Codeforces Round #710 (Div. 3) D. Epic Transformation

本文主要是介绍Codeforces Round #710 (Div. 3) D. Epic Transformation，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！

题目:
You are given an array a of length n consisting of integers. You can apply the following operation, consisting of several steps, on the array a zero or more times:

you select two different numbers in the array ai and aj;
you remove i-th and j-th elements from the array.
For example, if n=6 and a=[1,6,1,1,4,4], then you can perform the following sequence of operations:

select i=1,j=5. The array a becomes equal to [6,1,1,4];
select i=1,j=2. The array a becomes equal to [1,4].
What can be the minimum size of the array after applying some sequence of operations to it?

Input
The first line contains a single integer t (1≤t≤104). Then t test cases follow.

The first line of each test case contains a single integer n (1≤n≤2⋅105) is length of the array a.

The second line of each test case contains n integers a1,a2,…,an (1≤ai≤109).

It is guaranteed that the sum of n over all test cases does not exceed 2⋅105.

Output
For each test case, output the minimum possible size of the array after applying some sequence of operations to it.

Example
inputCopy
5
6
1 6 1 1 4 4
2
1 2
2
1 1
5
4 5 4 5 4
6
2 3 2 1 3 1
outputCopy
0
0
2
1
0
题解:
将每个数的个数储存在优先队列里,每次处理最大的

#include <bits/stdc++.h>
using namespace std;
int main()
{int t;cin>>t;while(t--){map<long long,int> a;int n;cin>>n;for(int i=0;i<n;i++){long long b;scanf("%lld",&b);a[b]++;}priority_queue<int,vector<int>,less<int>>q;for(auto x:a){q.push(x.second);}while(q.size()>1){int c=q.top();q.pop();int d=q.top();q.pop();c--;d--;if(c) q.push(c);if(d) q.push(d);}if(q.size()) cout<<q.top()<<endl;else cout<<0<<endl;}return 0;
}