hdu 5147 Sequence II 树状数组

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

思路：用树状数组求逆序对；

　　　pre[i]存前边比a[i]小的；

　　　nex[i]存a[i]开始的二元组；

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
const int N=5e4+,M=1e6+,inf=1e9+,mod=1e9+;
const ll INF=1e18+;
int tree[N];
int a[N];
int pre[N];
int nex[N];
void init()
{
    memset(tree,,sizeof(tree));
    memset(pre,,sizeof(pre));
    memset(nex,,sizeof(nex));
}
int lowbit(int x)
{
    return x&(-x);
}
void update(int x,int c)
{
    while(x<N)
    {
        tree[x]=tree[x]+c;
        x+=lowbit(x);
    }
}
int query(int x)
{
    int sum=;
    while(x)
    {
        sum+=tree[x];
        x-=lowbit(x);
    }
    return sum;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        init();
        int n;
        scanf("%d",&n);
        for(int i=;i<=n;i++)
        scanf("%d",&a[i]);
        for(int i=;i<=n;i++)
        {
            pre[i]=query(a[i]-);
            update(a[i],);
        }
        memset(tree,,sizeof(tree));
        for(int i=n;i>=;i--)
        {
            nex[i]=query(n)-query(a[i])+nex[i+];
            update(a[i],);
        }
        ll ans=;
        for(int i=;i<=n;i++)
            ans+=(ll)pre[i]*nex[i+];
        printf("%lld\n",ans);
    }
    return ;
}