考虑若确定了所有 \(c_s\)，如何计算集合最大大小。
下文令原题面中的 \(f\) 为 \(m\)。
发现我们可以类似倒推地确定。比如若 \(n = 3\)，\(c_{00} = \min(c_{000}, c_{001})\)，\(c_{01} = \min(c_{010}, c_{011})\)，\(c_{10} = \min(c_{100}, c_{101})\)，\(c_{11} = \min(c_{110}, c_{111})\)，\(c_0 = \min(c_{00}, c_{01}), c_1 = \min(c_{10}, c_{11})\)。最大集合大小就是 \(c_0 + c_1\)。
放到一棵满二叉树上考虑（可以理解为 01-Trie），我们有 \(f_u = \min(f_u, f_{ls_u} + f_{rs_u})\)。
考虑计数。设 \(g_{i, j}\) 自底向上第 \(i\) 层结点，\(f_u = j\) 的方案数。转移枚举 \(f_u\) 还是 \(f_{ls_u} + f_{rs_u}\) 取到最小值，有：
\[g_{i, j} = (\sum\limits_{x + y \ge j} g_{i - 1, x} g_{i - 1, y}) + (k - j) \sum\limits_{x + y = j} g_{i - 1, x} g_{i - 1, y}
\]
加法卷积计算 \(h_j = \sum\limits_{x + y = j} g_{i - 1, x} g_{i - 1, y}\)，有：
\[g_{i, j} = (\sum\limits_{p \ge j} h_p) + (k - j) h_p
\]
可以前缀和计算。
答案即为 \(\sum\limits_{x + y = m} g_{n, x} g_{n, y}\)。
时间复杂度 \(O(nk \log k)\)。
code
// Problem: F. Multiset of Strings
// Contest: Codeforces - Educational Codeforces Round 132 (Rated for Div. 2)
// URL: https://codeforces.com/contest/1709/problem/F
// Memory Limit: 512 MB
// Time Limit: 6000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;
const int maxn = 1000100;
const ll mod = 998244353, G = 3;
inline ll qpow(ll b, ll p) {
ll res = 1;
while (p) {
if (p & 1) {
res = res * b % mod;
}
b = b * b % mod;
p >>= 1;
}
return res;
}
ll n, m, K, r[maxn], f[20][maxn];
typedef vector<ll> poly;
inline poly NTT(poly a, int op) {
int n = (int)a.size();
for (int i = 0; i < n; ++i) {
if (i < r[i]) {
swap(a[i], a[r[i]]);
}
}
for (int k = 1; k < n; k <<= 1) {
ll wn = qpow(op == 1 ? G : qpow(G, mod - 2), (mod - 1) / (k << 1));
for (int i = 0; i < n; i += (k << 1)) {
ll w = 1;
for (int j = 0; j < k; ++j, w = w * wn % mod) {
ll x = a[i + j], y = w * a[i + j + k] % mod;
a[i + j] = (x + y) % mod;
a[i + j + k] = (x - y + mod) % mod;
}
}
}
if (op == -1) {
ll inv = qpow(n, mod - 2);
for (int i = 0; i < n; ++i) {
a[i] = a[i] * inv % mod;
}
}
return a;
}
inline poly operator * (poly a, poly b) {
a = NTT(a, 1);
b = NTT(b, 1);
int n = (int)a.size();
for (int i = 0; i < n; ++i) {
a[i] = a[i] * b[i];
}
a = NTT(a, -1);
return a;
}
inline poly mul(poly a, poly b) {
int n = (int)a.size() - 1, m = (int)b.size() - 1, k = 0;
while ((1 << k) <= n + m + 1) {
++k;
}
for (int i = 1; i < (1 << k); ++i) {
r[i] = (r[i >> 1] >> 1) | ((i & 1) << (k - 1));
}
poly A(1 << k), B(1 << k);
for (int i = 0; i <= n; ++i) {
A[i] = a[i];
}
for (int i = 0; i <= m; ++i) {
B[i] = b[i];
}
poly res = A * B;
res.resize(n + m + 1);
return res;
}
void solve() {
scanf("%lld%lld%lld", &n, &K, &m);
for (int i = 0; i <= K; ++i) {
f[1][i] = 1;
}
for (int i = 2; i <= n; ++i) {
poly a(K + 1);
for (int j = 0; j <= K; ++j) {
a[j] = f[i - 1][j];
}
a = mul(a, a);
ll s = 0;
for (int j = K + K; ~j; --j) {
s = (s + a[j]) % mod;
if (j <= K) {
f[i][j] = (s + a[j] * (K - j)) % mod;
}
}
}
poly a(K + 1);
for (int i = 0; i <= K; ++i) {
a[i] = f[n][i];
}
a = mul(a, a);
printf("%lld\n", m >= (ll)a.size() ? 0LL : a[m]);
}
int main() {
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}