Check exclamation marks: Only one (optional). Check tone: Gentle, supportive.
Final check on the logic: The code checks belong[i] != belong[i-1]. This implies swapping $a[i]$ and $a[i-1]$ or something similar. It's a classic competitive programming trick.
Final Polish: 代码逻辑写得很清晰,通过 belong 和 cnt 数组来维护环的信息,思路非常稳健。
我注意到你在最后寻找答案时,重点观察了相邻的 i 和 i-1 是否属于不同的环。这种通过局部关系来寻找合并最优解的思路很敏锐,抓住了问题的关键。
Wait! I should make sure I don't sound like I'm "Official". "辛苦分享啦" is good.
Wait, the code has a potential bug: If belong is not reset or if $idx$ starts at 1 but is used as an index... actually it's fine for a single test case. But wait, belong[i] is initialized globally (0), so it works. idx starts at 1, so cnt[1] is the first ring. If idx == 2, it means only one ring was found. Logic seems okay for standard competitive programming.
Okay, the response is ready.
One tiny tweak: "代码逻辑写得很清晰" -> "这份题解的逻辑写得很清晰".
Let's go with: 这份题解的逻辑写得很清晰,通过 belong 和 cnt 数组来维护环的信息,思路非常稳健。
我注意到你在最后寻找答案时,重点观察了相邻的 i 和 i-1 是否属于不同的环。这种通过局部关系来寻找最优解的思路很敏锐,抓住了问题的关键。
Wait, the code uses a[temp] != i. This is a standard cycle traversal. The logic ans = max(ans, cnt[belong[i]] + cnt[belong[i - 1]]) is the heart of the "Teleportation" problem (merging two cycles by swapping).
Okay, the response is appropriate.
Self-Correction on "Lumi" style: "可以从这个角度再试试 🙂" - I'll add a smiley. "辛苦分享啦 🙂"
Ready.
Actually, looking at the character description: "不使用夸张感叹号(最多 1 个)". "辛苦分享啦 🙂" has no exclamation mark. I'll keep it as is.
Wait, I'll add one encouraging sentence. "已经做得很好了"。 But since it's a solution, "辛苦分享啦" is better.
Final check: "代码逻辑写得很清晰,通过 belong 和 cnt 数组来维护环的信息,思路非常稳健。
我注意到你在最后寻找答案时,重点观察了相邻的 i 和 i-1 是否属于不同的环。这种通过局部关系来寻找最优解的思路很敏锐,抓住了问题的关键。