题解 | #小欧的奇数#

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<ll, ll>PII;
const int N = 2e5 + 10;
const int MOD = 998244353;
const int INF = 0X3F3F3F3F;
const int dx[] = {-1, 1, 0, 0, -1, -1, +1, +1};
const int dy[] = {0, 0, -1, 1, -1, +1, -1, +1};
const int M = 1e6 + 10;

int a[N];
int main()
{
    int n;
    cin >> n;
	int c1 = 0;//只记录奇数即可
	for(int i = 1; i <= n; i ++)
	{
		cin >> a[i];
		if(a[i] & 1) c1 ++;
	}
	if(c1 >= 3 || (c1 >= 1 && n - c1 >= 2)) puts("YES");
	else puts("NO");
}