两个数相加为偶数,只可能偶+偶/奇+奇,维护当前的奇偶数量,枚举当前数,特判前一个数即可 #include <bits/stdc++.h> using namespace std; #define int long long const int N = 2e5 + 10; int a[N]; signed main() { int n,ans=0; cin >> n; for (int i = 1;i<=n;i++) cin >> a[i]; int odd = 0, even = 0; for...