Educational Codeforces Round 14 E.Xor-sequences

时间：2023-03-08 17:20:27

题目链接

分析：K很大，以我现有的极弱的知识储备，大概应该是快速幂了。。。怎么考虑这个快速幂呢，用到了dp的思想。定义dp[i][j]表示从a[i]到a[j]的合法路径数。那么递推式就是dp[i][j]=∑k(dp[i][k]∗dp[k][j])。每次进行这样一次计算，那么序列的长度就会增加一，因此只要将这个式子做k次就行了。怎么满足相邻两个数异或值的1的个数为3倍数呢？这就是用到矩阵的时候了。枚举ij，建立一个N∗N的矩阵，当a[i]⊗a[j]为3的倍数，m[i][j]为1，否则为零。再考虑到矩阵的乘法其实和刚才的dp递推式是一样的？？！！因此只要将矩阵乘K−1次就行了。

/*****************************************************/

//#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <map>

#include <set>

#include <ctime>

#include <stack>

#include <queue>

#include <cmath>

#include <string>

#include <vector>

#include <cstdio>

#include <cctype>

#include <cstring>

#include <sstream>

#include <cstdlib>

#include <iostream>

#include <algorithm>

using namespace std;

#define   offcin        ios::sync_with_stdio(false)

#define   sigma_size    26

#define   lson          l,m,v<<1

#define   rson          m+1,r,v<<1|1

#define   slch          v<<1

#define   srch          v<<1|1

#define   sgetmid       int m = (l+r)>>1

#define   LL            long long

#define   ull           unsigned long long

#define   mem(x,v)      memset(x,v,sizeof(x))

#define   lowbit(x)     (x&-x)

#define   bits(a)       __builtin_popcount(a)

#define   mk            make_pair

#define   pb            push_back

#define   fi            first

#define   se            second

const int    INF    = 0x3f3f3f3f;

const LL     INFF   = 1e18;

const double pi     = acos(-1.0);

const double inf    = 1e18;

const double eps    = 1e-9;

const LL     mod    = 1e9+7;

const int    maxmat = 10;

const ull    BASE   = 31;

/*****************************************************/

const int maxn = 1e2 + 5;

LL p[maxn];

struct Mat {

    int n;

    LL a[105][105];

    Mat(int _n = 0) : n(_n) {mem(a, 0);}

    Mat operator *(const Mat &rhs) const {

        int n = rhs.n;

        Mat c(n);

        for (int i = 0; i < n; i ++)

            for (int j = 0; j < n; j ++)

                for (int k = 0; k < n; k ++)

                    c.a[i][j] = (c.a[i][j] + a[i][k] * rhs.a[k][j] % mod) % mod;

        return c;

    }

};

int count(LL k) {

    int ans = 0;

    while (k) {

        if (k & 1) ans ++;

        k >>= 1;

    }

    return ans;

}

Mat qpow(Mat A, LL b) {

    int n = A.n;

    Mat c(n);

    for (int i = 0; i < n; i ++) c.a[i][i] = 1;

    while (b) {

        if (b & 1) c = c * A;

        b >>= 1;

        A = A * A;

    }

    return c;

}

int main(int argc, char const *argv[]) {

    int N;

    LL K;

    cin>>N>>K;

    for (int i = 0; i < N; i ++) cin>>p[i];

    Mat A(N);

    for (int i = 0; i < N; i ++)

        for (int j = 0; j < N; j ++)

            if (count(p[i] ^ p[j]) % 3 == 0)

                A.a[i][j] = 1;

    A = qpow(A, K - 1);

    LL ans = 0;

    for (int i = 0; i < N; i ++)

        for (int j = 0; j < N; j ++)

            ans = (ans + A.a[i][j]) % mod;

    cout<<ans<<endl;

    return 0;

}

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