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);
}
}