【HDOJ6595】Everything Is Generated In Equal Probability（期望DP）

题意：给定一个N，随机从[1,N]里产生一个n，

然后随机产生一个n个数的全排列，求出n的逆序数对的数量并累加ans，

然后随机地取出这个全排列中的一个子序列，重复这个过程，直到为空，求ans在模998244353下的期望

思路：期望仅与长度有关，随手推一下式子

听说有通项公式

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef vector<int> VI;
#define N  110000
#define M  1100000
#define fi first
#define se second
#define MP make_pair
#define pi acos(-1)
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define lowbit(x) x&(-x)
#define Rand (rand()*(1<<16)+rand())
#define id(x) ((x)<=B?(x):m-n/(x)+1)
#define ls p<<1
#define rs p<<1|1

const ll MOD=998244353,inv2=(MOD+1)/2;
      double eps=1e-6;
      ll INF=1e14;

ll fac[N],inv[N],dp[N],mi[N];

ll pw(ll x,ll y)
{
    ll t=1;
    while(y)
    {
        if(y&1) t=t*x%MOD;
        x=x*x%MOD;
        y>>=1;
    }
    return t;
}

int read()
{
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}

ll calc(ll n)
{
    ll t=n*(n-1)/2;
    return t*inv2%MOD;
}

ll c(ll n,ll m)
{
    if(n<m||m<0) return 0;
    ll t=fac[n]*inv[m]%MOD*inv[n-m]%MOD;
    return t;
}

int main()
{
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    int n;
    fac[0]=1;
    rep(i,1,3000) fac[i]=fac[i-1]*i%MOD;
    inv[0]=inv[1]=1;
    rep(i,2,3000) inv[i]=inv[MOD%i]*(MOD-MOD/i)%MOD;
    rep(i,1,3000) inv[i]=inv[i-1]*inv[i]%MOD;
    mi[0]=1;
    rep(i,1,3000) mi[i]=mi[i-1]*inv2%MOD;
    rep(i,1,3000)
    {
        ll t=0;
        rep(j,1,i-1) t=(t+dp[j]*c(i,j)%MOD*mi[i]%MOD)%MOD;
        dp[i]=(calc(i)+t)%MOD*pw(1ll-mi[i]+MOD,MOD-2)%MOD;
    }
    rep(i,1,3000) dp[i]=(dp[i]+dp[i-1])%MOD;
    inv[0]=inv[1]=1;
    rep(i,2,3000) inv[i]=inv[MOD%i]*(MOD-MOD/i)%MOD;
    while(scanf("%d",&n)!=EOF)
    {
        printf("%I64d
",dp[n]*inv[n]%MOD);
    }
    return 0;
}