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我的简洁算法从tot/(C1+C2)到n枚举，最少tot/(C1+C2)趟，最多n趟，从小到达搜索，找到答案就退出，比如搜索在i趟内能不能完工，可以转换为在一趟内用两种型号的车各i辆能不能完工，然后用深搜就可以实现了，很简洁 Followed by: Post your reply here: |
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