Luogu4705 玩游戏 分治FFT

传送门


\(\begin{align*} Ans_k &= \sum\limits_{i=1}^n\sum\limits_{j=1}^m (a_i + b_j)^k \\ &= \sum\limits_{i=1}^n \sum\limits_{j=1}^m \sum\limits_{p=0}^k \binom{k}{p} a_i^p b_j^{k-p} \\ &= k! \sum\limits_{p=0}^k \frac{\sum\limits_{i=1}^n a_i^p}{p!}\frac{\sum\limits_{j=1}^m b_j^{k-p}}{(k-p)!} \end{align*}\)

最后的式子是一个卷积的形式,所以我们只需要求出多项式\(A(x) = \sum\limits_{p=0}^T \sum\limits_{i=1}^n a_i^p x^p\)\(B(x) = \sum\limits_{p=0}^T \sum\limits_{i=1}^m b_i^p x^p\)的各项系数,就可以通过一次卷积求出答案。

显然的一件事情是\(A(x)\)\(B(x)\)求法一样,所以我们只考虑\(A(x)\)

不难发现设\(F_i(x) = \sum\limits_{j=0}^\infty a_i^jx^j = \frac{1}{1-a_ix}\),那么\(A(x) = \sum\limits_{i=1}^n F_i(x)\)。直接暴力通分复杂度显然是\(O(n^2)\),但是不难发现和若干个二次多项式相乘一样,这个可以使用分治FFT优化通分,复杂度可以变为\(O(nlog^2n)\)

但是因为分治FFT优化通分还是太慢所以及其容易被卡常,可以考虑当分治区间小到某个值以下时暴力通分。

多项式Ln的做法学不来qwq

#include<bits/stdc++.h>
//this code is written by Itst
using namespace std;

int read(){
    int a = 0; char c = getchar();
    while(!isdigit(c)) c = getchar();
    while(isdigit(c)){
        a = a * 10 + c - 48; c = getchar();
    }
    return a;
}

const int _ = (1 << 18) + 7 , MOD = 998244353;

int poww(long long a , int b){
    int times = 1;
    while(b){
        if(b & 1) times = times * a % MOD;
        a = a * a % MOD; b >>= 1;
    }
    return times;
}

#define VI vector < int > 

namespace poly{
    const int G = 3 , INV = 332748118;
    int dir[_] , need , invnd;

    void init(int len){
        need = 1;
        while(need < len) need <<= 1;
        invnd = poww(need , MOD - 2);
        for(int i = 1 ; i < need ; ++i)
            dir[i] = (dir[i >> 1] >> 1) | (i & 1 ? need >> 1 : 0);
    }

    void NTT(VI &arr , int tp){
        arr.resize(need);
        for(int i = 1 ; i < need ; ++i)
            if(i < dir[i])
                arr[i] ^= arr[dir[i]] ^= arr[i] ^= arr[dir[i]];
        for(int i = 1 ; i < need ; i <<= 1){
            int wn = poww(tp == 1 ? G : INV , MOD / i / 2);
            for(int j = 0 ; j < need ; j += i << 1){
                long long w = 1;
                for(int k = 0 ; k < i ; ++k , w = w * wn % MOD){
                    int x = arr[j + k] , y = arr[i + j + k] * w % MOD;
                    arr[j + k] = x + y >= MOD ? x + y - MOD : x + y;
                    arr[i + j + k] = x < y ? x + MOD - y : x - y;
                }
            }
        }
        if(tp != 1){
            for(int i = 0 ; i < need ; ++i)
                arr[i] = 1ll * arr[i] * invnd % MOD;
            int len = arr.size();
            while(len && arr[len - 1] == 0) --len;
            arr.resize(len);
        }
    }

    void mul(VI a , VI b , VI &c){
        int l = (int)(a.size() + b.size()) - 1;
        init(l); NTT(a , 1); NTT(b , 1); c.resize(need);
        for(int i = 0 ; i < need ; ++i)
            c[i] = 1ll * a[i] * b[i] % MOD;
        NTT(c , -1); c.resize(l);
    }

    void getInv(VI a , VI &b , int len){
        if(len == 1){b.resize(a.size()); return (void)(b[0] = poww(a[0] , MOD - 2));}
        getInv(a , b , (len + 1) >> 1);
        vector < int > A = a , B = b; A.resize(len); B.resize(len);
        init(A.size() + B.size() + 1); NTT(A , 1); NTT(B , 1);
        for(int i = 0 ; i < need ; ++i)
            A[i] = 1ll * A[i] * B[i] % MOD * B[i] % MOD;
        NTT(A , -1);
        for(int i = 0 ; i < len ; ++i)
            b[i] = (2ll * b[i] - A[i] + MOD) % MOD;
    }
}
using namespace poly;

#define mid ((l + r) >> 1)
#define lch (x << 1)
#define rch (x << 1 | 1)
vector < int > arr[_ << 2][2];
int a[_] , b[_] , jc[_] , inv[_] , N , M , T;

void solve(int x , int l , int r , int *num){
    arr[x][0].clear(); arr[x][1].clear();
    if(r - l <= 30){
        vector < int > &A = arr[x][0] , &B = arr[x][1] , tpa , tpb;
        init((r - l + 1) * 2); A.push_back(0); B.push_back(1);
        NTT(A , 1); NTT(B , 1);
        for(int i = l ; i <= r ; ++i){
            tpa.clear(); tpb.clear();
            tpa.push_back(1); tpb.push_back(1); tpb.push_back(MOD - num[i]);
            NTT(tpa , 1); NTT(tpb , 1);
            for(int i = 0 ; i < need ; ++i){
                A[i] = (1ll * A[i] * tpb[i] + 1ll * B[i] * tpa[i]) % MOD;
                B[i] = 1ll * B[i] * tpb[i] % MOD;
            }
        }
        NTT(A , -1); NTT(B , -1);
    }
    else{
        solve(lch , l , mid , num); solve(rch , mid + 1 , r , num);
        int l = arr[lch][1].size() + arr[rch][1].size();
        init(l); vector < int > a , b , c , d;
        arr[x][0].resize(need); arr[x][1].resize(need);
        a = arr[lch][0]; b = arr[lch][1]; c = arr[rch][0]; d = arr[rch][1];
        NTT(a , 1); NTT(b , 1); NTT(c , 1); NTT(d , 1);
        for(int i = 0 ; i < need ; ++i){
            arr[x][0][i] = (1ll * a[i] * d[i] + 1ll * b[i] * c[i]) % MOD;
            arr[x][1][i] = 1ll * b[i] * d[i] % MOD;
        }
        NTT(arr[x][0] , -1); NTT(arr[x][1] , -1);
    }
}

void init(){
    jc[0] = 1;
    for(int i = 1 ; i <= T ; ++i)
        jc[i] = 1ll * jc[i - 1] * i % MOD;
    inv[T] = poww(jc[T] , MOD - 2);
    for(int i = T - 1 ; i >= 0 ; --i)
        inv[i] = inv[i + 1] * (i + 1ll) % MOD;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen("in","r",stdin);
    //freopen("out","w",stdout);
#endif
    N = read(); M = read();
    for(int i = 1 ; i <= N ; ++i) a[i] = read();
    for(int i = 1 ; i <= M ; ++i) b[i] = read();
    T = read(); init();
    vector < int > tp1 , tp2;
    solve(1 , 1 , N , a); arr[1][1].resize(T + 1); getInv(arr[1][1] , tp1 , T + 1);
    mul(tp1 , arr[1][0] , tp1); tp1.resize(T + 1);
    solve(1 , 1 , M , b); arr[1][1].resize(T + 1); getInv(arr[1][1] , tp2 , T + 1);
    mul(tp2 , arr[1][0] , tp2); tp2.resize(T + 1);
    for(int i = 0 ; i <= T ; ++i){
        tp1[i] = 1ll * tp1[i] * inv[i] % MOD;
        tp2[i] = 1ll * tp2[i] * inv[i] % MOD;
    }
    mul(tp1 , tp2 , tp1);
    int Inv = poww(1ll * N * M % MOD , MOD - 2);
    for(int i = 1 ; i <= T ; ++i)
        printf("%lld\n" , 1ll * tp1[i] * jc[i] % MOD * Inv % MOD);
    return 0;
}
posted @ 2019-06-08 09:41 CJOIer_Itst 阅读(...) 评论(...) 编辑 收藏