Codeforces Round #634 C. Two Teams Composing（排序）

题目描述

You have $a_i ai (different students can have the same skills).$

So, about the teams. Firstly, these two teams should have the same size. Two more constraints:

The first team should consist of students with distinct skills (i.e. all skills in the first team are unique).
The second team should consist of students with the same skills (i.e. all skills in the second team are equal).

Note that it is permissible that some student of the first team has the same skill as a student of the second team.

Consider some examples (skills are given):

Your task is to find the maximum possible size

You have to answer

输入格式

The first line of the input contains one integer $\le t \le 10^4 1≤t≤104 ) — the number of test cases. Then t t t test cases follow.$

The first line of the test case contains one integer $\le n \le 2 \cdot 10^5 1≤n≤2⋅105 ) — the number of students. The second line of the test case contains n n n integers a1,a2,…,an a_1, a_2, \dots, a_n a1,a2,…,an ( 1≤ai≤n 1 \le a_i \le n 1≤ai≤n ), where ai a_i ai is the skill of the i i i -th student. Different students can have the same skills.$

It is guaranteed that the sum of $\cdot 10^5 2⋅105 ( ∑n≤2⋅105 \sum n \le 2 \cdot 10^5 ∑n≤2⋅105 ).$

输出格式

For each test case, print the answer — the maximum possible size

输入输出样例

输入 #1

4
7
4 2 4 1 4 3 4
5
2 1 5 4 3
1
1
4
1 1 1 3

输出 #1

3
1
0
2
sort以后求出有多少不同的数diff以及出现最多的数的出现次数mmax,最终答案是max(min(mmax-1,diff),min(mmax,diff-1))因为可以这么分(1 1 1 1)(2 3 4) 也可以这么分(1 1 1)(1 2 3 4)这两种情况取最大即可。

#include <bits/stdc++.h>
using namespace std;
int a[200005];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,i;
        cin>>n;
        int pre=0,cnt=0,mmax=0,diff=0;//mmax:出现次数最多的数的次数 diff：多少个不同的数 cnt：当前数出现的次数 pre：上一个数 
        for(i=1;i<=n;i++) scanf("%d",&a[i]);
        sort(a+1,a+n+1);
        for(i=1;i<=n+1;i++)//得多写一个 
        {
            if(a[i]!=pre)
            {
                pre=a[i];
                diff++;
                mmax=max(mmax,cnt);
                cnt=1;
            }
            else cnt++;
        }
        diff--;//减去多的那个 
        cout<<max(min(mmax-1,diff),min(mmax,diff-1))<<endl;
    }
    return 0;
}

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