Codeforces 459D Pashmak and Parmida's problem【树状数组】-优快云博客

本文链接：https://blog.youkuaiyun.com/mengxiang000000/article/details/74979269

本文解析了一道算法题，题目要求计算特定条件下的数对数量。通过使用前缀和后缀计数的方法，并结合树状数组进行优化，实现了高效的解答。

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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 denotef(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.

Examples

Input

7
1 2 1 1 2 2 1

Output

Input

3
1 1 1

Output

Input

5
1 2 3 4 5

Output

题目大意：

给你N个数，定义F（L,R,X）=区间【L,R】内，a【i】==x的个数。

让你求一共有多少对（i，j）使得F（1，i，a【i】）>F（j，n，a【j】）；

思路：

用map统计前缀a【i】个数，设定为pre【i】。表示【1，i】中，a【i】的个数；

以及后缀a【i】个数，设定为back【i】。表示【i，n】中，a【i】的个数。

然后O（n）动态更新树然后查询i后边有多少个back【i】小于pre【i】，这里树状数组优化一下即可。

Ac代码：

#include<stdio.h>
#include<string.h>
#include<map>
using namespace std;
#define ll __int64
int back[1500000];
int pre[1500000];
int a[1500000];
int tree[1500005];//树
int n;
int lowbit(int x)//lowbit
{
    return x&(-x);
}
int sum(int x)//求和求的是比当前数小的数字之和，至于这里如何实现，很简单：int sum=sum(a[i])；
{
    int sum=0;
    while(x>0)
    {
        sum+=tree[x];
        x-=lowbit(x);
    }
    return sum;
}
void add(int x,int c)//加数据。
{
    while(x<=n)
    {
        tree[x]+=c;
        x+=lowbit(x);
    }
}
void Get_array()
{
    map<int ,int >s;
    for(int i=1;i<=n;i++)
    {
        s[a[i]]++;
        pre[i]=s[a[i]];
    }
    s.clear();
    for(int i=n;i>=1;i--)
    {
        s[a[i]]++;
        back[i]=s[a[i]];
    }
}
int main()
{
    while(~scanf("%d",&n))
    {
        memset(tree,0,sizeof(tree));
        ll output=0;
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        Get_array();
        for(int i=1;i<=n;i++)
        {
            add(back[i],1);
        }
        for(int i=1;i<=n;i++)
        {
            add(back[i],-1);
            output+=(ll)sum(pre[i]-1);
        }
        printf("%I64d\n",output);
    }
}