【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=,inv2=(MOD+)/;
       double eps=1e-;
       ll INF=1e14;

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

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

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

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

 ll c(ll n,ll m)
 {
     if(n<m||m<) return ;
     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[]=;
     rep(i,,) fac[i]=fac[i-]*i%MOD;
     inv[]=inv[]=;
     rep(i,,) inv[i]=inv[MOD%i]*(MOD-MOD/i)%MOD;
     rep(i,,) inv[i]=inv[i-]*inv[i]%MOD;
     mi[]=;
     rep(i,,) mi[i]=mi[i-]*inv2%MOD;
     rep(i,,)
     {
         ll t=;
         rep(j,,i-) t=(t+dp[j]*c(i,j)%MOD*mi[i]%MOD)%MOD;
         dp[i]=(calc(i)+t)%MOD*pw(1ll-mi[i]+MOD,MOD-)%MOD;
     }
     rep(i,,) dp[i]=(dp[i]+dp[i-])%MOD;
     inv[]=inv[]=;
     rep(i,,) inv[i]=inv[MOD%i]*(MOD-MOD/i)%MOD;
     while(scanf("%d",&n)!=EOF)
     {
         printf("%I64d\n",dp[n]*inv[n]%MOD);
     }
     return ;
 }