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先上题目：

D. Pashmak and Parmida's problem

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Parmida is a clever girl and she wants to participate in Olympiads this year. Of course she wants her partner to be clever too (although he's not)! Parmida has prepared the following test problem for Pashmak.

There is a sequence a that consists of n integers a1, a2, ..., an. Let's denote f(l, r, x) the number of indices k such that: l ≤ k ≤ r andak = x. His task is to calculate the number of pairs of indicies i, j (1 ≤ i < j ≤ n) such that f(1, i, ai) > f(j, n, aj).

Help Pashmak with the test.

Input

The first line of the input contains an integer n (1 ≤ n ≤ 106). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

Print a single integer — the answer to the problem.

Sample test(s)

input

7
1 2 1 1 2 2 1

output

8

input

3
1 1 1

output

1

input

5
1 2 3 4 5

output

0

 　　题意：给你n个数，定义一个函数f(l,r,x)表示[l,r]里面有多少个数是等于x的。现在要求有多少对i,j符合f(1,i,ai)>f(j,n,aj)

　　做法：求出某个ak它的前面和后面各有多少个数等于它本身。然后将这些数用树状数组计数，这里需要先对前面的数离散化(当然，这里用map来保存也没有问题)，然后像统计逆序对那样统计就可以得到答案了。

上代码：

#include <bits/stdc++.h>
#define lowbit(x) (x&(-x))
#define MAX 1000002
#define ll long long
using namespace std;

int a[MAX],h[MAX],tot;
ll num[MAX],anum[MAX],f[MAX],r[MAX];
ll c[MAX];
int n;

void add(int i){
    for(;i<=MAX-1;i+=lowbit(i)) c[i]++;
}

ll sum(int i){
    ll ans=0;
    for(;i>0;i-=lowbit(i)) ans+=c[i];
    return ans;
}

int main()
{
    ll ans,e;
    //freopen("data.txt","r",stdin);
    while(scanf("%d",&n)!=EOF){
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
            h[i]=a[i];
            num[i+1]=0;
        }
        sort(h,h+n);
        tot=unique(h,h+n)-h;
        for(int i=0;i<n;i++){
            a[i]=lower_bound(h,h+tot,a[i])-h+1;
        }
        memset(anum,0,sizeof(anum));
        for(int i=0;i<n;i++){
            anum[a[i]]++;
            f[i]=anum[a[i]];
        }
        for(int i=0;i<n;i++){
            r[i]=anum[a[i]]-f[i]+1;
        }
        ans=0;
        memset(c,0,sizeof(c));
        for(int i=n-1;i>=0;i--){
            e=sum(f[i]-1);
            ans+=e;
            add(r[i]);
        }
        printf("%I64d
",ans);
    }
    return 0;
}