When you publish content on LimeWire, you will receive 70% of all ad revenue from other users who view your images, music, and videos on the platform.
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。
。业内人士推荐safew官方下载作为进阶阅读
Scroll to load interactive demo
In microcode, the privilege check reduces to a single conditional jump: