Codeforces Round #261 (Div. 2) D. Pashmak and Parmida ' s problem (tree-like array)

D. Pashmak and Parmida ' s problem
Time limit per test 3 seconds memory limit per test megabytes input standard input output standard output

Parmida is a clever girl and she wants to participate in Olympiads the year. Of course she wants her partner to be clever too (although he's not)! Parmida have 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 and AK = x. His task was 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

Parse: F (L, r, x) indicates the number of l≤k≤r and AK = x, to be calculated for F (1, I, AI) > F (J, N, AJ), and I, J (1≤i < j≤n) logarithm.

The first preprocessing is F (1, I, AI) and F (j, N, AJ), and then it is particularly like an inverse number. Just use a tree-shaped array to get a little bit done.

AC Code:

#include <bits/stdc++.h> using namespace std;

const int N = 1000005;

int a[n], f[n], g[n], c[n];

Map<int, int> m;

int lowbit (int x) {return x & x;}

void Add (int x, int d) {for (; x < N; x + = Lowbit (x)) c[x] + = D;}
    int sum (int x) {int ret = 0;
    for (; x > 0; X-= Lowbit (x)) ret + = c[x];
return ret;
    } int main () {#ifdef sxk freopen ("In.txt", "R", stdin);
    #endif//SXK int n;
        while (~SCANF ("%d", &n)) {m.clear ();
            for (int i=1; i<=n; i++) {scanf ("%d", &a[i]);
            m[A[i]] + +;        F[i] = m[A[i]];
        F (1, I, AI)} m.clear ();
            for (int i=n; i>0; i--) {m[A[i]] + +;        G[i] = m[A[i]];
        F (j, N, AJ)} memset (c, 0, sizeof (c));
        A long long ans = 0;
            for (int i=n-1; i>0; i--) {Add (g[i+1], 1);
        Ans + = SUM (F[i]-1);

} printf ("%lld\n", ans);    } return 0;
 }