Codeforces Round 1052 A. Equal Occurrences (相同出现次数)

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A. Equal Occurrences

We call an array balanced if and only if the numbers of occurrences of any of its elements are the same.
For example, [1,1,3,3,6,6] and [2,2,2,2] are balanced,
but [1,2,3,3] is not balanced (the numbers of occurrences of elements 1 and 3 are different).
Note that an empty array is always balanced.

You are given a non-decreasing array a consisting of n integers.
Find the length of its longest balanced subsequence.

(A sequence b is a subsequence of a sequence a if b can be obtained from a
by the deletion of several (possibly, zero or all) element from arbitrary positions.)

Input
Each test contains multiple test cases. The first line contains the number of test cases t (1≤t≤500).
The description of the test cases follows.

The first line of each test case contains a single integer n(1≤n≤100) — the length of a.
The second line contains n integers a1,a2,…,an (1≤a1≤a2≤⋯≤an≤n) — the elements of a.

Output
For each test case, output a single integer — the length of the longest balanced subsequence of a.
Example
Input
4
5
1 1 4 4 4
2
1 2
15
1 1 1 1 1 2 2 2 2 3 3 3 4 4 5
5
3 3 3 3 3
Output
4
2
9
5

Note
In the first test case, the whole array a=[1,1,4,4,4] is not balanced because the number of occurrences of element 1 is 2,
while the number of occurrences of element 4 is 3, which are not equal.
The subsequence [1,1,4,4] is balanced because the numbers of occurrences of elements 1 and 4 are both 2.
Thus, the length of the longest balanced subsequence of a is 4.

In the second test case, the whole array a=[1,2] is already balanced,
so the length of the longest balanced subsequence of a is 2.

In the third test case, the longest balanced subsequence of a is [1,1,1,2,2,2,3,3,3].

In the fourth test case, the whole array a=[3,3,3,3,3] is already balanced,
so the length of the longest balanced subsequence of a is 5.

题意：

例如：1 1 1 1 1 2 2 2 2 3 3 3 4 4 5 中，求出现相同次数是最多的数。

第一步：统计每个不同的数，出现次数是多少：【5-4-3-2-1】

第二步：统计出现次数一样或者包含的总次数：【5-8-9-8】

例如：

1出现5次，其他都不包含5次，所以是5，

2出现了4次，1也出现了4+1次，那么就是 = 4+4=8次，

3出现3次，前面5，4都包含3次，那么总次数=3+3+3=9，

4出现2次，前面都包含，那么总次数就是：2+2+2+2=8,

5出现1次，前面都包含，总次数就是5，

那么最大是9，就是111 222 333，一共出现次数是3+3+3=9

理解题意后，然后开始编程

思路：

1.首先找到每个不同数字出现次数。

2.找到每个不同数字出现的次数与其他被包含（<=）的出现次数总和

3.从以上总次数里，找到最大次数，就是题意要求的结果。

耗费了3小时，主要是开始没有完全理解。

#include <iostream>
#include <vector>
using namespace std;

/*
2025-10-01
*/
int main()
{
	int t = 0;
	cin >> t;
	while (t--)
	{
		int n = 0;
		cin >> n;
		int len = n;
		int *list = new int[len];
		for (int i = 0; i < len; i++)
		{
			cin >> list[i];
		}

		//total:times of every number
		vector<pair<int, int>> clist;
		pair<int, int> p1(0, 1);
		for (int i = 0; i < len; i++)
		{
			p1.first = list[i];
			for (int j = i + 1; j < len; j++)
			{
				if (list[i] == list[j])
				{
					p1.second++;
				}
			}
			clist.push_back(p1);
			//cout << p1.first << "::" << p1.second << endl;
			i += (p1.second - 1);
			p1.second = 1;

		}
		//total:tims
		vector<int> sumlist;
		int cout1 = 0;
		int curr_cout = 0;
		for (int i = 0; i < clist.size(); i++)
		{
			curr_cout = clist[i].second;
			for (int j = 0; j < clist.size(); j++)
			{
				if (curr_cout <= clist[j].second)
				{
					cout1 += clist[i].second;
				}
			}
			sumlist.push_back(cout1);
			cout1 = 0;
		}

		//get max
		int max_count = 0;		
		for (int i = 0; i < sumlist.size(); i++)
		{
			if (max_count < sumlist[i])
			{
				max_count = sumlist[i];
				
			}
		}
		cout <<max_count << endl;
	}
	return 0;
}

/*
A. Equal Occurrences
We call an array balanced if and only if the numbers of occurrences of any of its elements are the same.
For example, [1,1,3,3,6,6] and [2,2,2,2] are balanced,
but [1,2,3,3] is not balanced (the numbers of occurrences of elements 1 and 3 are different).
Note that an empty array is always balanced.

You are given a non-decreasing array a consisting of n integers.
Find the length of its longest balanced subsequence.
(A sequence b is a subsequence of a sequence a if b can be obtained from a
by the deletion of several (possibly, zero or all) element from arbitrary positions.)

Input
Each test contains multiple test cases. 

The first line contains the number of test cases t (1≤t≤500).
The description of the test cases follows.

The first line of each test case contains a single integer n(1≤n≤100) — the length of a.
The second line contains n integers a1,a2,...,an (1≤a1≤a2≤...≤an≤n)

Output
For each test case, output a single integer — the length of the longest balanced subsequence of a.

Input
4
5
1 1 4 4 4
2
1 2
15
1 1 1 1 1 2 2 2 2 3 3 3 4 4 5
5
3 3 3 3 3

Output
4
2
9
5

Note
In the first test case, the whole array a=[1,1,4,4,4] is not balanced because the number of occurrences of element 1 is 2,
while the number of occurrences of element 4 is 3, which are not equal.
The subsequence [1,1,4,4] is balanced because the numbers of occurrences of elements 1 and 4 are both 2.
Thus, the length of the longest balanced subsequence of a is 4.

In the second test case, the whole array a=[1,2] is already balanced,
so the length of the longest balanced subsequence of a is 2.

In the third test case, the longest balanced subsequence of a is [1,1,1,2,2,2,3,3,3].

In the fourth test case, the whole array a=[3,3,3,3,3] is already balanced,
so the length of the longest balanced subsequence of a is 5.
*/