A. Odd Selection【BruteForce】

A. Odd Selection

The question

Given n and x（x <= n) . about n Select the number x Number , Make the selected number sum The sum is odd .

Cause of error

Even one does not , It is required to select an even number , The output should be no. Less discussion on classification .

analysis

For this question , What I can know is that odd numbers play a key role , The number of odd numbers must be odd ！ In fact, you can enumerate directly according to this feature .

But I want to classify , Want to get the answer directly in one step , This discussion is easy to make mistakes and takes a lot of brains , So you can enumerate directly according to the characteristics .

be-all brute force The wrong questions are the loopholes in the classified discussion .

First, according to the range of data and the sense of problem , Judgment is brute force The subject of ！

So the data range can brute force The topic , The first time should not be to discuss the classification and get the answer directly , Instead, look for enumerated objects .

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;

int main(){
	std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
 	int t;
 	cin >> t;
 	while(t -- ) {
 		int n, x;
 		cin >> n >> x;
 		std::vector<int> a(n);
 		int cnt[2] = {0, 0};
 		for (int i = 0; i < n; i ++ ) {
 			cin >> a[i];
 			cnt[0] += a[i] & 1;
 			cnt[1] += !(a[i] & 1);
 		}	
 		bool ok = false;
 		for (int i = 1; i <= cnt[0] && i <= x; i += 2)
 		{
 			int need = x - i;
 			if(cnt[1] >= need) ok = true;
 		}
 		cout << (ok ? "Yes" : "No") << '\n';
 	}   
	return 0;
}