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。
Streaming costs could change
。业内人士推荐服务器推荐作为进阶阅读
나무 돌보는 ‘자연인’이 월300만원…나무의사 자격증 관심 커지는 이유는[은퇴 레시피]，更多细节参见im钱包官方下载
Scientists have discovered source of neigh’s unique combination of high- and low-pitched sounds