luogu P4705 玩游戏

https://www.luogu.com.cn/problem/P4705

又碰到一个自己不会的套路，麻了

首先把要计算的式子写出来

$$\sum_{i=1}^n\sum_{j=1}^m(a_i+b_j)^k$$

首先二项式展开

$\large \sum\limits_{i=1}^n\sum\limits_{j=1}^m\sum\limits_{t=0}^k\binom{k}{t}a_i^tb_j^{k-t}$

交换一下求和顺序,把组合数拆开

$\large k!\sum\limits_{t=0}^k\sum\limits_{i=1}^n\frac{a_i^t}{t!}\sum\limits_{j=1}^m\frac{b_j^{k-t}}{(k-t)!}$

考虑生成函数
设$\large A(x)=\sum\limits_{t=0} \frac{x^t}{t!} \sum\limits_{i=1}^n a_i^t$
$\large B(x)=\sum\limits_{t=0} \frac{x^t}{t!} \sum\limits_{i=1}^n b_i^t$

答案的$EGF=F(x)$

容易发现$F(x)=A(x)*B(x)$

然后我们来考虑$A(x)$怎么求？

主要是后面那坨$\sum\limits_{i=1}^n a_i^t$没办法快速计算
我们交换一波求和顺序

$\large \sum\limits_{i=1}^n\sum\limits_{t=0}a_i^tx^t$

很容易得到这个东西为

$\large \sum\limits_{i=1}^n\frac{1}{1-a_ix}$
然而我推到这一步就不会了 /kk

这时候我们需要一个经典套路

注意到$(ln(1-a_ix))'=\frac{-a_i}{1-a_ix}$
我们上面那条式子显然可以变为

$\large \sum\limits_{i=1}^n1-\frac{-a_ix}{1-a_ix}$
$= \large n-x\sum\limits_{i=1}^n\frac{-a_i}{1-a_ix}$

$= \large n-x\sum\limits_{i=1}^n(ln(1-a_ix))'$

因为导数的和=和的导数
所以
$= \large n-x (ln(\prod(1-a_ix)))'$

里面那个$\prod(1-a_ix)$显然可以分治$ntt$

然后大力计算即可

code:

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#include<bits/stdc++.h>
#define N 800050
#define sz(x) ((int)x.size())
#define poly vector<int>
#define mod 998244353
using namespace std;
int add(int x, int y) { x += y;
    if(x >= mod) x -= mod;
    return x;
}
int sub(int x, int y) { x -= y;
    if(x < 0) x += mod;
    return x;
}
int mul(int x, int y) {
    return 1ll * x * y % mod;
}
int qpow(int x, int y) {
    int ret = 1;
    for(; y; y >>= 1, x = mul(x, x)) if(y & 1) ret = mul(ret, x);
    return ret;
}
const int G = 3;
const int Ginv = qpow(G, mod - 2);
int rev[N << 1];
void ntt(int *a, int n, int o) {
    for(int i = 0; i < n; i ++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * (n >> 1));
    for(int i = 0; i < n; i ++) if(rev[i] > i) swap(a[i], a[rev[i]]);
    for(int len = 2; len <= n; len <<= 1) {
        int w0 = qpow((o == 1)? G : Ginv, (mod - 1) / len);
        for(int j = 0; j < n; j += len) {
            int wn = 1;
            for(int k = j; k < j + (len >> 1); k ++, wn = mul(wn, w0)) {
                int X = a[k], Y = mul(a[k + (len >> 1)], wn);
                a[k] = add(X, Y), a[k + (len >> 1)] = sub(X, Y);
            }   
        }
    }
    int ninv = qpow(n, mod - 2);
    if(o == -1)
        for(int i = 0; i < n; i ++) a[i] = mul(a[i], ninv);
}
poly operator + (const poly &A, const poly & B) {
    poly C = A; C.resize(max(sz(A), sz(B)));
    for(int i = 0; i < sz(B); i ++) C[i] = add(C[i], B[i]);
    return C;
}
poly operator - (const poly &A, const poly & B) {
    poly C = A; C.resize(max(sz(A), sz(B)));
    for(int i = 0; i < sz(B); i ++) C[i] = sub(C[i], B[i]);
    return C;
}
#define clr(a, n) (memset(a, 0, sizeof(int) * n))
int a[N << 1], b[N << 1], lim;
poly operator * (const poly & A, const poly & B) {
    for(int i = 0; i < sz(A); i ++) a[i] = A[i];
    for(int i = 0; i < sz(B); i ++) b[i] = B[i];
    poly C; C.resize(min(lim, sz(A) + sz(B) - 1));
    int len = 1;
    for(; len <= sz(A) + sz(B) - 1; len <<= 1);
    ntt(a, len, 1), ntt(b, len, 1);
    for(int i = 0; i < len; i ++) a[i] = mul(a[i], b[i]);
    ntt(a, len, -1);
    for(int i = 0; i < sz(C); i ++) C[i] = a[i];
    clr(a, len), clr(b, len);
    return C;
}
poly operator * (const int & a, const poly & A) {
    poly C; C.resize(sz(A));
    for(int i = 0; i < sz(A); i ++) C[i] = mul(A[i], a);
    return C;
}
void pINV(poly &A, poly &B, int n) {
    if(n == 1) B.push_back(qpow(A[0], mod - 2));
    else {
        pINV(A, B, (n + 1) / 2);
        poly C = A; C.resize(n);
        B = 2 * B - B * B * C;
        B.resize(n);
    }
    
}
poly INV(poly A) {
    poly B; pINV(A, B, sz(A));
    return B;
}
int inv[N], fac[N], ifac[N];
void init(int n) {
    inv[1] = 1;
    for(int i = 2; i <= n; i ++)
        inv[i] = sub(0, mul(mod / i, inv[mod % i]));
    fac[0] = 1;
    for(int i = 1; i <= n; i ++) fac[i] = mul(fac[i - 1], i);
    ifac[n] = qpow(fac[n], mod - 2);
    for(int i = n - 1; i >= 0; i --) ifac[i] = mul(ifac[i + 1], i + 1);
}
poly qiudao(const poly A) {
    poly B;
    for(int i = 1; i < sz(A); i ++) B.push_back(mul(i, A[i]));
    B.pop_back();
    return B;
}
poly jifen(const poly A) {
    poly B; B.resize(sz(A));
    for(int i = 1; i < sz(A); i ++) B[i] = mul(A[i - 1], inv[i]);
    return B;
}
poly ln(const poly A) {
    return jifen(qiudao(A) * INV(A));
}
void pexp(poly &A, poly & B, int n) {
    if(n == 1) B.push_back(1);
    else {
        pexp(A, B, (n + 1) / 2);
        poly lnB; lnB = B; lnB.resize(n);
        lnB = ln(lnB);
        for(int i = 0; i < sz(lnB); i ++) lnB[i] = sub(A[i], lnB[i]);
        lnB[0] = add(lnB[0], 1);
        B = B * lnB;
        B.resize(n);
    }
}
poly exp(poly A) {
    poly C; pexp(A, C, sz(A));
    return C;
}
poly cdq(poly &a, int l, int r) {
   // printf("%d %d\n", l, r);
    if(l == r) {
        poly b;  b.push_back(1); b.push_back(sub(0, a[l]));
     //   printf("** %d %d\n", l, a[l]);
        return b;
    }
    int mid = (l + r) >> 1;
    poly f = cdq(a, l, mid), g = cdq(a, mid + 1, r);
    return f * g;
}
int t;
poly get(poly &a, int n) {
    poly f = cdq(a, 1, n); f.resize(t + 5); 

    //f = INV(f) * qiudao(f);
    f = ln(f); 
    f = qiudao(f);
    // for(int i = 0; i <= t; i ++) printf("%d ", f[i]); printf("\n");
    for(int i = t - 1; i >= 0; i --) 
        f[i + 1] = sub(0, f[i]);
    f[0] = n;

    for(int i = 0; i <= t; i ++) f[i] = mul(f[i], ifac[i]);
    return f;
}
poly f, g;
int n, m;
int main() {
    scanf("%d%d", &n, &m); f.resize(n + 3), g.resize(m + 3);
    for(int i = 1; i <= n; i ++) scanf("%d", &f[i]);
    for(int i = 1; i <= m; i ++) scanf("%d", &g[i]);//, printf("fuck%d  %d  ", b[i], i);
    scanf("%d", &t); t ++; lim = 2 * max(n, max(m, t));
  // f.resize(lim + 1), g.resize(lim + 1);
    init(lim);
    f = get(f, n), g = get(g, m);
    //for(int i = 1; i <= m; i ++) printf("** %d ", b[i]); printf("\n");
    // for(int i = 0; i <= t; i ++) printf("%d ", f[i]); printf("\n");
    // for(int i = 0; i <= t; i ++) printf("%d ", g[i]); printf("\n");
    f = f * g;
    int nminv = mul(inv[n], inv[m]);
    for(int i = 1; i < t; i ++) {
        f[i] = mul(f[i], fac[i]);
        printf("%d\n", mul(f[i], nminv));
    }
    return 0;
}