题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5792

Description

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 integers A1,A2⋯An.
1≤n≤50000
0≤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

Hint

题意

给你n个数，问你有多少个四元组，满足a!=b,b!=c,c!=d,a!=d,a!=c,b!=d，且A[a]<A[b],a<b,A[c]>A[d],c<d的关系的。

题解：

考虑容斥，我们知道(a,b)的对数，（c,d）的对数，两个乘起来，再减去不合法的就好了。不合法的，显然就是长度为3的哪些重复计算的，只要减去就好了。维护l[i]表示前面有多少个比他小，l1[i]前面有多少个比他大r[i]后面有多少个比他小，r1[i]有多少个比他大然后莽一波容斥。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 5e4+7;
int d[maxn],n,a[maxn],l[maxn],r[maxn],l1[maxn],r1[maxn];;
int lowbit(int x){return x&(-x);
}
void update(int x,int v){for(int i=x;i<maxn;i+=lowbit(i))d[i]+=v;
}
int get(int x){int ans = 0;for(int i=x;i;i-=lowbit(i))ans+=d[i];return ans;
}
map<int,int>H;
vector<int> V;
int main(){while(scanf("%d",&n)!=EOF){memset(l1,0,sizeof(l1));memset(r1,0,sizeof(r1));memset(l,0,sizeof(l));memset(r,0,sizeof(r));memset(d,0,sizeof(d));V.clear();H.clear();for(int i=1;i<=n;i++){scanf("%d",&a[i]);V.push_back(a[i]);}sort(V.begin(),V.end());V.erase(unique(V.begin(),V.end()),V.end());for(int i=0;i<V.size();i++)H[V[i]]=i+1;for(int i=1;i<=n;i++)a[i]=H[a[i]];long long ans1=0,ans2=0,ans3=0;for(int i=1;i<=n;i++){l[i]=get(a[i]-1);l1[i]=get(maxn-1)-get(a[i]);update(a[i],1);}memset(d,0,sizeof(d));for(int i=n;i>=1;i--){r[i]=get(a[i]-1);r1[i]=get(maxn-1)-get(a[i]);update(a[i],1);}for(int i=1;i<=n;i++){ans3+=1ll*l[i]*r[i];ans3+=1ll*l1[i]*r1[i];ans3+=1ll*l[i]*l1[i];ans3+=1ll*r[i]*r1[i];ans1+=l[i],ans2+=r[i];}cout<<ans1*ans2-ans3<<endl;}
}