链接

推导过程

$\sum_{i=1}^n \sum_{j=1}^m (a_i+b_j)^t \\ = \sum_{i=1}^n \sum_{j=1}^m \sum_{k=0}^t \binom{t}{k} a_i^k b_j^{k-t} \\ = \sum_{k=0}^t \binom{t}{k} ( \sum_{i=1}^n a_i^k ) ( \sum_{j=1}^m b_j^{t-k} ) \\ = t! \sum_{k=0}^t \frac{ ( \sum_{i=1}^n a_i^k ) } {k!} \frac{ ( \sum_{j=1}^m b_j^{t-k} ) } {(t-k)!}$

令
$A(x) = \sum_{i=1}^\infin \frac{\sum_{j=1}^n a_j^i}{i!}x^i \\ B(x) = \sum_{i=1}^\infin \frac{\sum_{j=1}^m b_j^i}{i!}x^i$

那么我只需求 $P(x) = A(x)B(x)$ ，再把第 $t$ 项乘以 $t!$ 就可以得出答案

现在需要求 $A(x)$

考虑求
$A_1(x) = \sum_{i=1}^\infin ( \sum_{j=0}^n a_j^i ) x^i$

把式子写开，能更形象一点：

$A_1(x) = n + (a_1+a_2+\dots+a_n)x + (a_1^2+a_2^2 + \dots + a_n^2)x^2 + \dots$

然后可以写成

$A_1(x) = (1+a_1x+a_1^2x^2+\dots) + (1+a_2x+a_2^2x^2+\dots) + \dots$

推到这里就会突然手足无措，别急，根据这道题的结论，我知道
$ln(1-x^i)=-\sum_{k=1}^\infin \frac{x^{ki}}{k}$

然后我用 $a_ix$ 替代 $x^i$ ，得到 $ln(1-a_ix) = - \sum_{k=1}^\infin \frac{a_i^k}{k}$

然后就可以发现，我只要求一下 $\sum_{i=1}^n ln(1-a_ix)= \ln(\prod (1-a_ix))$ ，然后再把第 $k$ 项系数乘以 $-k$ 给常数项赋值 $n$ ，就得到 $A_1(x)$ 了，然后第 $k$ 项系数除以 $k!$ 就能得到 $A(x)$

同样的方法算出 $B(x)$ ，最后就能求出 $F(x)$ ，然后就有答案了

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#define drep(i,a,b) for(i=a;i>=b;i--)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
struct EasyMath
{
    ll prime[maxn], phi[maxn], mu[maxn];
    bool mark[maxn];
    ll fastpow(ll a, ll b, ll c)
    {
        ll t(a%c), ans(1ll);
        for(;b;b>>=1,t=t*t%c)if(b&1)ans=ans*t%c;
        return ans;
    }
    void exgcd(ll a, ll b, ll &x, ll &y)
    {
        if(!b){x=1,y=0;return;}
        ll xx, yy;
        exgcd(b,a%b,xx,yy);
        x=yy, y=xx-a/b*yy;
    }
    ll inv(ll x, ll p)  //p是素数
    {return fastpow(x%p,p-2,p);}
    ll inv2(ll a, ll p)
    {
        ll x, y;
        exgcd(a,p,x,y);
        return (x+p)%p;
    }
    void shai(ll N)
    {
        ll i, j;
        for(i=2;i<=N;i++)mark[i]=false;
        *prime=0;
        phi[1]=mu[1]=1;
        for(i=2;i<=N;i++)
        {
            if(!mark[i])prime[++*prime]=i, mu[i]=-1, phi[i]=i-1;
            for(j=1;j<=*prime and i*prime[j]<=N;j++)
            {
                mark[i*prime[j]]=true;
                if(i%prime[j]==0)
                {
                    phi[i*prime[j]]=phi[i]*prime[j];
                    break;
                }
                mu[i*prime[j]]=-mu[i];
                phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
    ll CRT(vector<ll> a, vector<ll> m) //要求模数两两互质
    {
        ll M=1, ans=0, n=a.size(), i;
        for(i=0;i<n;i++)M*=m[i];
        for(i=0;i<n;i++)(ans+=a[i]*(M/m[i])%M*inv2(M/m[i],m[i]))%=M;
        return ans;
    }
}em;
#define mod 998244353ll
struct NTT
{
    ll n;
    vector<ll> R;
    void init(ll bound)    //bound是积多项式的最高次幂
    {
        ll L(0);
        for(n=1;n<=bound;n<<=1,L++);
        R.resize(n);
        for(ll i=0;i<n;i++)R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));
    }
    void ntt(vector<ll>& a, int opt)
    {
        ll i, j, k, wn, w, x, y, inv(em.fastpow(n,mod-2,mod));
        for(i=0;i<n;i++)if(i>R[i])swap(a[i],a[R[i]]);
        for(i=1;i<n;i<<=1)
        {
            if(opt==1)wn=em.fastpow(3,(mod-1)/(i<<1),mod);
            else wn=em.fastpow(3,(mod-1-(mod-1)/(i<<1)),mod);
            for(j=0;j<n;j+=i<<1)
                for(w=1,k=0;k<i;k++,w=w*wn%mod)
                {
                    x=a[k+j], y=a[k+j+i]*w%mod;
                    a[k+j]=(x+y)%mod, a[k+j+i]=(x-y)%mod;
                }
        }
        if(opt==-1)for(i=0;i<n;i++)(a[i]*=inv)%=mod;
    }
};
struct Poly
{
    vector<ll> v;
    ll n;	//n是最高次项的次数
    Poly(ll N){v.resize(N+1);n=N;}
    Poly(const Poly& p){v=p.v; n=p.n;}
    void resize(ll N){n=N; v.resize(N+1);}
    ll& operator[](ll id){return v[id];}
    void show()
    {
        printf("n=%lld\n",n);
        ll i; rep(i,0,n-1)printf("%lldx^%lld + ",(v[i]+mod)%mod,i);
        printf("%lldx^%lld\n",(v[n]+mod)%mod,n);
    }
};
Poly operator+(Poly A, Poly B)
{
    ll i;
    Poly C(max(A.n,B.n));
	A.resize(C.n), B.resize(C.n);
    rep(i,0,C.n)C[i]=(A[i]+B[i])%mod;
    return C;
}
Poly operator*(ll x, Poly A)
{
    x%=mod;
    ll i; rep(i,0,A.n)(A[i]*=x)%=mod;
    return A;
}
Poly operator*(Poly A, Poly B)
{
    NTT ntt;
    ll i, n=A.n+B.n;
    ntt.init(A.n+B.n);
    A.resize(ntt.n-1), B.resize(ntt.n-1);
    ntt.ntt(A.v,1), ntt.ntt(B.v,1);
    Poly C(ntt.n-1);
    rep(i,0,C.n)C[i]=(A[i]*B[i])%mod;
    ntt.ntt(C.v,-1);
    C.resize(n);
    return C;
}
Poly operator~(Poly F)
{
    Poly H(0), f(0);
    ll i, j;
    H[0]=em.inv(F[0],mod);
    for(i=2;(i>>1)<=F.n;i*=2)
    {
        f.resize(i-1);
        rep(j,0,min(i-1,F.n))f[j]=F[j];
        Poly t=H*H*f; t.resize(i-1);
        H=2*H+(-1)*t;
    }
    H.resize(F.n);
    return H;
}
Poly ln(Poly A)	//常数项必须为1
{
	Poly B(A.n);
	ll i;
	rep(i,1,A.n)B[i-1]=(A[i]*i)%mod;
	B=B*~A; B.resize(A.n);
	drep(i,A.n,1)B[i]=(B[i-1]*em.inv(i,mod))%mod;
	B[0]=0;
	return B;
}
Poly solve(vector<ll>& a, ll l, ll r)
{
    ll mid(l+r>>1);
    if(l==r)
    {
        Poly ret(1);
        ret[0]=1;
        ret[1]=-a[l];
        return ret;
    }
    Poly L=solve(a,l,mid), R=solve(a,mid+1,r);
    return L*R;
}
int main()
{
    ll n=read(), m=read(), i, j;
    vector<ll> a(n), b(m);
    for(auto &x:a)x=read();
    for(auto &x:b)x=read();
    auto t=read();
    auto A = solve(a,0,n-1), B = solve(b,0,m-1);
    A.resize(t), B.resize(t);
    A = ln(A), B = ln(B);
    rep(i,0,A.n)A[i]=(-A[i]*i)%mod;
    rep(i,0,B.n)B[i]=(-B[i]*i)%mod;
    ll fact=1;
    rep(i,1,A.n)(fact*=i)%=mod, (A[i]*=em.inv(fact,mod))%=mod;
    fact=1;
    rep(i,1,B.n)(fact*=i)%=mod, (B[i]*=em.inv(fact,mod))%=mod;
    A[0]=n, B[0]=m;
    auto F = A*B;
    fact=1;
    rep(i,1,t)
    {
        (fact*=i)%=mod;
        ll ans = F[i]*fact%mod * em.inv(n*m%mod,mod) %mod;
        printf("%lld\n",(ans+mod)%mod);
    }
    return 0;
}

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链接

推导过程

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