具体就是枚举每个裁判,左边比裁判小的个数乘以右边比裁判大的个数,以及左边比裁判大的个数乘以右边小的个数,总和即为结果.
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define ll long long
#define lbt(x) (x&(-x))
#define IOS ios::sync_with_stdio(false)
using namespace std;
ll e,n;
void adds(ll *a,ll x,ll d){
while(x <= e){
a[x]+=d;x+=lbt(x);
}
}
ll sdsd(ll *a,ll x){
ll ret=0;
while(x>0){
ret+=a[x]; x-=lbt(x);
}
return ret;
}
ll l[202010],z[202010],a[202010];
int main(){IOS;
int t;
cin>>t;
while(t--){
fill(z,z+202010,0);
cin>>n;
for(int i=1;i<=n;i++){
cin>>a[i];e=max(e,a[i]);
}
for(int i=1;i<=n;i++){
adds(z,a[i],1);
l[i]=sdsd(z,a[i]-1);
}
fill(z,z+202010,0);
ll ans=0;
for(int i=n;i>0;i--){
adds(z,a[i],1);
int d=sdsd(z,a[i]-1);
ans+=(l[i] * (n - i - d)) + d * (i - 1 - l[i]);
}
cout<<ans<<endl;
}
return 0;
}