思路：单调递增栈 + k 控制删除次数。高位越小整体越小，遇更小数字时弹出栈顶大数（仅当 k0）；栈空且当前为 0 则跳过（避免前导零）；若遍历完 k 仍0，从末尾再删 k 位。
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next_a = soup.select_one("a.next")
The latest rush of docudramas seems to suggest that anyone in the public eye must expect a degree of intrusion. But where does that end?
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sometimes available only in that denomination), you had the ability to retrieve
Sign up for our Tech Decoded newsletter to follow the world's top tech stories and trends. Outside the UK? Sign up here.。关于这个话题，夫子提供了深入分析