Codeforces Round 459 D. Pashmak and Parmida's problem 树状数组求逆序数变形

最新推荐文章于 2017-05-19 11:12:00 发布

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最新推荐文章于 2017-05-19 11:12:00 发布

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分类专栏：回顾问题 Codeforces

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本篇介绍了一个算法挑战题目，需计算特定整数序列中满足条件的索引对数量。通过树状数组优化求解过程，实现高效算法。

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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 andak = 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

分析：贴一个详细的解释点击打开链接

这个代码像谜一样啊，待我再研究几天树状数组再回来看看吧

#include <cmath>
#include <iostream>
#include<cstdio>
#include<map>
using namespace std;
const int MAX=1000017;
int n;
int a[MAX],c[MAX],f[MAX];
long long res;
map<int,int>freq;
int lowbit(int x)
{
    return x&(-x);
}
void add(int i,int x)
{
    while(i<=n){
        c[i]+=x;
        i+=lowbit(i);
    }
}
int sum(int x)
{
    int sum=0;
    while(x>0){
        sum+=c[x];
        x-=lowbit(x);
    }
    return sum;
}
int main()
{
        scanf("%d",&n);
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        for(int i=n-1;i>=0;i--)
            f[i]=++freq[a[i]];
        freq.clear();
        for(int i=0;i<n;i++){
            res+=i-sum(f[i]);
            add(++freq[a[i]],1);
        }
        printf("%I64d\n",res);
        return 0;
}