CF1251F Red-White Fence

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给$n$个白木板，$k$个红木板，放置木板。
$Q$个询问。

  • 周长为$Q_i$
  • 有一块红木板，其余都是白木板，成以红为中心单峰。即前面上升，之后下降

求方案数，$n,l_i,l_k,q\leq 3\times 10^5,k\leq 5$

周长$=2\times (红木板长度+木板数量)$

枚举每一块红木板。考虑如何放置($a$白木板，$b$红木板)

  • 如果$a[j]<b[i],cnt[a[j]=1$ ,表示可以放左边也可以放右边。
  • 如果$a[j]1$ ,只存在一个木板放在左边，一个放在右边。（这样就不出现重复了)
  • $n1$表示$a[j]<b[i],cnt[a[j]=1$的数量，
  • $n2$表示$a[j]1$的数量。

还需要放$k$个白木板的就是$\sum_{i+j=k} (2^i)C_{n1}^iC_{n2}^j$
这里使用$NTT$卷积即可。复杂度$O(k(n+nlogn)+kq)$

代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 3e5 + 1000;
const int mod = 998244353;

const int NT = 2e6 + 1000;
const int P = 998244353, gi = 3;
const double pi = acos(-1.0);

int qpow(int a, int x, int mo)
{
    int res = 1;
    while (x)
    {
        if (x & 1)
            res = 1ll * res * a % mod;
        a = 1ll * a * a % mod;
        x >>= 1;
    }
    return res;
}
int fac[N], facinv[N];
void prepare()
{
    fac[0] = 1;
    facinv[0] = 1;
    for (int i = 1; i < N; i++)
        fac[i] = 1ll * fac[i - 1] * i % mod;
    facinv[N - 1] = qpow(fac[N - 1], mod - 2, mod);
    for (int i = N - 2; i >= 1; i--)
        facinv[i] = 1ll * facinv[i + 1] * (i + 1) % mod;
}
int C(int x, int i)
{
    if (x <= 0)
        return 0;
    if (i > x)
        return 0;
    if (i == 0)
        return 1;
    return 1ll * fac[x] * facinv[i] % mod * facinv[x - i] % mod;
}
int inc(int x, int y, int mo)
{
    if (x + y >= mo)
        return (x + y - mo);
    else
        return (x + y);
}

int del(int x, int y, int mo)
{
    if (x - y < 0)
        return (x - y + mo);
    else
        return (x - y);
}
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}

int rev[NT];
void NTT(int *A, int n, int inv)
{
    for (int i = 0; i < n; i++)
        if (i < rev[i])
            swap(A[i], A[rev[i]]);
    for (int l = 1; l < n; l <<= 1)
    {
        int tt = qpow(gi, (P - 1) / (l << 1), P);
        int temp = (inv == 1 ? tt : qpow(tt, P - 2, P));
        for (int i = 0; i < n; i += (l << 1))
        {
            int omega = 1;
            for (int j = 0; j < l; j++, omega = 1ll * omega * temp % P)
            {
                int x = A[i + j], y = 1ll * omega * A[i + j + l] % P;
                A[i + j] = inc(x, y, P);
                A[i + j + l] = del(x, y, P);
            }
        }
    }
    int invv = qpow(n, P - 2, P);
    if (inv == -1)
        for (int i = 0; i < n; i++)
            A[i] = 1ll * A[i] * invv % P;
}

void NTTX(int *a, int n, int *b, int m, int *ans)
{
    int ML = 1, bit = 0;
    while (ML < n + m)
        ML <<= 1, bit++;
    for (int i = 0; i < ML; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    NTT(a, ML, 1);
    NTT(b, ML, 1);
    for (int i = 0; i < ML; i++)
        ans[i] = 1ll * a[i] * b[i] % P;
    NTT(ans, ML, -1);
}
int mp[N];
int a[N], c[N], p[6][NT], b[N], f[NT], g[NT];
int main()
{
    prepare();
    int n = read(), k = read();
    for (int i = 1; i <= n; i++)
        a[i] = read();
    for (int i = 1; i <= k; i++)
        b[i] = read();
    for (int t = 1; t <= k; t++)
    {
        memset(mp, 0, sizeof(mp));
        memset(f, 0, sizeof(f));
        memset(g, 0, sizeof(g));
        for (int j = 1; j <= n; j++)
            if (a[j] < b[t])
                mp[a[j]]++;
        int n1 = 0, n2 = 0;
        for (int i = 1; i < N; i++)
        {
            if (mp[i] == 1)
                n1++;
            if (mp[i] >= 2)
                n2 += 2;
        }
        g[0] = 1;
        f[0] = 1;
        for (int i = 1; i <= n1; i++)
            f[i] = 1ll * qpow(2, i, mod) * C(n1, i) % mod;
        for (int i = 1; i <= n2; i++)
            g[i] = C(n2, i);
        NTTX(f, n1, g, n2, p[t]);
        //cout << n1 << " " << n2 << endl;
        // for (int i = 1; i <= n2 + n1; i++)
        //     cout << p[t][i] << " ";
        // cout << endl;
    }
    int q;
    scanf("%d", &q);
    for (int i = 1; i <= q; i++)
        c[i] = read() / 2;
    for (int i = 1; i <= q; i++)
    {
        int ans = 0;
        for (int j = 1; j <= k; j++)
        {
            int s = c[i] - b[j] - 1;
            if (s < 0)
                continue;
            if (s == 0)
                ans = inc(ans, 1, mod);
            else
            {
                ans = inc(ans, p[j][s], mod);
            }
        }
        printf("%d\n", ans);
    }
}