luogu P6031 CF1278F Cards 加强版
https://www.luogu.com.cn/problem/P6031
首先发现每次洗牌都是独立的，所以概率
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P=\frac{1}{m}
P=m1
要计算的实际上是
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\sum_{i=0}^n\binom{n}{i}p^i(1-p)^{n-i}i^k
i=0∑n(in)pi(1−p)n−iik
按照套路拆开
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\sum\limits_{j=0}^{k}S(k,j)j!\sum\limits_{i=j}^k\binom{i}{j}\binom{n}{i}p^i(1-p)^{n-i}
j=0∑kS(k,j)j!i=j∑k(ji)(in)pi(1−p)n−i
用二项式定理把右边那个东西合并一下就可以得到
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\sum_{j=0}^kS(k,j)j!\binom{n}{j}p^j
j=0∑kS(k,j)j!(jn)pj
这样就可以做简单版了
显然可以用ntt优化到 O ( n l o g 2 n ) O(nlog_2n) O(nlog2n)，但还是不够
把第二类斯特林数再拆开，推一家式子就可以得到线性的做法了
code:
#include<bits/stdc++.h>
#define ll long long
#define N 10000050
#define mod 998244353
using namespace std;
ll qpow(ll x, ll y) {
ll ret = 1;
for(; y; y >>= 1, x = x * x % mod) if(y & 1) ret = ret * x % mod;
return ret;
}
int vis[N], prime[N], sz, k;
int idk[N];
void get(int n) {
idk[1] = 1;
for(int i = 2; i <= n; i ++) {
if(!vis[i]) {
idk[i] = qpow(i, k);
prime[++ sz] = i;
}
for(int j = 1; j <= sz && i * prime[j] <= n; j ++) {
vis[prime[j] * i] = 1; idk[prime[j] * i] = 1ll * idk[prime[j]] * idk[i] % mod;
if(i % prime[j] == 0) break;
}
}
}
int inv[N], S[N];
void init(int n) {
inv[1] = 1;
for(int i = 2; i <= n; i ++) inv[i] = 1ll * (mod - mod / i) * inv[mod % i] % mod;
}
int n, m;
int main() {
scanf("%d%d%d", &n, &m, &k);
get(k + 2), init(k + 2);
ll P = qpow(m, mod - 2), Pm = 1, Cm = 1;
S[0] = 1;
for(int i = 1; i <= k; i ++) {
Pm = Pm * (mod - P) % mod;
Cm = Cm * ((n - k + i - 1 + mod) % mod) % mod * inv[i] % mod;
S[i] = (1ll * (1 - P + mod) % mod * S[i - 1] % mod + Pm * Cm % mod) % mod;
}
//for(int i = 0; i <= k; i ++) printf("%lld ", S[i]); printf("\n");
ll ans = 0; Cm = 1, Pm = 1;
for(int i = 0; i <= k; i ++) {
(ans += idk[i] * Cm % mod * Pm % mod * S[k - i] % mod) %= mod;
Cm = Cm * ((n - i + mod) % mod) % mod * inv[i + 1] % mod;
Pm = Pm * P % mod;
}
printf("%lld", ans);
return 0;
}
