【CTS2019Day1】珍珠

传送门.

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不知道怎么说这题，考场时完全没有想过生成函数，随便想想就有70+分了啊。

考虑d个中如果有k个是奇数，则$(n-k)/2>=m$
即$k<=n-2m$，当然$k<=D$

设$f[k]$表示k个奇数的方案。

考虑用生成函数强解：
$f[k]=C_D^k\ast D!\ast (\frac{e^x-e^{-x}}{2}{)}^k\ast (\frac{e^x+e^{-x}}{2}{)}^{D-k}[x^n]$
发现这个真的不会拆了，但是随便写写就有70分了。

不好拆的关键主要在于有两个东西，不妨容斥一下

$g[k]$表示至少k个的方案
$g[k]={\sum }_{i=k}^D{C}_i^k\ast f[i]$
$f[k]={\sum }_{i=k}^D{C}_i^k\ast g[i]\ast (-1{)}^{i-k}$

$g[k]=C_D^k\ast D!\ast e^{D-k}\ast (\frac{e^x-e^{-x}}{2}{)}^k[x^n]$
$g[k]=C_D^k\ast D!\ast 2^{-k}\ast e^{(D-k)x}\ast {\sum }_{i=0}^k(-1{)}^i\ast e^{ix+(k-i)x}[x^n]\ast C_k^i$
$g[k]=C_D^k\ast D!\ast 2^{-k}\ast {\sum }_{i=0}^k(-1{)}^i\ast e^{ix+(k-i)x+(D-k)x}[x^n]\ast C_k^i$
$g[k]=C_D^k\ast D!\ast 2^{-k}\ast {\sum }_{i=0}^k(-1{)}^i\ast e^{(D-2i)x}[x^n]\ast C_k^i$
$g[k]=C_D^k\ast D!\ast 2^{-k}\ast {\sum }_{i=0}^k(-1{)}^i\ast (D-2i{)}^n\ast C_k^i$

到这一步已经非常显然的可以$NTT$优化了。

Code：

#include<bits/stdc++.h>
#define fo(i, x, y) for(int i = x, B = y; i <= B; i ++)
#define ff(i, x, y) for(int i = x, B = y; i <  B; i ++)
#define fd(i, x, y) for(int i = x, B = y; i >= B; i --)
#define ll long long
#define pp printf
#define hh pp("\n")
#define ui unsigned int
using namespace std;

const int mo = 998244353;

ll ksm(ll x, ll y) {
	ll s = 1;
	for(; y; y /= 2, x = x * x % mo)
		if(y & 1) s = s * x % mo;
	return s;
}

const int N = 1e5 + 4;

ll a[N * 4], b[N * 4]; int r[N * 4];
void dft(ll *a, int n, int f) {
	ff(i, 0, n) {
		r[i] = r[i / 2] / 2 + (i & 1) * n / 2;
		if(i < r[i]) swap(a[i], a[r[i]]);
	}
	for(int h = 1; h < n; h *= 2) {
		ll wn = ksm(ksm(3, (mo - 1) / 2 / h), f == 1 ? 1 : mo - 2);
		for(int j = 0; j < n; j += 2 * h) {
			ll w = 1, b, *l = a + j, *r = a + j + h;
			ff(i, 0, h) {
				b = *r * w, *r = (*l - b) % mo, *l = (*l + b) % mo;
				w = w * wn % mo, l ++, r ++;
			}
		}
	}
	if(f == -1) {
		ll v = ksm(n, mo - 2);
		ff(i, 0, n) a[i] = (a[i] + mo) * v % mo;
	}
}
void fft(ll *a, ll *b, int n) {
	dft(a, n, 1); dft(b, n, 1);
	ff(i, 0, n) a[i] = a[i] * b[i] % mo;
	dft(a, n, -1);
}

int n, n2, m;
ll fac[N], nf[N];

#define mem(a) memset(a, 0, sizeof a)

int main() {
	scanf("%d %d %d", &n, &n2, &m);
	fac[0] = 1; fo(i, 1, n) fac[i] = fac[i - 1] * i % mo;
	nf[n] = ksm(fac[n], mo - 2); fd(i, n, 1) nf[i - 1] = nf[i] * i % mo;
	int tp = 0; while(1 << ++ tp <= 2 * n);
	fo(i, 0, n) {
		a[i] = ksm(n - 2 * i, n2) * (i % 2 ? -1 : 1) * nf[i] % mo;
		b[i] = nf[i];
	}
	fft(a, b, 1 << tp);
	ff(i, n + 1, 1 << tp) a[i] = 0;
	mem(b);
	fo(i, 0, n) {
		a[i] = a[i] * ksm((mo + 1) / 2, i) % mo * fac[n] % mo * nf[n - i] % mo * fac[i] % mo;
		b[n - i] = nf[i] % mo * (i % 2 ? -1 : 1);
	}
	fft(a, b, 1 << tp);
	fo(i, 0, n) a[i] = a[i + n] * nf[i] % mo;
	ll ans = 0;
	fo(i, 0, min(n2 - 2 * m, n)) ans += a[i];
	pp("%lld\n", (ans % mo + mo) % mo);
}