【条件】奇偶判断 - 题解 - 蒙青创OJ

2
余梓铭ml LV 6 @ 2026-5-23 15:25:32
#include <bits/stdc++.h> using namespace std;

int main() { long long a; cin>>a; if(a%2==0){ cout<<"even"; } else { cout<<"odd"; }
```
return 0;
```
}
1
陆子麟ly LV 3 @ 2026-5-23 16:40:01

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

int main() { long long a; cin>>a; if(a%2==0){ cout<<"even"; } else { cout<<"odd"; }

return 0; }
0
张智皓bw LV 2 @ 2026-5-23 9:14:21

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

int main() { int n; cin >> n; if (n % 2 == 0) { cout << "even"; } else { cout << "odd"; } return 0; }

ID: 30926
时间: 1000ms
内存: 256MiB
难度: 6
标签: (无)
递交数: 217
已通过: 71
上传者: Hydro