題目鏈接

洛谷T30212

題解

式子很容易推出來，二項式定理展開後對於\(k\)的答案即可化簡為如下：
\[k!(\sum\limits_{i = 0}^{k} \frac{\sum\limits_{x = 1}^{n} a_x^{i}}{i!} \centerdot \frac{\sum\limits_{x = 1}^{n} b_x^{k - i}}{(k - i)!})\]
是一個卷積的形式
我們只需對所有\(k\)預處理出\(\sum\limits_{i = 1}^{n} a_i^{k}\)，\(b\)也是類似的

月賽時並不會，暴力預處理便滾粗了，，

考慮泰勒展開，有這樣一個式子：
\[ln(1 + x) = \sum\limits_{i = 0}^{\infty} (-1)^{i} \frac{x^{i + 1}}{i + 1}\]

分治\(NTT\) + 多項式求\(ln\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
 

#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 400005,maxm = 100005,INF = 1000000000;
inline int read(){
    int out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return out * flag;
}
const int G = 3,P = 998244353;
int R[maxn];
inline int qpow(int a,LL b){
    int re = 1;
    for (; b; b >>= 1,a = 1ll * a * a % P)
        if (b & 1) re = 1ll * re * a % P;
    return re;
}
void NTT(int* a,int n,int f){
    for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
    for (int i = 1; i < n; i <<= 1){
        int gn = qpow(G,(P - 1) / (i << 1));
        for (int j = 0; j < n; j += (i << 1)){
            int g = 1,x,y;
            for (int k = 0; k < i; k++,g = 1ll * g * gn % P){
                x = a[j + k],y = 1ll * g * a[j + k + i] % P;
                a[j + k] = (x + y) % P,a[j + k + i] = ((x - y) % P + P) % P;
            }
        }
    }
    if (f == 1) return;
    int nv = qpow(n,P - 2); reverse(a + 1,a + n);
    for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}
int n,m,a[maxn],b[maxn],c[maxn],A[maxn],B[maxn],cv[maxn],N;
int fac[maxn],fv[maxn],inv[maxn];
void init(){
    fac[0] = fac[1] = fv[0] = fv[1] = inv[0] = inv[1] = 1;
    for (int i = 2; i <= 100000; i++){
        fac[i] = 1ll * fac[i - 1] * i % P;
        inv[i] = 1ll * (P - P / i) * inv[P % i] % P;
        fv[i] = 1ll * fv[i - 1] * inv[i] % P;
    }
}
int F[30][maxn],deg[maxn],cnt;
void solve(int l,int r){
    if (l == r){
        deg[++cnt] = 1;
        F[cnt][0] = 1; F[cnt][1] = c[l];
        return;
    }
    int mid = l + r >> 1;
    solve(l,mid); solve(mid + 1,r);
    int n = 1,L = 0,a = cnt - 1,b = cnt,m = deg[a] + deg[b];
    while (n <= m) n <<= 1,L++;
    for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    for (int i = deg[a] + 1; i < n; i++) F[a][i] = 0;
    for (int i = deg[b] + 1; i < n; i++) F[b][i] = 0;
    NTT(F[a],n,1); NTT(F[b],n,1);
    for (int i = 0; i < n; i++) F[a][i] = 1ll * F[a][i] * F[b][i] % P;
    NTT(F[a],n,-1);
    cnt--;
    deg[cnt] = m;
    for (int i = m + 1; i < n; i++) F[cnt][i] = 0;
}
void Der(int* a,int n){
    for (int i = 0; i < n; i++) a[i] = 1ll * a[i + 1] * (i + 1) % P;
    a[n] = 0;
}
void Int(int* a,int n){
    for (int i = n + 1; i; i--) a[i] = 1ll * a[i - 1] * inv[i] % P;
    a[0] = 0;
}
void Inv(int* a,int* b,int deg){
    if (deg == 1){b[0] = qpow(a[0],P - 2); return;}
    Inv(a,b,(deg + 1) >> 1);
    int n = 1,L = 0;
    while (n < (deg << 1)) n <<= 1,L++;
    for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    for (int i = 0; i < deg; i++) c[i] = a[i];
    for (int i = deg; i < n; i++) c[i] = 0;
    NTT(c,n,1); NTT(b,n,1);
    for (int i = 0; i < n; i++)
        b[i] = 1ll * ((2ll - 1ll * b[i] * c[i] % P) % P + P) % P * b[i] % P;
    NTT(b,n,-1);
    for (int i = deg; i < n; i++) b[i] = 0;
}
void Getln(int* a,int* b,int deg){
    Inv(a,cv,deg);
    Der(a,deg);
    int n = 1,L = 0;
    while (n <= (deg << 1)) n <<= 1,L++;
    for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    for (int i = deg; i < n; i++) a[i] = 0;
    NTT(a,n,1); NTT(cv,n,1);
    for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * cv[i] % P;
    NTT(a,n,-1);
    for (int i = 0; i <= deg; i++) b[i] = a[i];
    Int(b,deg);
}
int main(){
    init();
    n = read(); m = read(); int v = qpow(1ll * n * m % P,P - 2);
    REP(i,n) a[i] = read();
    REP(i,m) b[i] = read();
    N = read();
    REP(i,n) c[i] = a[i];
    solve(1,n);
    //REP(i,n) printf("%d ",F[1][i]); puts("");
    for (int i = n + 1; i <= N; i++) F[1][i] = 0;
    Getln(F[1],A,N);
    for (int i = 1; i <= N; i++){
        if (!(i & 1)) A[i] = P - A[i];
        A[i] = 1ll * A[i] * i % P * fv[i] % P;
    }
    //REP(i,N) printf("%d ",A[i]); puts("");
    A[0] = n;
    cls(cv);
    REP(i,m) c[i] = b[i]; cnt = 0;
    solve(1,m);
    for (int i = m + 1; i <= N; i++) F[1][i] = 0;
    Getln(F[1],B,N);
    for (int i = 1; i <= N; i++){
        if (!(i & 1)) B[i] = P - B[i];
        B[i] = 1ll * B[i] * i % P * fv[i] % P;
    }
    //REP(i,N) printf("%d ",B[i]); puts("");
    B[0] = m;
    int n = 1,L = 0;
    while (n <= (N << 1)) n <<= 1,L++;
    for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
    for (int i = N + 1; i < n; i++) A[i] = B[i] = 0;
    NTT(A,n,1); NTT(B,n,1);
    for (int i = 0; i < n; i++) A[i] = 1ll * A[i] * B[i] % P;
    NTT(A,n,-1);
    for (int i = 1; i <= N; i++)
        printf("%lld\n",1ll * A[i] * fac[i] % P * v % P);
    return 0;
}

洛谷5月月賽T30212 玩遊戲 【分治NTT + 多項式求ln】