World is Exploding HDU

Given a sequence A with length n,count how many quadruple (a,b,c,d) satisfies: a≠b≠c≠d,1≤a<b≤n,1≤c<d≤n,Aa<Ab,Ac>Ad.
Input
The input consists of multiple test cases.
Each test case begin with an integer n in a single line.

The next line contains n integersA1,A2⋯AnA1,A2⋯An.
1≤n≤500001≤n≤50000
0≤Ai≤1e90≤Ai≤1e9
Output
For each test case,output a line contains an integer.
Sample Input
4
2 4 1 3
4
1 2 3 4
Sample Output
1
0
最近总是做到这个类型的题，发现离散化和树状数组捆绑的很紧密，如果发现题目的要求之和数于数之间的大小有关系而非具体的值，那么可以离散化了，
这题的思路是找出数列里的所有的正序对（num1）和逆序对的（num2）个数相乘，但是里面会有重不合法的情况，不能满足4个index都不相等，会有4种情况

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>#define N 50005using namespace std;int n;int a[N];int id[N];int c[N];int Lmin[N],Rmin[N],Lmax[N],Rmax[N];int lowBit(int x){    return x&-x;}int sum(int x){    int ans=0;    while(x>0)    {        ans+=c[x];        x-=lowBit(x);    }    return ans;}void change(int x,int p){    while(x<=n)    {        c[x]+=p;        x+=lowBit(x);    }}int main(){    while(scanf("%d",&n)==1)    {        memset(Lmin,0,sizeof(Lmin));        memset(Rmin,0,sizeof(Rmin));        memset(Lmax,0,sizeof(Lmax));        memset(Rmax,0,sizeof(Rmax));        for(int i=0;i<n;i++)        {            scanf("%d",a+i);            id[i]=a[i];        }        sort(id,id+n);        int num=unique(id,id+n)-id;        for(int i=0;i<n;i++)        a[i]=lower_bound(id,id+num,a[i])-id+1;        int num1=0,num2=0;        memset(c,0,sizeof(c));        for(int i=0;i<n;i++)        {            Lmin[i]=sum(a[i]-1);            num1+=Lmin[i];            Lmax[i]=sum(n)-sum(a[i]);            num2+=Lmax[i];            change(a[i],1);        }        memset(c,0,sizeof(c));        for(int i=n-1;i>=0;i--)        {            Rmin[i]=sum(a[i]-1);            Rmax[i]=sum(n)-sum(a[i]);                   change(a[i],1);        }        long long ans=(long long)num1*num2;        for(int i=0;i<n;i++)        {            ans-=Lmin[i]*Rmin[i];            ans-=Lmax[i]*Rmax[i];            ans-=Lmax[i]*Lmin[i];            ans-=Rmax[i]*Rmin[i];        }        printf("%lld\n",ans);    }}