逆序对分三类:
1.已知对已知
树状数组直接处理即可
2.未知对未知
设未知数的位置数为\(m\),则有\(m(m-1)/2\)个数对。一个数对是逆序对的期望是\(0.5\)(一个逆序对与一个非逆序对对应)。因为期望的可加性,总期望为\(m(m-1)/4\)
3.已知对未知
处理出对于每个数\(i\),比它大且可填入原序列的数的个数\(a_i\)和比它小且可填入原序列的数的个数\(b_i\)
如果未知数在已知数\(i\)的左边,期望为\(a_i/m\),否则为\(b_i/m\),全加起来就行了
代码:
#include <bits/stdc++.h>
#define mod 998244353ll
#define ll long long
#define rep(i,x,y) for(i=x;i<=y;++i)
#define des(i,x,y) for(i=x;i>=y;--i)
#define rd(x) scanf("%d",&x)
#define N 200005
using namespace std;
int a[N],bg[N],sm[N],n;
ll c[N],t[N];
bool vis[N];
inline ll ksm(ll x,ll y){
ll z=1;
while(y){
if(y&1) (z*=x)%=mod;
(x*=x)%=mod,y>>=1;
}
return z;
}
inline int lowbit(int x){ return x&(-x);}
inline void add(ll *a,int x,int y){
for(int i=x;i<=n;i+=lowbit(i)) (a[i]+=y)%=mod;
}
inline ll query(ll *a,int x){
ll tmp=0;
for(int i=x;i>0;i-=lowbit(i))
(tmp+=a[i])%=mod;
return tmp;
}
int main(){
int i,tot=0;
ll ans=0,inv;
rd(n);
rep(i,1,n){
rd(a[i]);
if(a[i]==-1) tot++;
else vis[a[i]]=1;
}
inv=ksm(1ll*tot,mod-2);
(ans+=1ll*tot*(tot-1)%mod*ksm(4ll,mod-2)%mod)%=mod;
bg[n]=0,sm[1]=0;
des(i,n-1,1) bg[i]=bg[i+1]+(!vis[i+1]);
rep(i,2,n) sm[i]=sm[i-1]+(!vis[i-1]);
rep(i,1,n){
if(~a[i]) add(c,a[i],sm[a[i]]*inv%mod);
else (ans+=query(c,n))%=mod;
}
memset(c,0,sizeof(c));
des(i,n,1){
if(~a[i]){
(ans+=query(t,a[i]))%=mod;
add(t,a[i],1),add(c,a[i],bg[a[i]]*inv%mod);
} else (ans+=query(c,n))%=mod;
}
printf("%I64d",ans);
}