题面
题解
设 (F(n, m) = sum_{k=0}^n k^mx^kinom nk)，于是答案就是 (sum_i a_iF(n, i))。
那么有：
[egin{aligned}
F(n, m) &= sum_{k=0}^n k^m x^k inom nk\
&= nsum_{k=0}^n k^{m-1}x^kinom {n-1}{k-1}
end{aligned}
]
然后：
[egin{aligned}
F(n, m) &= sum_{k=0}^n k^mx^kinom nk\
&= sum_{k=0}^n k^mx^kleft(inom {n-1}k + inom{n-1}{k-1}
ight)\
&= F(n-1,m) + frac 1n F(n, m + 1)
end{aligned}
]
也就是说：(F(n, m) = n(F(n, m - 1) - F(n - 1, m - 1)))。
边界为 (F(n, 0) = (x + 1)^n)。
代码
#include <bits/stdc++.h>
inline int read()
{
int data = 0, w = 1; char ch = getchar();
while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
if (ch == '-') w = -1, ch = getchar();
while (ch >= '0' && ch <= '9') data = data * 10 + (ch ^ 48), ch = getchar();
return data * w;
}
const int N(1010);
int n, x, P, m, f[N], g[N], a[N];
int fastpow(int x, int y)
{
int ans = 1;
for (; y; y >>= 1, x = 1ll * x * x % P)
if (y & 1) ans = 1ll * ans * x % P;
return ans;
}
int main()
{
n = read(), x = read(), P = read(), m = read();
for (int i = 0; i <= m; i++) a[i] = read();
f[0] = fastpow(x + 1, n - m);
for (int i = m - 1; ~i; i--)
{
g[0] = 1ll * f[0] * (x + 1) % P;
for (int j = 1; j <= m - i; j++)
g[j] = 1ll * (g[j - 1] - f[j - 1] + P) % P * (n - i) % P;
for (int j = 0; j <= m - i; j++) f[j] = g[j];
}
int ans = 0;
for (int i = 0; i <= m; i++) ans = (ans + 1ll * f[i] * a[i]) % P;
printf("%d
", ans);
return 0;
}