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。
features ATMs have today. It also didn't sell. I haven't been able to find any
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This article originally appeared on Engadget at https://www.engadget.com/mobile/everything-announced-at-samsung-unpacked-the-galaxy-s26-ultra-galaxy-buds-4-and-more-180000530.html?src=rss