2021牛客暑期多校训练营4 B.Sample Game(期望平方递推)

Posted 2021-09-04 issue是fw

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定义$f[i]$表示当前最大的数字为$i$,游戏结束还需要的游戏次数

$f[i]=1+\sum _{j=i+1}^{n}f[j]\ast p_j+p_i\ast f[i]$

$f[i]=\frac{1+\sum _{j=i+1}^n{p}_j\ast f[j]}{1-p_i}$

令$g[i]$为当前最大数字为$i$,还能获得的期望分数

因为$E[(x+1{)}^2]=E[x^2]+2\ast E[x]+1$

$g[x]=\sum _{i=1}^{x-1}{p}_i+\sum _{i=x}^n(g[i]+2\ast f[i]+1)\ast p_i$

$g[x]=\frac{1+\sum _{i=x+1}^n{p}_i\ast (g[i]+2\ast f[i])+2\ast p_i\ast f_i}{1-p_i}$

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int mod = 998244353;
const int maxn = 3e5+10;
int n,f[maxn],g[maxn],sum[maxn],p[maxn];
int quick(int x,int n)
{
	x = ( x%mod+mod ) % mod;
	int ans = 1;
	for( ; n ; n>>=1,x=x*x%mod )
		if( n&1 )	ans = ans*x%mod;
	return ans;
}
void upd(int &x,int y){ x = ( x+y )%mod; }
signed main()
{
	cin >> n;
	for(int i=1;i<=n;i++)	cin >> p[i], sum[i] = sum[i-1]+p[i];
	int inv = quick( sum[n],mod-2 );
	for(int i=1;i<=n;i++)	p[i] = p[i]*inv%mod, sum[i] = ( sum[i-1]+p[i] )%mod;
	int sf = 0, sg = 0;
	for(int i=n;i>=0;i--)
	{
		upd( f[i],1+sf );
		f[i] = f[i]*quick( 1-p[i],mod-2 )%mod;
		upd( g[i],1+sg+2*p[i]*f[i]%mod );
		g[i] = g[i]*quick( 1-p[i],mod-2 )%mod;
		upd( sf, f[i]*p[i]%mod ); upd( sg,p[i]*( g[i]+2*f[i] )%mod );
	}
	cout << ( g[0]%mod+mod )%mod;
}