cf459D Pashmak and Parmida's problem

最新推荐文章于 2023-08-26 20:26:38 发布

转载最新推荐文章于 2023-08-26 20:26:38 发布 · 131 阅读

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原文链接：http://www.cnblogs.com/zhber/p/4035917.html

本文提供了一种解决特定序列问题的算法实现方法，通过使用树状数组进行数据统计，并结合二分查找优化效率，成功解决了给定序列中特定元素出现次数的比较问题。实例演示了如何快速计算满足条件的元素对数量。

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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 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

题意是给定a数组，令s[i]表示从1到i中a[i]出现的次数，然后求二元组(i,j)的个数，使得1<=i<j<=n && s[i]>s[n-j+1]

求s[]用hash搞之最快了，但是cf 嘛……不敢用hash……一用肯定被人出数据hack

然后退而求其次二分搞之

统计用树状数组随便搞一下

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<set>
#include<cstdlib>
#include<map>
#include<ctime>
#define inf 1000000000
#define ll long long
#define pa pair<ll,int>
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
ll ans;
int n;
int a[1000005],disc[1000005],t[1000005],sum[1000005],now[1000005];
int find(int x)
{
	int l=1,r=n;
	while(l<=r)
	{
		int mid=(l+r)>>1;
		if(disc[mid]<x)l=mid+1;
		else if(disc[mid]==x)return mid;
		else r=mid-1;
	}
}
int lowbit(int x)
{
	return x&(-x);
}
void add(int x,int v)
{
	for(int i=x;i<=n;i+=lowbit(i))
		t[i]+=v;
}
int query(int x)
{
	int sum=0;
	for(int i=x;i;i-=lowbit(i))
		sum+=t[i];
	return sum;
}
int main()
{
	n=read();
	for(int i=1;i<=n;i++)
		a[i]=disc[i]=read();
	sort(disc+1,disc+n+1);
	for(int i=1;i<=n;i++)
		a[i]=find(a[i]),sum[a[i]]++;
	for(int i=1;i<n;i++)
	{
		now[a[i]]++;
		add(now[a[i]],1);
		sum[a[i]]--;
		int t=query(n)-query(sum[a[i+1]]);
		ans+=t;
	}
	printf("%lld",ans);
	return 0;
}

转载于:https://www.cnblogs.com/zhber/p/4035917.html