf("%s%d%lld", word, &m, &mod)) {
        memset(TB, 0, sizeof(TB));
        DFS(0, 0, 0);

        BN n = BN(word);
        Mat s(1 << m, 1 << m, TB), fuck(1 << m, 1);
        for(int i = 0; i < (1 << m); i++) {
            fuck.S[i][0] = 1;
        }
        Mat res = mat_mul(mat_pow(s, n - 1), fuck);

        LL ans = 0;
        for(int i = 0; i < (1 << m); i++) {
            ans = (ans + res.S[i][0]) % mod;
        }
        printf("%lld\n", ans);
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。