LuoGu: CF1278F Cards
F. Cards
同样是将幂转化为下降幂和第二类斯特林数。设 $p = \frac 1 m, q = 1 - p$
$$
\begin {aligned}
ANS & = \sum _ {i = 0} ^ n \binom n i p ^ i q ^ {n - i} i ^ k \\
& = \sum _ {i = 0} ^ n \binom n i p ^ i q ^ {n - i} \sum _ {j = 0} ^ k {k \brace j} i ^ {\underline j} \\
& = \sum _ {j = 0} ^ k {k \brace j} j! \sum _ {i = 0} ^ n \binom n i \binom i j p ^ i q ^ {n - i}\\
& = \sum _ {j = 0} ^ k {k \brace j} j! \sum _ {i = 0} ^ n \binom n j \binom {n - j} {i - j} p ^ i q ^ {n - i}\\
& = \sum _ {j = 0} ^ k {k \brace j} j! \binom n j \sum _ {i = 0} ^ n \binom {n - j} {i - j} p ^ i q ^ {n - i}\\
& = \sum _ {j = 0} ^ k {k \brace j} j! \binom n j \sum _ {i = 0} ^ {n - j} \binom {n - j} i p ^ {i + j} q ^ {n - i - j}\\
& = \sum _ {j = 0} ^ k {k \brace j} n ^ {\underline j} p ^ j\sum _ {i = 0} ^ {n - j} \binom {n - j} i p ^ i q ^ {n - j - i}\\
& = \sum _ {j = 0} ^ k {k \brace j} n ^ {\underline j} p ^ j (p + q) ^ {n - j} \\
& = \sum _ {j = 0} ^ k {k \brace j} n ^ {\underline j} p ^ j
\end{aligned}
$$
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')
c == '-' && (flag = true);
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(x % 10 + '0');
}
typedef long long LL;
const int N = 5e3 + 10, mod = 998244353;
int n, p, m, s[N][N];
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 init()
{
s[0][0] = 1;
for (int i = 1; i <= m; ++i)
for (int j = 1; j <= m; ++j)
s[i][j] = (s[i - 1][j - 1] + (LL)j * s[i - 1][j]) % mod;
}
int main ()
{
read(n, p, m);
p = binpow(p);
init();
int res = 0;
for (int i = 0, t = 1; i <= min(n, m); t = (LL)t * (n - i++) % mod * p % mod)
res = (res + (LL)t * s[m][i]) % mod;
write(res);
return 0;
}
|