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-;
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 = ;
    while (isdigit(c)) { x = x *  + c - ''; c = getchar(); }
    return x;
}

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

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

Matrix X, Y, Z;

Matrix Matrix::operator*(Matrix b)
{
    Matrix c;
    memset(c.A, , sizeof(c.A));
    int i, j, k;
    for (i = ; i <= R; i++)
        for (j = ; j <= C; j++)
            for (k = ; 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=;
    while(x) { if(x%==) sz++; x=x/; }
    if(sz%==) return ; return ;
}

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

    Z.R = ; Z.C = n;

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

    for(int i=;i<=n;i++)
        for(int j=;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 %  == ) Y = Y*X; k = k >> ,X = X*X; } Z = Z*Y;
    LL ans=;
    for(int i=;i<=n;i++) ans=(ans+Z.A[][i])%mod;
    printf("%lld\n", ans);
}

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