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对N的最大修改位数<=K的位数+1，想优化的去改代码吧设K的位数为D，通过修改N的最后(D+1)位，至少可以生成(K+1)个连续的新数,在这(K+1)个数中一定有一个数可以整除K. 所以对任何修改超过(D+1)位的尝试都可以忽略. Followed by: Post your reply here: |
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