Educational Codeforces Round 14 E.Xor-sequences

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;
}