牛客 133D 挑选队友 (分治FFT)

大意: $n$个人, 分别属于$m$个组, 要求选出$k$个人, 使得每组至少有一人, 求方案数.

显然答案为$\prod((1+x)^{a_i}-1)$的第$k$项系数, 分治$FFT$即可.

#include <iostream>
#include <cstdio>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define mid ((l+r)>>1)
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<int*,int> poly;
const int N = 4e5+10, P = 998244353, G = 3, Gi = 332748118;
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
int lim,l,A[N],B[N],R[N];
void init(int n) {
    for (lim=1,l=0; lim<=n; lim<<=1,++l) ;
    REP(i,0,lim-1) R[i]=(R[i>>1]>>1)|((i&1)<<(l-1));
}

void NTT(int *J, int tp=1) {
    REP(i,0,lim-1) if (i<R[i]) swap(J[i],J[R[i]]);
    for (int j=1; j<lim; j<<=1) {
        ll T = qpow(tp==1?G:Gi,(P-1)/(j<<1));
        for (int k=0; k<lim; k+=j<<1) {
            ll t = 1;
            for (int l=0; l<j; ++l,t=t*T%P) {
                int y = t*J[k+j+l]%P;
                J[k+j+l] = (J[k+l]-y)%P;
                J[k+l] = (J[k+l]+y)%P;
            }
        }
    }
    if (tp==-1) {
        ll inv = qpow(lim, P-2);
        REP(i,0,lim-1) J[i]=(ll)inv*J[i]%P;
    }
}

poly mul(poly a, poly b) {
	init(a.y+b.y);
	REP(i,0,lim-1) A[i]=B[i]=0;
	REP(i,0,a.y) A[i]=a.x[i];
	REP(i,0,b.y) B[i]=b.x[i];
    NTT(A),NTT(B);
	poly c(new int[lim],lim-1);
    REP(i,0,lim-1) c.x[i]=(ll)A[i]*B[i]%P;
    NTT(c.x,-1);
	return c;
}

int n,m,k,a[N],fac[N],ifac[N];
int C(int n, int m) {
	if (m>n) return 0;
	return (ll)fac[n]*ifac[m]%P*ifac[n-m]%P;
}
poly solve(int l, int r) {
	if (l==r) {
		poly ret(new int[a[l]+1],a[l]);
		REP(i,1,a[l]) ret.x[i]=C(a[l],i);
		return ret;
	}
	return mul(solve(l,mid),solve(mid+1,r));
}

int main() {
	fac[0]=ifac[0]=1;
	REP(i,1,N-1) fac[i]=(ll)fac[i-1]*i%P;
	ifac[N-1]=qpow(fac[N-1],P-2);
	PER(i,1,N-2) ifac[i]=(ll)ifac[i+1]*(i+1)%P;
	scanf("%d%d%d", &n, &m, &k);
	REP(i,1,m) scanf("%d",a+i);
	int ans = solve(1,m).x[k];
	if (ans<0) ans += P;
	printf("%d\n", ans);
}