POJ3928Ping pong[树状数组 仿逆序对]

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int N=2e4+5,M=1e5+5;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();}
    return x*f;
}
int T,n,m,a[N],bit[M],f[N],g[N];
inline int lowbit(int x){return x&-x;}
inline void add(int p,int v,int bit[]){
    for(int i=p;i<=m;i+=lowbit(i)) bit[i]+=v; 
}
inline int sum(int p,int bit[]){
    int ans=0;
    for(int i=p;i>0;i-=lowbit(i)) ans+=bit[i];
    return ans;
}
int main(){
    T=read();
    while(T--){
        n=read();
        for(int i=1;i<=n;i++) a[i]=read(),m=max(m,a[i]);
        memset(bit,0,sizeof(bit));
        for(int i=1;i<=n;i++){
            add(a[i],1,bit);
            f[i]=sum(a[i]-1,bit);
        }
        memset(bit,0,sizeof(bit));
        for(int i=n;i>=1;i--){
            add(a[i],1,bit);
            g[i]=sum(a[i]-1,bit);
        }
        ll ans=0;
        for(int i=2;i<=n-1;i++) ans+=f[i]*(n-i-g[i])+g[i]*(i-1-f[i]);//,printf("t%d %d %d\n",i,f[i],g[i]);
        printf("%lld\n",ans);
    }
}