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/(Σwi+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(-xwi+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<'0'||c>'9')) c=getchar();return c;} int gcd(int n,int m){return m==0?n:gcd(m,n%m);} int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } int n,a[N],s[N],r[N*3],inv[N*3],f[N*3],t,ans; int ksm(int a,int k) { int s=1; for (;k;k>>=1,a=1ll*a*a%P) if (k&1) s=1ll*s*a%P; return s; } void DFT(int *a,int n,int g) { for (int i=0;i<n;i++) r[i]=(r[i>>1]>>1)|(i&1)*(n>>1); for (int i=0;i<n;i++) if (i<r[i]) swap(a[i],a[r[i]]); for (int i=2;i<=n;i<<=1) { int wn=ksm(g,(P-1)/i); for (int j=0;j<n;j+=i) { int w=1; for (int k=j;k<j+(i>>1);k++,w=1ll*w*wn%P) { int x=a[k],y=1ll*w*a[k+(i>>1)]%P; a[k]=(x+y)%P,a[k+(i>>1)]=(x-y+P)%P; } } } } void mul(int *a,int *b,int n) { DFT(a,n,3),DFT(b,n,3); for (int i=0;i<n;i++) a[i]=1ll*a[i]*b[i]%P; DFT(a,n,inv[3]); for (int i=0;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[0]=1,f[a[l]]=P-1;return;} int a[n]={0},mid=l; for (int i=l;i<=r;i++) if (s[i]-s[l-1]>s[r]-s[i]) {mid=i;break;} if (mid==r) mid--; int t1=1;while (t1<=(s[mid]-s[l-1]<<1)) t1<<=1; solve(l,mid,f,t1); t1=1;while (t1<=(s[r]-s[mid]<<1)) t1<<=1; solve(mid+1,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=1;i<=n;i++) s[i]=s[i-1]+(a[i]=read()); t=1;while (t<=(s[n]<<1)) t<<=1; inv[1]=1;for (int i=2;i<N*3;i++) inv[i]=P-1ll*(P/i)*inv[P%i]%P; solve(2,n,f,t); for (int i=0;i<=s[n];i++) ans=(ans+1ll*a[1]*inv[i+a[1]]%P*f[i])%P; cout<<ans; return 0; }