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\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

P6620 [省选联考 2020 A 卷] 组合数问题

P6620 [省选联考 2020 A 卷] 组合数问题

多项式是若干个幂的和,注意到有组合数,考虑将其拆为斯特林数和下降幂。那么

$$
a _ i k ^ i = a _ i \sum _ {j = 0} ^ i {i \brace j} k ^ {\underline j}
$$

将 $a _ i$ 贡献到每个位置,得到

$$
b _ i = \sum _ {j = i} ^ m {j \brace i} a _ j
$$

那么

$$
\begin {aligned}
ANS & = \sum _ {k = 0} ^ n \binom n k x ^ k\sum _ {i = 0} ^ m b _ i k ^ {\underline i} \\
& = \sum _ {i = 0} ^ m b _ i i!\sum _ {k = 0} ^ n \binom n k \binom k i x ^ k \\
& = \sum _ {i = 0} ^ m b _ i i! \sum _ {k = 0} ^ n \binom n i \binom {n - i} {k -i} x ^ k \\
& = \sum _ {i = 0} ^ m b _ i i! \binom n i \sum _ {k = 0} ^ {n - i} \binom {n - i} k x ^ {k + i} \\
& = \sum _ {i = 0} ^ m b _ i n ^ {\underline i} x ^ i \sum _ {k = 0} ^ {n - i} \binom {n - i} k x ^ k \\
& = \sum _ {i = 0} ^ m b _ i n ^ {\underline i} x ^ i (x + 1) ^ {n - i} \\
\end {aligned}
$$

查看代码
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#include <cstdio>
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;
int n, x, p, m, f[N], g[N], s[N][N];
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]) % p;
}
int binpow (int b, int k)
{
int res = 1;
for (; k; k >>= 1, b = (LL)b * b % p)
if (k & 1) res = (LL)res * b % p;
return res;
}
int main ()
{
read(n, x, p, m);
init();
for (int i = 0; i <= m; ++i) read(f[i]);
for (int i = 0; i <= m; ++i)
for (int j = i; j <= m; ++j)
g[i] = (g[i] + (LL)s[j][i] * f[j]) % p;
int res = 0;
for (int i = 0, t = 1; i <= m; t = (LL)t * (n - i++) % p * x % p)
res = (res + (LL)g[i] * t % p * binpow(x + 1, n - i)) % p;
write((res + p) % p);
return 0;
}