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BFS好像也是指数级别的复杂度,可以用最短路n^2搞定预处理f(r)=(10^r)%19 在长度为19的数轴上,建立边,变长为f(r),若从0到19有通路,则通路上的边对应的r就是取1,否则取0 Followed by: Post your reply here: |
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