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 ", ans); }