Description

N(3N20000) ping pong players live along a west-east street(consider the street as a line segment). Each player has a unique skill rank. To improve their skill rank, they often compete with each other. If two players want to compete, they must choose a referee among other ping pong players and hold the game in the referee’s house. For some reason, the contestants can’t choose a referee whose skill rank is higher or lower than both of theirs. The contestants have to walk to the referee’s house, and because they are lazy, they want to make their total walking distance no more than the distance between their houses. Of course all players live in different houses and the position of their houses are all different. If the referee or any of the two contestants is different, we call two games different. Now is the problem: how many different games can be held in this ping pong street?

Input

The first line of the input contains an integer T(1T20) , indicating the number of test cases, followed by T lines each of which describes a test case.

Every test case consists of N + 1 integers. The first integer is N , the number of players. Then N distinct integers a1, a2…aN follow, indicating the skill rank of each player, in the order of west to east ( 1ai100000 , i= 1…N ).

Output

For each test case, output a single line contains an integer, the total number of different games.

Sample Input

1
3 1 2 3

思路:考慮第i個人當裁判的情形，假設a1到a[i-1]中有ci個比ai小，那麼就有(i-1)-ci個比ai大，同理，假設a[i+1]到an中有di個比ai小，那麼就有(n-i)-di個比ai大，然後根據乘法原理和加法原理，i當裁判有ci(n-i-di)+(i-ci-1)*di，這樣問題就轉化為求c,d

程式碼:

#include<cstdio>
#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
int a[20005];
//b[i]表示i左邊比i小的個數，d[i]表示i右邊比i小的個數
int b[20005];
int d[20005];
int C[100005];
int T,n;
int max_a;
int Sum(int x)
{
      int ret=0;
      while(x>0)
      {
            ret+=C[x];
            x-=x&(-x);
      }
      return ret;
}
int Add(int x,int m)
{
      while(x<=100000)
      {
            C[x]+=m;
            x+=(x&(-x));
      }
}
int main()
{
    scanf("%d",&T);
    while(T--)
    {
        max_a=0;
        memset(C,0,sizeof(C));
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
           scanf("%d",&a[i]);
        for(int i=1;i<=n;i++)
        {
           Add(a[i],1);
           b[i]=Sum(a[i]-1);
        }
        memset(C,0,sizeof(C));
        for(int i=n;i>=1;i--)
        {
           Add(a[i],1);
           d[i]=Sum(a[i]-1);
        }
        long long  int ans=0;
        for(int i=2;i<n;i++)
        {
           ans+=b[i]*(n-i-d[i])+(i-1-b[i])*d[i];
        }
        printf("%lld\n",ans);
    }
}