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  • Project Euler 110:Diophantine reciprocals II 丢番图倒数II

    Diophantine reciprocals II

    In the following equation x, y, and n are positive integers.

    For n = 4 there are exactly three distinct solutions:

     

    It can be verified that when n = 1260 there are 113 distinct solutions and this is the least value of n for which the total number of distinct solutions exceeds one hundred.

    What is the least value of n for which the number of distinct solutions exceeds four million?

    NOTE: This problem is a much more difficult version of Problem 108 and as it is well beyond the limitations of a brute force approach it requires a clever implementation.


    丢番图倒数II

    在如下方程中,x、y、n均为正整数。

    对于n = 4,上述方程恰好有3个不同的解:

    可以验证当n = 1260时,恰好有113种不同的解,这也是不同的解的总数超过一百种的最小n值。

    不同的解的总数超过四百万种的最小n值是多少?

    注意:这是第108题一个极其困难的版本,而且远远超过暴力解法的能力范围,因此需要更加聪明的手段。

     解题

    先占坑

     待完善

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  • 原文地址:https://www.cnblogs.com/theskulls/p/5155229.html
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