树阵:

每个号码的前面维修比其数数少,和大量的这后一种数比他的数字

再枚举每一个位置组合一下

Sequence II

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Long long ago, there is a sequence A with length n. All numbers in this sequence is no smaller than 1 and no bigger than n, and all numbers are different in this sequence.

Please calculate how many quad (a,b,c,d) satisfy:

1. 1≤a<b<c<d≤n

2. Aa<Ab

3. Ac<Ad

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case begins with a line contains an integer n.

The next line follows n integers A1,A2,…,An.

[Technical Specification]

1 <= T <= 100

1 <= n <= 50000

1 <= Ai <=
n

Output

For each case output one line contains a integer,the number of quad.

Sample Input

1
5
1 3 2 4 5

Sample Output

4

Source

BestCoder Round #23

/* ***********************************************
Author        :CKboss
Created Time  :2014年12月20日 星期六 21时38分00秒
File Name     :HDOJ5147.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;

const int maxn=55000;

int a[maxn];
int n;

LL sum1[maxn],sum2[maxn];
int t1[maxn],t2[maxn];

int lowbit(int x) { return x&(-x); }

/// 1 找比当前数小的 2 找比当前数大的

void init()
{
	memset(sum1,0,sizeof(sum1));
	memset(sum2,0,sizeof(sum2));
	memset(t1,0,sizeof(t1));
	memset(t2,0,sizeof(t2));
}

void add(int kind,int p)
{
	if(kind==1) for(int i=p;i<maxn;i+=lowbit(i)) t1[i]+=1;
	else if(kind==2) for(int i=p;i;i-=lowbit(i)) t2[i]+=1;
}

int sum(int kind,int p)
{
	int ret=0;
	if(kind==1) for(int i=p;i;i-=lowbit(i)) ret+=t1[i];
	else if(kind==2) for(int i=p;i<maxn;i+=lowbit(i)) ret+=t2[i];
	return ret;
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);

	int T_T;
	scanf("%d",&T_T);
	while(T_T--)
	{
		scanf("%d",&n);
		for(int i=1;i<=n;i++) scanf("%d",a+i);

		init();

		/// from left to right
		for(int i=1;i<=n;i++)
		{
			int ss=sum(1,a[i]);
			sum1[i]=ss;
			add(1,a[i]);
		}
		/// from right to left
		for(int i=n;i>=1;i--)
		{
			int ss=sum(2,a[i]);
			sum2[i]=sum2[i+1]+ss;
			add(2,a[i]);
		}

		LL ans=0;
		for(int i=2;i<=n-1;i++)
		{
			/// X...i i+1...X
			ans+=sum1[i]*sum2[i+1];
		}

		cout<<ans<<endl;
	}

    return 0;
}

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