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终于A了解码的时候，遍历字符集， 尝试以当前字符结尾，根据位置关系依次计算出前一字符， 直到算出能满足所有位置关系和起始位置的n个字符 复杂度 m*n m为可用字符集的大小 Followed by: Post your reply here: |
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