Chiori and Doll Picking (hard version) CodeForces - 1336E2

This is the hard version of the problem. The only difference between easy and hard versions is the constraint of $m$

Chiori loves dolls and now she is going to decorate her bedroom!

$\text{[math]}$

As a doll collector, Chiori has got $n$

Let $x$

Tell her the number of picking ways with value $i$

Input

The first line contains two integers $n$

The second line contains $n$

Output

Print $m + 1$

Examples

Input

4 4
3 5 8 14

Output

2 2 6 6 0

Input

6 7
11 45 14 9 19 81

Output

1 2 11 20 15 10 5 0 

tutorial：

Build linear basis

Algorithm 1

Enumerate each base of

Algorithm 2

Let's assume the highest

Then, let

At last, we add up

So far, the easy version can be passed if you write a solution which runs Algorithm 1 or Algorithm 2 by the value of

Algorithm 3

We can regard

By XOR Fast Walsh-Hadamard Transform, we calculate

Lemma 1:

Proof: The linear space satisfies closure, which means

Lemma 2: The

Proof: XOR Fast Walsh-Hadamard Transform tells us, the

Lemma 3: The indices of

Proof: See Lemma 2. If

Suppose

Divide key bits for

Look at the following picture. Do you notice that the non-key bit matrices (green areas) are symmetrical along the diagonal?

$\text{[math]}$

The proof is intuitive.

In order to get

Lemma 4: The

Proof: The

Let

Note that

It takes

Finally, let's consider the sum of every number in

Just like Algorithm 1. We can enumerate each base of

#include <bits/stdc++.h>
using namespace std;
using i64 = long long;

const int maxN = 223456;
const int P = 998244353;
int n, m, rnk;
i64 base[maxN], p[maxN], dp[60][60][2];
int cnt[maxN], ans[maxN];

void dfs1(int d, i64 x)
{
    if (d == rnk)
    {
        cnt[__builtin_popcountll(x)]++;//__builtin_popcountll()返回‘1’的个数
    }
    else
    {
        dfs1(d + 1, x);
        dfs1(d + 1, x ^ p[d]);
    }
}
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; i++)
    {
        i64 x;
        scanf("%lld", &x);
        for (int j = m - 1; j >= 0; j--)
        {
            if (base[j])
            {
                x = min(x, base[j] ^ x);
            }
            else if (x & (1ll << j))
            {
                base[j] = x;
                p[rnk++] = x;
                break;
            }
        }
    }
    if (rnk <= m / 2)
    {
        dfs1(0, 0);
        i64 multi = 1;
        for (int i = 0; i < n - rnk; i++)
            multi = multi * 2 % P;
        for (int i = 0; i <= m; i++)
        {
            printf("%lld ", cnt[i] * multi % P);
        }
    }
    else
    {
        for (int i = m - 1; i >= 0; i--)
        {
            for (int j = i - 1; j >= 0; j--)
                base[i] = min(base[i], base[i] ^ base[j]);
        }
        rnk = 0;
        for (int i = 0; i < m; i++)
            if (base[i] == 0)
            {
                i64 x = (1ll << i);
                for (int j = i + 1; j < m; j++)
                    if (base[j] & (1ll << i))
                    {
                        x ^= (1ll << j);
                    }
                p[rnk++] = x;
            }
        dfs1(0, 0);
        for (int x = 0; x <= m; x++)
        {

            memset(dp, 0, sizeof(dp));
            dp[0][0][0] = 1;
            for (int j = 0; j < m; j++)
                for (int k = 0; k <= m; k++)
                    for (int par = 0; par <= 1; par++)
                    {
                        dp[j + 1][k + 1][par ^ (j < x)] += dp[j][k][par];
                        dp[j + 1][k][par] += dp[j][k][par];
                    }
            for (int k = 0; k <= m; k++)
            {
                i64 w = dp[m][k][0] - dp[m][k][1];
                w %= P;
                if (w < 0)
                    w += P;
                ans[k] = (ans[k] + cnt[x] * w) % P;
                
            }
        }
        i64 multi = 1;
        for (int i = 0; i < n; i++)
            multi = multi * 2 % P;
        for (int i = 0; i < m; i++)
            multi = multi * (P + 1) / 2 % P;
        for (int i = 0; i <= m; i++)
        {
            printf("%lld ", ans[i] * multi % P);
        }
    }
}

Chiori and Doll Picking (hard version) CodeForces - 1336E2

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