hdu 6270 Marriage

题解：常见的二项式反演，用启发式合并（每次选最小的两个集合作NTT，类比下线段树）加速NTT即可。复杂度（n logn l ogn）

#include"bits/stdc++.h"
using namespace std;
typedef long long LL;
const int MX = 4e5+7;
const int mod = 998244353;
const int P = 998244353, G = 3;
const int NUM = 20;
LL  wn[NUM];
LL  va[MX],vb[MX];
LL mul(LL x,LL y)//乘法超ll用快速乘，主函数也需要用
{
    LL ans=(x*y-(LL)((long double)x/mod*y+1e-8)*mod);
    return ans<0?ans+mod:ans;
}
LL quick_mod(LL a, LL x, LL mod) {
    LL ans = 1;
    a %= mod;
    while(x) {
        if(x & 1)ans = ans * a % mod;
        x >>= 1;
        a = a * a % mod;
    }
    return ans;
}
//在程序的开头就要放
void GetWn() {
    for(int i = 0; i < NUM; i++) {
        int t = 1 << i;
        wn[i] = quick_mod(G, (P - 1) / t, P);
    }
}
void Rader(LL F[], int len) {
    int j = len >> 1;
    for(int i = 1; i < len - 1; i++) {
        if(i < j) swap(F[i], F[j]);
        int k = len >> 1;
        while(j >= k)j -= k,k >>= 1;
        if(j < k) j += k;
    }
}
void NTT(LL F[], int len, int t) {
    Rader(F, len);
    int id = 0;
    for(int h = 2; h <= len; h <<= 1) {
        id++;
        for(int j = 0; j < len; j += h) {
            LL E = 1;
            for(int k = j; k < j + h / 2; k++) {
                LL u = F[k];
                LL v = E * F[k + h / 2] % P;
                F[k] = (u + v) % P;
                F[k + h / 2] = (u - v + P) % P;
                E = E * wn[id] % P;
            }
        }
    }
    if(t == -1) {
        for(int i = 1; i < len / 2; i++)swap(F[i], F[len - i]);
        LL inv = quick_mod(len, P - 2, P);
        for(int i = 0; i < len; i++)F[i] = F[i] * inv % P;
    }
}
void Conv(LL a[], LL b[], int len) {
    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);
}
int n;

LL fac[MX],inv[MX];
void init()
{
    int n = 1e5;
    fac[0] = fac[1] = inv[0] = inv[1] = 1;
    for(int i = 2; i <= n; i++){
        fac[i] = fac[i-1]*i%mod;
        inv[i] = (mod-mod/i)*inv[mod%i]%mod;
    }
    for(int i = 2; i <= n; i++) inv[i] = inv[i-1]*inv[i]%mod;
}

LL C(int n, int m)
{
    if(n < m) return 0;
    return fac[n]*inv[m]%mod*inv[n-m]%mod;
}

vector<LL> v[MX];
struct node{
    int id,sz;
    node(){}
    node(int id, int sz) : id(id),sz(sz){}
    bool operator < (const node &a) const{
        return sz > a.sz;
    }
};
priority_queue<node> q;
void work(vector<LL> &a, vector<LL> &b, node &c)
{
    int mx = (int)a.size() + (int)b.size() - 1;
    int len = 1;
    while(len <= mx) len <<= 1;
    for(int i = 0; i < len; i++) va[i] = vb[i] = 0;
    for(int i = 0; i < a.size(); i++) va[i] = a[i];
    for(int i = 0; i < b.size(); i++) vb[i] = b[i];
    Conv(va,vb,len);
    a.clear();
    for(int i = 0; i < mx; i++) a.push_back(va[i]);
    c.sz = a.size();
}
int main()
{
#ifdef LOCAL
    freopen("input.txt","r",stdin);
#endif // LOCAL
    int T;
    init();
    GetWn();
    scanf("%d",&T);
    while(T--){
        int sum = 0;
        scanf("%d",&n);
        for(int i = 1, man,fem; i <= n; i++){
            scanf("%d%d",&man,&fem);
            sum += man;
            v[i].clear();
            for(int j = 0; j <= min(man,fem); j++){
                v[i].push_back(C(man,j)*C(fem,j)%mod*fac[j]);
            }
            q.push(node(i,v[i].size()));
        }

        while(q.size() >= 2){
            node a = q.top(); q.pop();
            node b = q.top(); q.pop();
            node c = node(a.id,0);
            work(v[a.id],v[b.id],c);
            q.push(c);
        }
        node a = q.top(); q.pop();
        while(q.size()) q.pop();
        LL ans = 0;

        for(int i = 0; i < a.sz; i++){
            ans += quick_mod(-1,i,P)*v[a.id][i]%mod*fac[sum-i];
            ans = (ans%mod+mod)%mod;
        }
        printf("%lld\n",ans);
    }
}