思路：单调递减栈，找「上一个比当前价格大的元素索引」。正序遍历，弹出所有 ≤ 当前价格的索引；跨度 = 当前索引 - 栈顶索引（栈空则为 当前索引 + 1）。
2026年2月26日のヘッドラインニュース，推荐阅读搜狗输入法下载获取更多信息
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Sign anonymously. Your name will not be published.。关于这个话题，爱思助手下载最新版本提供了深入分析
I completely ignored Anthropic’s advice and wrote a more elaborate test prompt based on a use case I’m familiar with and therefore can audit the agent’s code quality. In 2021, I wrote a script to scrape YouTube video metadata from videos on a given channel using YouTube’s Data API, but the API is poorly and counterintuitively documented and my Python scripts aren’t great. I subscribe to the SiIvagunner YouTube account which, as a part of the channel’s gimmick (musical swaps with different melodies than the ones expected), posts hundreds of videos per month with nondescript thumbnails and titles, making it nonobvious which videos are the best other than the view counts. The video metadata could be used to surface good videos I missed, so I had a fun idea to test Opus 4.5: