CF #261 div2 D. Pashmak and Parmida's problem （树状数组版）

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, ..., a_n.
Let's denote f(l, r, x) the number of indices
k such that: l ≤ k ≤ r and
a_k = x. His task is to calculate the number of pairs of indicies
i, j (1 ≤ i < j ≤ n) such that
f(1, i, a_i) > f(j, n, a_j).

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, ..., a_n
(1 ≤ a_i ≤ 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

树状数组写起来更方便。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <set>
#include <stack>
#include <cctype>
#include <algorithm>
#define lson o<<1, l, m
#define rson o<<1|1, m+1, r
using namespace std;
typedef long long LL;
const int mod = 99999997;
const int MAX = 0x3f3f3f3f;
const int maxn = 1000010;
int n, a, b;
int in[maxn], f[maxn], vis[maxn], l[maxn], r[maxn], tt[maxn];
int c[maxn];
int bs(int v, int x, int y) {
    while(x < y) {
        int m = (x+y) >> 1;
        if(in[m] >= v) y = m;
        else x = m+1;
    }
    return x;
}
int main()
{
    cin >> n;
    for(int i = 0; i < n; i++) {
        scanf("%d", &in[i]);
        tt[i] = in[i];
    }
    sort(in, in+n);
    int m = unique(in, in+n) - in;
    for(int i = 0; i < n; i++) {
        f[i] = bs(tt[i], 0, m-1) + 1;
        vis[ f[i] ]++;
        l[i+1] = vis[ f[i] ];
    }
    memset(vis, 0, sizeof(vis));
    for(int i = n-1; i >= 0; i--) {
        f[i] = bs(tt[i], 0, m-1) + 1;
        vis[ f[i] ]++;
        r[i+1] = vis[ f[i] ];
    }
    LL sum = 0;
    for(int i = 1; i <= n; i++) {
        for(int j = r[i]+1; j <= maxn; j += j&-j) sum += c[j];
        for(int j = l[i]; j > 0; j -= j&-j) c[j]++;
    }
    cout << sum << endl;
    return 0;
}