hdu 5147 Sequence II（树状数组，前缀和，后缀和）

Sequence II

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1004    Accepted Submission(s): 410

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

题意：在给定序列里面求子序列的个数，子序列满足A1 思路：很容易想到用数组数组维护前缀和，可是如何找到当前点后面有多少个点对满足（Ai，Aj）（Ai 我们可以逆序插入序列里的数据，这样可以求得i点后面有几个点比a[i]小，再用i点后面的点的个数减去可以得到i点后面有几个点比a[i]大，那么以i点为点对的第一个点，是不是就有sum[i]个点对呢？   那么我们只需要维护后缀和，然后每次遍历到i，用qian[i](i点前面比a[i]小的数的个数)乘以hou[i+1]（i点后面的满足条件的点对）即可

代码：

#include 
#include 
#include 
#include 
#include 
using namespace std;
#define N 50050
int a[N],n;
int c[N];
long long qian[N],hou[N];
int lowbit(int x)
{
    return x&-x;
}
void add(int x,int v)
{
    for(int i=x;i<=n;i+=lowbit(i))
        c[i]+=v;
}
int sum(int x)
{
    int ans=0;
    for(int i=x;i;i-=lowbit(i))
        ans+=c[i];
    return ans;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++)
            {
                scanf("%d",&a[i]);
                qian[i]=sum(a[i]);
                add(a[i],1);
            }
        memset(c,0,sizeof(c));
        for(int i=n;i>=1;i--)
        {
            hou[i]=n-i-sum(a[i]);
            add(a[i],1);
        }
        for(int i=n-1;i>=1;i--)
            hou[i]=hou[i]+hou[i+1];
        long long ans=0;
        for(int i=1;i