树状数组Codeforces Round #261 (Div. 2)D

最新推荐文章于 2025-02-10 17:11:28 发布

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最新推荐文章于 2025-02-10 17:11:28 发布

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分类专栏： codeforces 文章标签： codeforces 树状数组

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本文介绍了一种使用动态规划（DP）算法解决特定组合问题的方法，包括输入输出规范、问题描述、实例分析以及代码实现。重点在于理解如何通过DP算法高效地计算满足特定条件的组合数量。

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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, ..., an. Let's denote f(l, r, x) the number of indices k such that: l ≤ k ≤ r and ak = x. His task is 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

和求逆序对差不多,需要先预处理一下

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<map>
#include<iostream>
#include<algorithm>
#include<math.h>
using namespace std;
typedef long long LL;
#define maxn (int)1e6+5
int A[maxn],c[maxn],b[maxn],s[maxn];
int n;
void add(int x,int d)
{
    while(x<=n)
    {
        s[x]+=d;
        x+=(x&-x);
    }
}
LL Query(int x)
{
    LL sum=0;
    while(x>0)
    {
        sum+=s[x];
        x-=(x&-x);
    }
    return sum;
}
typedef long long LL;
int main()
{
    while(~scanf("%d",&n))
    {
        map<int, int> q;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",A+i);
            q[A[i]]++;
            b[i]=q[A[i]];
        }
        q.clear();
        memset(s,0,sizeof(s));
        for(int i=n;i>=2;i--)
        {
            q[A[i]]++;
            c[i]=q[A[i]];
            add(c[i],1);
        }
        LL ans=0;
        for(int i=1;i<n;i++)
        {
            ans+=Query(b[i]-1);
            add(c[i+1],-1);
        }
        cout<<ans<<endl;
    }
}