[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=300010,mod=998244353;
int n,top,ans,w[N],a[N],rev[N],S[51],tmp[51][N];

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

void NTT(int a[],int len,int f){
    int n=1,L=0;
    for (; n<len; n<<=1) L++;
    for (int i=0; i<n; i++) rev[i]=(rev[i>>1]>>1)|((i&1)<<(L-1));
    for (int i=0; i<n; i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
    for (int i=1; i<n; i<<=1){
        int wn=ksm(3,(f==1) ? (mod-1)/(i<<1) : (mod-1)-(mod-1)/(i<<1));
        for (int p=i<<1,j=0; j<n; j+=p){
            int w=1;
            for (int k=0; 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==1) return;
    int inv=ksm(n,mod-2);
    for (int i=0; i<n; i++) a[i]=1ll*a[i]*inv%mod;
}

int solve(int l,int r,int a[]){
    if (l==r){ a[0]=1; a[w[l]]=mod-1; return w[l]; }
    int mid=(l+r)>>1,l1,l2,ls,rs,n;
    ls=S[top--]; l1=solve(l,mid,tmp[ls]);
    rs=S[top--]; l2=solve(mid+1,r,tmp[rs]);
    for (n=1; n<=l1+l2; n<<=1);
    NTT(tmp[ls],n,1); NTT(tmp[rs],n,1);
    for (int i=0; i<n; i++) a[i]=1ll*tmp[ls][i]*tmp[rs][i]%mod;
    NTT(a,n,-1); S[++top]=ls; S[++top]=rs;
    for (int i=0; i<n; i++) tmp[ls][i]=tmp[rs][i]=0;
    return l1+l2;
}

int main(){
    freopen("kill.in","r",stdin);
    freopen("kill.out","w",stdout);
    scanf("%d",&n);
    rep(i,1,n) scanf("%d",&w[i]);
    rep(i,1,49) S[++top]=i;
    int len=solve(2,n,a);
    rep(i,0,len) ans=(ans+1ll*a[i]*ksm(w[1]+i,mod-2))%mod;
    printf("%lld
",1ll*ans*w[1]%mod);
    return 0;
}