题面
解析
答案显然是$\frac{\sum_{i=1}^n\sum_{j=1}^m (a_i+b_j)^k}{n*m}$
因此只需要求出$\sum_{i=1}^n\sum_{j=1}^m (a_i+b_j)^k$即可
暴力展开：$$\begin{align*}\sum_{i=1}^n\sum_{j=1}^m (a_i+b_j)^k&=\sum_{i=1}^n\sum_{j=1}^m\sum_{p=0}^k\binom{k}{p}a_i^p*b_j^{k-p}\\ &=k!\sum_{p=0}^k\sum_{i=1}^n\frac{a_i^p}{p!}\sum_{j=1}^m\frac{b_j^{k-p}}{(k-p)!}\\&=k!\sum_{p=0}^k\frac{\sum_{i=1}^na_i^p}{p!}\frac{\sum_{j=1}^mb_j^{k-p}}{(k-p)!}\end{align*}$$
现在就是要求对于任一$1\leqslant p \leqslant k$，$\sum_{i=1}^na_i^p$（求$\sum_{j=1}^mb_j^{k-p}$是类似的）
这个比较常见，我在生成函数小结里有写，这里直接给出结论：$$\begin{align*}F(x)&=\sum_{j=0}^{\infty}\sum_{i=1}^na_i^jx^j\\&=n-x\ln'(\prod_{i=1}^n(1-a_ix))\end{align*}$$
$\prod_{i=1}^n(1-a_ix)$可以分治$NTT$
对$a$、$b$分别求出它们的$F(x)$，第$i$项系数除以$i!$后卷积起来。卷积后的第$i$项系数乘以$i!$再除以$n*m$就是答案。
$O(N\log^2N)$
代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
#define ls (x << 1)
#define rs ((x << 1) | 1)
using namespace std;
typedef long long ll;
const int maxn = 200005, mod = 998244353, g = 3;
inline int read()
{
int ret, f=1;
char c;
while((c=getchar())&&(c<'0'||c>'9'))if(c=='-')f=-1;
ret=c-'0';
while((c=getchar())&&(c>='0'&&c<='9'))ret=(ret<<3)+(ret<<1)+c-'0';
return ret*f;
}
int add(int x, int y)
{
return x + y < mod? x + y: x + y - mod;
}
int rdc(int x, int y)
{
return x - y < 0? x - y + mod: x - y;
}
ll qpow(ll x, int y)
{
ll ret = 1;
while(y)
{
if(y&1)
ret = ret * x % mod;
x = x * x % mod;
y >>= 1;
}
return ret;
}
int n, m, a[maxn], b[maxn], lim, bit, rev[maxn<<1];
ll fac[maxn], fnv[maxn];
ll ginv, c[maxn<<1], d[maxn<<1], A[maxn<<1], B[maxn<<1], t[maxn<<1], iv[maxn<<1];
void init()
{
ginv = qpow(g, mod - 2);
fac[0] = 1;
for(int i = 1; i <= 100001; ++i)
fac[i] = fac[i-1] * i % mod;
fnv[100001] = qpow(fac[100001], mod - 2);
for(int i = 100000; i >= 0; --i)
fnv[i] = fnv[i+1] * (i + 1) % mod;
}
void NTT_init(int x)
{
lim = 1;
bit = 0;
while(lim <= x)
{
lim <<= 1;
++ bit;
}
for(int i = 1; i < lim; ++i)
rev[i] = (rev[i>>1] >> 1) | ((i & 1) << (bit - 1));
}
void NTT(ll *x, int y)
{
for(int i = 1; i < lim; ++i)
if(i < rev[i])
swap(x[i], x[rev[i]]);
ll wn, w, u, v;
for(int i = 1; i < lim; i <<= 1)
{
wn = qpow((y == 1)? g: ginv, (mod - 1) / (i << 1));
for(int j = 0; j < lim; j += (i << 1))
{
w = 1;
for(int k = 0; k < i; ++k)
{
u = x[j+k];
v = x[j+k+i] * w % mod;
x[j+k] = add(u, v);
x[j+k+i] = rdc(u, v);
w = w * wn % mod;
}
}
}
if(y == -1)
{
ll linv = qpow(lim, mod - 2);
for(int i = 0; i < lim; ++i)
x[i] = x[i] * linv % mod;
}
}
void get_inv(ll *x, ll *y, int len)
{
if(len == 1)
{
x[0] = qpow(y[0], mod - 2);
return ;
}
get_inv(x, y, (len + 1) >> 1);
for(int i = 0; i < len; ++i)
c[i] = y[i];
NTT_init(len << 1);
NTT(x, 1);
NTT(c, 1);
for(int i = 0; i < lim; ++i)
{
x[i] = rdc(add(x[i], x[i]), (c[i] * x[i] % mod) * x[i] % mod);
c[i] = 0;
}
NTT(x, -1);
for(int i = len; i < lim; ++i)
x[i] = 0;
}
void get_ln(ll *x, ll *y, int len)
{
for(int i = 0; i < len; ++i)
x[i] = y[i+1] * (i + 1) % mod;
get_inv(iv, y, len);
NTT_init(len << 1);
NTT(x, 1);
NTT(iv, 1);
for(int i = 0; i < lim; ++i)
{
x[i] = x[i] * iv[i] % mod;
iv[i] = 0;
}
NTT(x, -1);
for(int i = len - 1; i >= 1; --i)
x[i] = x[i-1] * qpow(i, mod - 2) % mod;
x[0] = 0;
for(int i = len; i < lim; ++i)
x[i] = 0;
}
vector<int> G[maxn<<1];
void solve(int x, int l, int r, int *y)
{
G[x].clear();
if(l == r)
{
G[x].push_back(1);
G[x].push_back(rdc(0, y[l]));
return ;
}
int mid = (l + r) >> 1;
solve(ls, l, mid, y);
solve(rs, mid + 1, r, y);
for(int i = 0; i <= mid - l + 1; ++i)
c[i] = G[ls][i];
for(int i = 0; i <= r - mid; ++i)
d[i] = G[rs][i];
NTT_init(r - l + 1);
NTT(c, 1);
NTT(d, 1);
for(int i = 0; i < lim; ++i)
{
c[i] = c[i] * d[i] % mod;
d[i] = 0;
}
NTT(c, -1);
for(int i = 0; i <= r - l + 1; ++i)
{
G[x].push_back(c[i]);
c[i] = 0;
}
for(int i = r - l + 2; i < lim; ++i)
c[i] = 0;
}
int main()
{
init();
n = read(); m = read();
for(int i = 1; i <= n; ++i)
a[i] = read();
for(int i = 1; i <= m; ++i)
b[i] = read();
int q = read();
solve(1, 1, n, a);
for(int i = 0; i <= n; ++i)
t[i] = G[1][i];
get_ln(A, t, max(q, n) + 1);
for(int i = 0; i <= max(q, n); ++i)
A[i] = A[i+1] * (i + 1) % mod;
for(int i = max(q, n); i >= 1; --i)
A[i] = rdc(0, A[i-1]);
A[0] = n;
for(int i = 0; i <= max(q, n); ++i)
A[i] = A[i] * fnv[i] % mod;
solve(1, 1, m, b);
memset(t, 0, sizeof(t));
for(int i = 0; i <= m; ++i)
t[i] = G[1][i];
get_ln(B, t, max(q, m) + 1);
for(int i = 0; i <= max(q, m); ++i)
B[i] = B[i+1] * (i + 1) % mod;
for(int i = max(q, m); i >= 1; --i)
B[i] = rdc(0, B[i-1]);
B[0] = m;
for(int i = 0; i <= max(q, m); ++i)
B[i] = B[i] * fnv[i] % mod;
NTT_init(max(q, n) + max(q, m));
NTT(A, 1);
NTT(B, 1);
for(int i = 0; i < lim; ++i)
A[i] = A[i] * B[i] % mod;
NTT(A, -1);
ll mul = qpow(1LL * n * m % mod, mod - 2);
for(int i = 1; i <= q; ++i)
printf("%lld\n", (A[i] * fac[i] % mod) * mul % mod);
return 0;
}
来源：https://www.cnblogs.com/Joker-Yza/p/12640512.html