2024ICPC网络赛第二场K

K. Match

位运算逐位考虑！

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

#define ll long long
#define ull unsigned long long
#define int ll

ll read()
{
    ll x = 0; bool f = false; char c = getchar();
    while(c < '0' || c > '9') f |= (c == '-'), c = getchar();
    while(c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c & 15), c = getchar();
    return f ? -x : x;
}

const ll mod = 998244353;
const int N = 205;
int n;
ll K;
vector<ll> a, b;

ll qpow(ll a, ll b, ll mod)
{
    ll ans = 1;
    while(b)
    {
        if(b & 1) ans = ans * a % mod;
        b >>= 1;
        a = a * a % mod; 
    }
    return ans;
}

ll jc[N], inv[N];

void init()
{
    jc[0] = jc[1] = inv[0] = inv[1] = 1;
    for(int i = 2; i <= 200; ++i) jc[i] = jc[i - 1] * i % mod;
    inv[200] = qpow(jc[200], mod - 2, mod);
    for(int i = 199; i >= 2; --i) inv[i] = inv[i + 1] * (i + 1) % mod;
}

ll C(int n, int m)
{
    if(n < 0 || m < 0 || n < m) return 0;
    return jc[n] * inv[m] % mod * inv[n - m] % mod;
}

vector<ll> dfs(vector<ll> &a, vector<ll> &b, int k)
{
    vector<ll> c;
    if(a.size() == 0 || b.size() == 0) return c;
    
    vector<ll> a0, a1, b0, b1;

    for(auto A : a)
        if((A >> k) & 1) a1.emplace_back(A);
        else a0.emplace_back(A);

    for(auto B : b)
        if((B >> k) & 1) b1.emplace_back(B);
        else b0.emplace_back(B);

    if(k == 0)
    {
        if((K >> k) & 1)
        {
            int size1 = min(a1.size(), b0.size()), size2 = min(a0.size(), b1.size());
            c.resize(size1 + size2);
            for(int i = 0; i <= size1; ++i)
                for(int j = 0; j <= size2; ++j)
                    if(i + j - 1 >= 0)
                        c[i + j - 1] = (c[i + j - 1] + C(a1.size(), i) * C(b0.size(), i) % mod * C(a0.size(), j) % mod * C(b1.size(), j) % mod * jc[i] % mod * jc[j]) % mod;

            return c;
        }else
        {
            int size1 = a1.size() + a0.size(), size2 = b0.size() + b1.size(), SIZE = min(size1, size2);
            c.resize(SIZE);
            for(int i = 1; i <= SIZE; ++i)
                c[i - 1] = (c[i - 1] + C(size1, i) * C(size2, i) % mod * jc[i]) % mod;

            return c;
        }
    }

    if((K >> k) & 1)
    {
        vector<ll> tmp1 = dfs(a0, b1, k - 1), tmp2 = dfs(a1, b0, k - 1);
        int size1 = tmp1.size(), size2 = tmp2.size();
        c.resize(size1 + size2);

        for(int i = 0; i <= size1; ++i)
            for(int j = 0; j <= size2; ++j)
                if(i + j > 0)
                    c[i + j - 1] = (c[i + j - 1] + (i > 0 ? tmp1[i - 1] : 1ll) * (j > 0 ? tmp2[j - 1] : 1ll)) % mod;
        
        return c;
    }else
    {
        vector<ll> tmp1 = dfs(a0, b0, k - 1), tmp2 = dfs(a1, b1, k - 1);
        int size1 = tmp1.size(), size2 = tmp2.size();
        int size3 = min(a0.size(), b1.size()), size4 = min(a1.size(), b0.size());
        int sizea0 = a0.size(), sizea1 = a1.size(), sizeb0 = b0.size(), sizeb1 = b1.size();
        c.resize(size1 + size2 + size3 + size4);

        for(int i = 0; i <= size1; ++i)
            for(int j = 0; j <= size2; ++j)
                for(int k = min(sizea0 - i, sizeb1 - j); k >= 0; --k)
                    for(int t = min(sizea1 - j, sizeb0 - i); t >= 0; --t)
                        if(i + j + k + t > 0)
                            c[i + j + k + t - 1] = (c[i + j + k + t - 1] + (i > 0 ? tmp1[i - 1] : 1ll) * (j > 0 ? tmp2[j - 1] : 1ll) % mod * C(sizea0 - i, k) % mod * C(sizea1 - j, t) % mod * C(sizeb0 - i, t) % mod * C(sizeb1 - j, k) % mod * jc[k] % mod * jc[t]) % mod;
        
        while(c.size() && (*--c.end()) == 0) c.pop_back();
        return c;
    }
}

signed main()
{
    init();
    n = read(), K = read();
    a.resize(n), b.resize(n);
    for(int i = 1; i <= n; ++i) a[i - 1] = read();
    for(int i = 1; i <= n; ++i) b[i - 1] = read();
    vector<ll> c = dfs(a, b, 59);
    for(int i = 0; i < c.size(); ++i) printf("%lld\n", c[i]);
    int m = n - c.size();
    for(int i = 1; i <= m; ++i) printf("0\n");
    return 0;
}

/*

4 40
24 72 49 26 
88 6 32 2 

10
29
24
4

*/