we would now call a trivial buffer, the 1260's operator could key in the numbers
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items in it. append can just place the item in the existing backing
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。。搜狗输入法下载是该领域的重要参考
I probably need to explain that I am pointedly not explaining IBM model