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所有线路的bus两两匹配，如果可以碰到加入同一集合，最后看是否只有1个集合设i线路周期为ti，j线路周期为tj，ij某个公共站k i线某bus到k要走a个站 j线某bus到k要走b个站 ti * x + a = tj * y + b 只要有xy非负整数解，则这2个bus肯定可以在k站碰面 有解条件为 abs(a-b) % gcd(x,y) == 0 Followed by: Post your reply here: |
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