Codeforces 691E Xor-sequences

矩阵快速幂。递推式：dp[k][i]=sum(dp[k-1][j]*f[i][j])，dp[k][i]表示的意义是序列中有k个元素，最后一个元素是i的方案数，f[i][j]=1表示i与j能放在一起，反之表示不能放在一起。因为k较大，所以可以构造矩阵进行加速。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi = acos(-1.0), eps = 1e-8;
void File()
{
    freopen("D:\in.txt", "r", stdin);
    freopen("D:\out.txt", "w", stdout);
}
inline int read()
{
    char c = getchar();  while (!isdigit(c)) c = getchar();
    int x = 0;
    while (isdigit(c)) { x = x * 10 + c - '0'; c = getchar(); }
    return x;
}

int n; LL k,a[105],mod=1e9+7;;

struct Matrix
{
    long long A[105][105];
    int R, C;
    Matrix operator*(Matrix b);
};

Matrix X, Y, Z;

Matrix Matrix::operator*(Matrix b)
{
    Matrix c;
    memset(c.A, 0, sizeof(c.A));
    int i, j, k;
    for (i = 1; i <= R; i++)
        for (j = 1; j <= C; j++)
            for (k = 1; k <= C; k++)
                c.A[i][j] = (c.A[i][j] + (A[i][k] * b.A[k][j])%mod)%mod;
    c.R=R; c.C=b.C;
    return c;
}

int check(LL x)
{
    int sz=0;
    while(x) { if(x%2==1) sz++; x=x/2; }
    if(sz%3==0) return 1; return 0;
}

void init()
{
    memset(X.A,0,sizeof X.A);
    memset(Y.A,0,sizeof Y.A);
    memset(Z.A,0,sizeof Z.A);

    Z.R = 1; Z.C = n;

    for(int i=1;i<=n;i++) Z.A[1][i]=Y.A[i][i]=1; Y.R = n; Y.C = n;

    for(int i=1;i<=n;i++)
        for(int j=1;j<=n;j++)
            X.A[i][j]=X.A[j][i]=check(a[i]^a[j]);
    X.R=n; X.C=n;
}

void work()
{
    k--;
    while (k) { if (k % 2 == 1) Y = Y*X; k = k >> 1,X = X*X; } Z = Z*Y;
    LL ans=0;
    for(int i=1;i<=n;i++) ans=(ans+Z.A[1][i])%mod;
    printf("%lld
", ans);
}

int main()
{
    scanf("%d%lld",&n,&k);
    for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
    init(); work();
    return 0;
}