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

最新推荐文章于 2023-08-26 20:26:38 发布

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最新推荐文章于 2023-08-26 20:26:38 发布

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分类专栏： ACM__数据结构 ACM__Hash ACM__树状数组

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本文讨论了一个关于序列中元素出现频率比较的问题，并通过引入树状数组优化了求解过程。详细解释了如何使用树状数组进行高效计算。

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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

Input

3
1 1 1

Output

Input

5
1 2 3 4 5

Output

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;
}