这题太玄学了，蒟蒻写篇题解来让之后复习 = =

假设第 \(i\) 个颜色有 \(cnt_i\) 个珍珠。

\(\sum\limits_{i=1}^{n} \left\lfloor\frac{cnt_i}{2}\right\rfloor \ge m\)

\(\sum\limits_{i=1}^{n} cnt_i - cnt_i \mod 2 \ge 2m\)

\(n - \sum\limits_{i=1}^{n} cnt_i \mod 2 \ge 2m\)

\(\sum\limits_{i=1}^{n} cnt_i \mod 2 \le n-2m\)

即 \(cnt\) 数组中的奇数个数小于等于 \(n-2m\)。

如果 \(n-2m\ge d\)，那么答案显然为 \(d^n\) (没有任何限制了)

如果 \(n-2m\le 0\) 那么答案为 \(0\)。

令 \(g = n - 2 m\)

\(odd_i\) : \(cnt\) 数组中恰好 \(i\) 个奇数的方案数

\(f_i\) : \(cnt\) 数组中钦定 \(i\) 个奇数的方案数

\[ans = \sum\limits_{i=0}^{g} odd_i
\]

\[f_x=\sum\limits_{i=x}^{d} C_i^x odd_i \Leftrightarrow odd_x=\sum\limits_{i=x}^{d} (-1)^{i-x} C_i^x f_i
\]

所以考虑通过算出 f 来算出 odd 数组, 从而求得答案。

\[f_i=n!(C_{D}^{i})(\sum\limits_{i=0} \frac{x^i}{i!})^{D-i}(\sum\limits_{i=0} [i \mod 2 = 1] \frac{x^i}{i!})^i [x^n]
\]

\[f_i=n!(C_{D}^{i})e^{x(D-i)} (\frac{e^{x} - e^{-x}}{2})^{i}[x^n]
\]

二项式展开：

\[f_i=n!(C_{D}^{i})\frac{1}{2^i} \sum\limits_{j=0}^{i} C_i^j e^{x(D-i)} e^{jx} (-e^{-x})^{i-j} [x^n]
\]

\[f_i=n!(C_{D}^{i})\frac{1}{2^i} \sum\limits_{j=0}^{i} (-1)^j C_i^j e^{x(D-i)} e^{(i-j)x} e^{-xj} [x^n]
\]

\[f_i=n!(C_{D}^{i})\frac{1}{2^i} \sum\limits_{j=0}^{i} (-1)^j C_i^j e^{x(D-2j)} [x^n]
\]

\[f_i=n!(C_{D}^{i})\frac{1}{2^i} \sum\limits_{j=0}^{i} (-1)^j C_i^j \frac{(D-2j)^n}{n!}
\]

\[f_i=n!(C_{D}^{i})\frac{1}{2^i} \sum\limits_{j=0}^{i} (-1)^j \frac{i!}{j!(i-j)!} \frac{(D-2j)^n}{n!}
\]

\[f_i=(C_{D}^{i})\frac{1}{2^i} \sum\limits_{j=0}^{i} (-1)^j \frac{i!}{j!(i-j)!} (D-2j)^n
\]

\[P_i=(-1)^i \frac{1}{i!} (D-2i)^n
\]

\[f_i=i!(C_{D}^{i})\frac{1}{2^i} \sum\limits_{j=0}^{i} P_j \frac{1}{(i-j)!}
\]

\[这就是个卷积式了！然后就求出了 f_i
\]

\[odd_x=\sum\limits_{i=x}^{d} (-1)^{i-x} C_i^x f_i
\]

\[odd_x=(-1)^x \sum\limits_{i=x}^{d} (-1)^i \frac{i!}{x!(i-x)!} f_i
\]

\[odd_x=\frac{1}{x!}(-1)^x \sum\limits_{i=x}^{d} (-1)^i \frac{i!}{(i-x)!} f_i
\]

\[G_i =f_i(-1)^ii!
\]

\[odd_x=\frac{1}{x!}(-1)^x \sum\limits_{i=x}^{d} \frac{1}{(i-x)!} G_i
\]

\[odd_x=\frac{1}{x!}(-1)^x \sum\limits_{i=x}^{d} \frac{1}{(i-x)!} G_i
\]

\[odd_x=\frac{1}{x!}(-1)^x \sum\limits_{i=0}^{d-x} \frac{1}{i!} G_{i+x}
\]

\[翻转 odd 和 G
\]

\[odd_{d-x}=\frac{1}{x!}(-1)^x \sum\limits_{i=0}^{d-x} \frac{1}{i!} G_{d-x-i}
\]

然后再卷一卷答案就求出来了！！！

#include<bits/stdc++.h>
using namespace std;
#define L(i, j, k) for(int i = (j), i##E = (k); i <= i##E; i++)
#define R(i, j, k) for(int i = (j), i##E = (k); i >= i##E; i--)
#define ll long long
#define db double
#define make_pair mp
#define first x
#define second y
#define pb push_back
#define mod 998244353
#define iG 3
#define invG 332748118
#define sz(x) (int)(x.size())
const int N = 4e5 + 7;
int qpow(int x, int y) {
    if(x == 0) return 0;
    int res = 1;
    for(; y; x = 1ll * x * x % mod, y >>= 1) if(y & 1) res = 1ll * res * x % mod;
    return res;
}
int ny(int x) { return qpow(x, mod - 2); }
int pp[N];
void fft(int *f, int len, int flag) {
    for(int i = 0; i < len; i++) if(i < pp[i]) swap(f[pp[i]], f[i]);
    for(int i = 2; i <= len; i <<= 1) {
        int l = (i >> 1);
        for(int j = 0; j < len; j += i) {
            int ch = qpow(flag == 1 ? iG : invG, (mod - 1) / i), now = 1;
            for(int k = j; k < j + l; k++) {
                int ta = f[k], tb = 1ll * f[k + l] * now % mod;
                f[k] = (ta + tb) % mod;
                f[k + l] = (ta - tb + mod) % mod;
                now = 1ll * now * ch % mod;
            }
        }
    }
    if(flag == -1) {
        int invn = ny(len);
        for(int i = 0; i < len; i++) f[i] = 1ll * f[i] * invn % mod;
    }
}
int n, m, d, g, ans, f[N], P[N], ml;
int jc[N], njc[N];
int C(int x, int y) { return 1ll * jc[x] * njc[y] % mod * njc[x - y] % mod; }
int main() {
    scanf("%d%d%d", &d, &n, &m);
    if(n - 2 * m < 0) return printf("0\n"), 0;
    if(n - 2 * m >= d) return printf("%d\n", qpow(d, n)), 0;
    jc[0] = njc[0] = 1;
    L(i, 1, d) jc[i] = 1ll * jc[i - 1] * i % mod, njc[i] = ny(jc[i]);
    for(ml = 1; ml <= d * 2; ml <<= 1);
    L(i, 0, d) P[i] = 1ll * qpow((d - 2 * i + mod) % mod, n) * (i % 2 == 0 ? 1 : mod - 1) % mod * njc[i] % mod, f[i] = njc[i];
    for(int i = 0; i < ml; i++) pp[i] = ((pp[i >> 1] >> 1) | ((i & 1) * (ml >> 1)));
    fft(P, ml, 1), fft(f, ml, 1);
    for(int i = 0; i < ml; i++) f[i] = 1ll * P[i] * f[i] % mod;
    fft(f, ml, -1);
    for(int i = d + 1; i < ml; i++) f[i] = 0;
    int now = 1;
    L(i, 0, d) f[i] = 1ll * f[i] * now % mod * jc[d] % mod * njc[d - i] % mod, now = 1ll * now * 499122177 % mod;
    L(i, 0, d) f[i] = 1ll * f[i] * (i % 2 == 0 ? 1 : mod - 1) % mod * jc[i] % mod;
    reverse(f, f + d + 1);
    for(int i = 0; i < ml; i++) P[i] = 0;
    L(i, 0, d) P[i] = njc[i];
    fft(f, ml, 1), fft(P, ml, 1);
    for(int i = 0; i < ml; i++) f[i] = 1ll * f[i] * P[i] % mod;
    fft(f, ml, -1);
    reverse(f, f + d + 1);
    L(i, 0, d) f[i] = 1ll * f[i] * (i % 2 == 0 ? 1 : mod - 1) % mod * njc[i] % mod;
    g = n - 2 * m;
    L(i, 0, g) (ans += f[i]) %= mod;
    printf("%d\n", ans);
    return 0;
}