[LOJ2541][PKUWC2018]猎人杀(容斥+分治+FFT)

https://blog.csdn.net/Maxwei_wzj/article/details/80714129

n个二项式相乘可以用分治+FFT的方法，使用空间回收可以只开log个数组。

 #include<cstdio>
 #include<algorithm>
 #define rep(i,l,r) for (int i=(l); i<=(r); i++)
 using namespace std;

 const int N=,mod=;
 int n,top,ans,w[N],a[N],rev[N],S[],tmp[][N];

 int ksm(int a,int b){
     int res=;
     for (; b; a=1ll*a*a%mod,b>>=)
         if (b & ) res=1ll*res*a%mod;
     return res;
 }

 void NTT(int a[],int len,int f){
     int n=,L=;
     for (; n<len; n<<=) L++;
     for (int i=; i<n; i++) rev[i]=(rev[i>>]>>)|((i&)<<(L-));
     for (int i=; i<n; i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
     for (int i=; i<n; i<<=){
         int wn=ksm(,(f==) ? (mod-)/(i<<) : (mod-)-(mod-)/(i<<));
         for (int p=i<<,j=; j<n; j+=p){
             int w=;
             for (int k=; k<i; k++,w=1ll*w*wn%mod){
                 int x=a[j+k],y=1ll*w*a[i+j+k]%mod;
                 a[j+k]=(x+y)%mod; a[i+j+k]=(x-y+mod)%mod;
             }
         }
     }
     if (f==) return;
     int inv=ksm(n,mod-);
     for (int i=; i<n; i++) a[i]=1ll*a[i]*inv%mod;
 }

 int solve(int l,int r,int a[]){
     if (l==r){ a[]=; a[w[l]]=mod-; return w[l]; }
     int mid=(l+r)>>,l1,l2,ls,rs,n;
     ls=S[top--]; l1=solve(l,mid,tmp[ls]);
     rs=S[top--]; l2=solve(mid+,r,tmp[rs]);
     for (n=; n<=l1+l2; n<<=);
     NTT(tmp[ls],n,); NTT(tmp[rs],n,);
     for (int i=; i<n; i++) a[i]=1ll*tmp[ls][i]*tmp[rs][i]%mod;
     NTT(a,n,-); S[++top]=ls; S[++top]=rs;
     for (int i=; i<n; i++) tmp[ls][i]=tmp[rs][i]=;
     return l1+l2;
 }

 int main(){
     freopen("kill.in","r",stdin);
     freopen("kill.out","w",stdout);
     scanf("%d",&n);
     rep(i,,n) scanf("%d",&w[i]);
     rep(i,,) S[++top]=i;
     int len=solve(,n,a);
     rep(i,,len) ans=(ans+1ll*a[i]*ksm(w[]+i,mod-))%mod;
     printf("%lld\n",1ll*ans*w[]%mod);
     return ;
 }