Codeforces Round #261 (Div. 2) D. Pashmak and Parmida's problem_parmida is a clever girl and she wants to particip-优快云博客

本文介绍了一种使用动态规划（DP）算法解决特定序列问题的方法，包括输入输出规范、问题描述、实例分析以及代码实现。重点在于通过树状数组（线段树的一种实现方式）优化查询效率，适用于处理涉及区间查询和更新的问题。

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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 indicesk such that: l ≤ k ≤ r anda_k = x. His task is to calculate the number of pairs of indiciesi, j (1 ≤ i < j ≤ n) such thatf(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 containsn 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

题意：给出数组A，定义f(l,r,x)为A[]的下标l到r之间，等于x的元素数。i和j符合f(1,i,a[i])>f(j,n,a[j])，求i和j的种类数。
题解：使用树状数组统计小于某数的元素数量。
STL（map）+树状数组
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <limits.h>
#include <math.h>
#include <iostream>
#include <map>
#include <algorithm>
using namespace std;

int a[1000005];

int data[1000005];

int mark[1000005];

int lowbit(int x)
{
	return x&(-x);
}
void add(int p,int num)
{
	if(p<=0)return;
	while(p<=1000000)
	{
		a[p]+=num;
		p+=lowbit(p);
	}
}

int sum(int p)
{
	int ans=0;
	while(p>0)
	{
		ans+=a[p];
		p-=lowbit(p);
	}
	return ans;
}

int main()
{
	map<int,int> dicta,dictb;
	int n;
	scanf("%d",&n);
	for(int i=0;i<n;i++)
	{
		scanf("%d",&data[i]);
	}
	for(int i=n-1;i>=0;i--)
	{
		dictb[data[i]]++;
		mark[i]=dictb[data[i]];
		add(dictb[data[i]],1);
	}
	long long ans=0;
	for(int i=0;i<n;i++)
	{
		dicta[data[i]]++;
		add(mark[i],-1);
		ans+=sum(dicta[data[i]]-1);
	}
	printf("%I64d\n",ans);
	return 0;
}