玩游戏——生成函数

题面

　　洛谷P4705

解析

　　答案显然是$frac{sum_{i=1}^nsum_{j=1}^m (a_i+b_j)^k}{n*m}$

　　因此只需要求出$sum_{i=1}^nsum_{j=1}^m (a_i+b_j)^k$即可

　　暴力展开：$$egin{align*}sum_{i=1}^nsum_{j=1}^m (a_i+b_j)^k&=sum_{i=1}^nsum_{j=1}^msum_{p=0}^kinom{k}{p}a_i^p*b_j^{k-p}\ &=k!sum_{p=0}^ksum_{i=1}^nfrac{a_i^p}{p!}sum_{j=1}^mfrac{b_j^{k-p}}{(k-p)!}\&=k!sum_{p=0}^kfrac{sum_{i=1}^na_i^p}{p!}frac{sum_{j=1}^mb_j^{k-p}}{(k-p)!}end{align*}$$

　　现在就是要求对于任一$1leqslant p leqslant k$，$sum_{i=1}^na_i^p$（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：$$egin{align*}F(x)&=sum_{j=0}^{infty}sum_{i=1}^na_i^jx^j\&=n-xln'(prod_{i=1}^n(1-a_ix))end{align*}$$

　　$prod_{i=1}^n(1-a_ix)$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。

　　$O(Nlog^2N)$

　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
", (A[i] * fac[i] % mod) * mul % mod);
    return 0;
}