POJ 3928 tree-like array

The topic only n individuals, each person has an ID and a skill value, a game requires two players and a referee, only when the referee's ID and skill value between the two players can play a game, now ask how many games can be organized.

Because after the order, the first insert must be small, so the left and right sides of greater than less can be determined, with a tree-like array to maintain the player's ID

Sample Input
1
3 1 2 3
Sample Output
1

1#include <cstdio>2#include <iostream>3#include <algorithm>4#include <cstring>5#include <cmath>6#include <queue>7 using namespacestd;8 intn,m,t;9 Const intmaxn=100000;Ten __int64 C[MAXN]; One structnode A { -     intX,id; -}p[maxn+2]; the BOOLCMP (node A,node b) - { -     returna.x<=b.x; - } + intLowbit (inti) - { +     returni& (-i); A } at voidAddintI,__int64 D) - { -      while(i<=MAXN) -     { -c[i]+=D; -i+=lowbit (i); in     } - } to__int64 sum (inti) + { -__int64 ret=0; the      while(i) *     { $ret+=C[i];Panax Notoginsengi-=lowbit (i); -     } the     returnret; + } A intMain () the { +     inti,j,k; -__int64 lmin,lmax,rmin,rmax,ans=0; $     //freopen ("1.in", "R", stdin); $scanf"%d",&t); -      while(t--) -     { the          for(i=0; i<=maxn;i++) c[i]=0; -scanf"%d",&n);Wuyi          for(i=1; i<=n;i++) scanf ("%d", &p[i].x), p[i].id=i; theSort (p+1, p+1+n,cmp); -Add (p[1].id,1); Wu          for(i=2; i<n;i++) -         { AboutLmin=sum (p[i].id-1);//The count is from 1 to i-1, so subtract one more 1 $lmax=p[i].id-1-lmin; -             //printf ("%i64d%i64d", Lmin,lmax); -Rmin=sum (N)-sum (p[i].id); -rmax=n-p[i].id-rmin; A             //printf ("%i64d%i64d \ n", Rmin,rmax); +ans+= (Lmin*rmax+rmin*lmax); Add (P[i].id,1); the             //printf ("%i64d\n", ans); -         } $printf"%i64d\n", ans); theans=0; the     } the     return 0; the}

POJ 3928 tree-like array