Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P4921 [MtOI2018]情侣?给我烧了!

P4921 [MtOI2018]情侣?给我烧了!

考虑将恰好 $k$ 对转化为不少于 $k$ 对,钦定 $k$ 对后,剩下的随便选取。那么有 $F(k) = \binom n k ^ 2 k! 2 ^ k (2n - 2k)!$ ,又有 $F(k) = \sum _ {i = k} ^ n \binom i k G(i)$ 。直接上二项式反演,有 $G(k) = \sum _ {i = k} ^ n \binom i k (-1) ^ {i - k} F(i)$ 。

对于所有的 $k$ 都要计算,考虑用多项式卷积。复杂度 $O(T n\log n)$ 。

查看代码
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#include <cstdio>
#include <algorithm>
using namespace std;
template <class Type>
void read (Type &x)
{
char c;
bool flag = false;
while ((c = getchar()) < '0' || c > '9')
flag |= c == '-';
x = c - '0';
while ((c = getchar()) >= '0' && c <= '9')
x = (x << 3) + (x << 1) + c - '0';
if (flag) x = ~x + 1;
}
template <class Type, class ...Rest>
void read (Type &x, Rest &...y) { read(x); read(y...); }
template <class Type>
void write (Type x)
{
if (x < 0) putchar('-'), x = ~x + 1;
if (x > 9) write(x / 10);
putchar(char('0' + x % 10));
}
typedef long long LL;
const int N = 5e3 + 10, mod = 998244353;
int n;
int rev[N];
int inv[N], fact[N], ifact[N], p2[N];
void init ()
{
inv[1] = 1;
for (int i = 2; i < N; ++i)
inv[i] = -(LL)(mod / i) * inv[mod % i] % mod;
fact[0] = ifact[0] = 1;
for (int i = 1; i < N; ++i)
{
fact[i] = (LL)fact[i - 1] * i % mod;
ifact[i] = (LL)ifact[i - 1] * inv[i] % mod;
}
p2[0] = 1;
for (int i = 1; i < N; ++i)
p2[i] = p2[i - 1] * 2ll % mod;
}
int C (int a, int b) { return a < b ? 0 : (LL)fact[a] * ifact[b] % mod * ifact[a - b] % mod; }
int binpow (int b, int k = mod - 2)
{
int res = 1;
for (; k; k >>= 1, b = (LL)b * b % mod)
if (k & 1) res = (LL)res * b % mod;
return res;
}
void ntt (int *x, int bit, int op)
{
int tot = 1 << bit;
for (int i = 0; i < tot; ++i)
if ((rev[i] = rev[i >> 1] >> 1 | (i & 1) << bit - 1) > i)
swap(x[i], x[rev[i]]);
for (int mid = 1; mid < tot; mid <<= 1)
{
int w1 = binpow(3, (mod - 1) / (mid << 1));
if (!~op) w1 = binpow(w1);
for (int i = 0; i < tot; i += mid << 1)
for (int j = 0, k = 1; j < mid; ++j, k = (LL)k * w1 % mod)
{
int p = x[i + j], q = (LL)k * x[i + j + mid] % mod;
x[i + j] = (p + q) % mod, x[i + j + mid] = (p - q) % mod;
}
}
if (~op) return;
int itot = binpow(tot);
for (int i = 0; i < tot; ++i)
x[i] = (LL)x[i] * itot % mod;
}
void PolyMul (int *x, int *y, int la, int lb)
{
int bit = 0;
while (la + lb - 1 > 1 << bit) ++bit;
int tot = 1 << bit;
for (int i = la; i < tot; ++i) x[i] = 0;
for (int i = lb; i < tot; ++i) y[i] = 0;
ntt(x, bit, 1), ntt(y, bit, 1);
for (int i = 0; i < tot; ++i)
x[i] = (LL)x[i] * y[i] % mod;
ntt(x, bit, -1);
}
int main ()
{
init();
int T; read(T);
while (T--)
{
read(n);
static int A[N], B[N];
for (int i = 0; i <= n; ++i)
{
A[n - i] = (i & 1 ? -1 : 1) * ifact[i];
B[i] = (LL)p2[i] * ifact[n - i] % mod * ifact[n - i] % mod * fact[n - i << 1] % mod;
}
for (int i = n + 1; i <= n << 1; ++i) A[i] = B[i] = 0;
PolyMul(A, B, n << 1 | 1, n << 1 | 1);
for (int i = 0; i <= n; ++i)
write(((LL)A[n + i] * ifact[i] % mod * fact[n] % mod * fact[n] % mod + mod) % mod), puts("");
}
return 0;
}