Sequence II

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 375    Accepted Submission(s): 173

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

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

1
5
1 3 2 4 5

4

/*
HDU 5147 树状数组
*/

#include<iostream>
#include<stdio.h>
using namespace std;
#define N 50010
int c[N],f[N],g[N],a[N];
int n;
//所有大于x的加上d 
void add(int x,int d)
{
	while(x<=n)
	{
		c[x]+=d;
		x+=x&(-x);
	}
}

//统计小于x的个数 
int sum(int x)
{
	int ans=0;
	while(x>0)
	{
		ans+=c[x];
		x-=x&(-x);
	}
	return ans;
}

int main()
{
	
	int t,i;
	__int64 ans,s;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(i=1;i<=n;i++)
			scanf("%d",&a[i]);
		
		memset(c,0,sizeof(c));
		for(i=1;i<=n;i++)
		{
			add(a[i],1);
			f[i]=sum(a[i])-1;//左边小于a[i]的个数 
		}
		memset(c,0,sizeof(c));
		for(i=n;i>=1;i--)
		{
			add(a[i],1);
			g[i]=n-i+1-sum(a[i]);//右边大于a[i]的个数 
		}
		
		ans=0;
		s=0;
		for(i=1;i<=n;i++)
		{
			ans+=s*g[i];//与左边的相乘 
			s+=f[i];//统计右边的和 
		} 
		printf("%I64d\n",ans); 
	}
	return 0;
}