LOJ2541 PKUWC2018猎人杀（概率期望+容斥原理+生成函数+分治NTT）

　　考虑容斥，枚举一个子集S在1号猎人之后死。显然这个概率是w₁/(Σw_i+w₁) (i∈S)。于是我们统计出各种子集和的系数即可，造出一堆形如(-x^w_i+1)的生成函数，分治NTT卷起来就可以了。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
#define N 100010
#define P 998244353
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
    int x=,f=;char c=getchar();
    while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
    while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
    return x*f;
}
int n,a[N],s[N],r[N*],inv[N*],f[N*],t,ans;
int ksm(int a,int k)
{
    int s=;
    for (;k;k>>=,a=1ll*a*a%P) if (k&) s=1ll*s*a%P;
    return s;
}
void DFT(int *a,int n,int g)
{
    for (int i=;i<n;i++) r[i]=(r[i>>]>>)|(i&)*(n>>);
    for (int i=;i<n;i++) if (i<r[i]) swap(a[i],a[r[i]]);
    for (int i=;i<=n;i<<=)
    {
        int wn=ksm(g,(P-)/i);
        for (int j=;j<n;j+=i)
        {
            int w=;
            for (int k=j;k<j+(i>>);k++,w=1ll*w*wn%P)
            {
                int x=a[k],y=1ll*w*a[k+(i>>)]%P;
                a[k]=(x+y)%P,a[k+(i>>)]=(x-y+P)%P;
            }
        }
    }
}
void mul(int *a,int *b,int n)
{
    DFT(a,n,),DFT(b,n,);
    for (int i=;i<n;i++) a[i]=1ll*a[i]*b[i]%P;
    DFT(a,n,inv[]);
    for (int i=;i<n;i++) a[i]=1ll*a[i]*inv[n]%P;
}
void solve(int l,int r,int *f,int n)
{
    if (l==r) {f[]=,f[a[l]]=P-;return;}
    int a[n]={},mid=l;
    for (int i=l;i<=r;i++) if (s[i]-s[l-]>s[r]-s[i]) {mid=i;break;}
    if (mid==r) mid--;
    int t1=;while (t1<=(s[mid]-s[l-]<<)) t1<<=;
    solve(l,mid,f,t1);
    t1=;while (t1<=(s[r]-s[mid]<<)) t1<<=;
    solve(mid+,r,a,t1);
    mul(f,a,n);
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("loj2541.in","r",stdin);
    freopen("loj2541.out","w",stdout);
    const char LL[]="%I64d\n";
#else
    const char LL[]="%lld\n";
#endif
    n=read();
    for (int i=;i<=n;i++) s[i]=s[i-]+(a[i]=read());
    t=;while (t<=(s[n]<<)) t<<=;
    inv[]=;for (int i=;i<N*;i++) inv[i]=P-1ll*(P/i)*inv[P%i]%P;
    solve(,n,f,t);
    for (int i=;i<=s[n];i++)
    ans=(ans+1ll*a[]*inv[i+a[]]%P*f[i])%P;
    cout<<ans;
    return ;
}