POJ - 2109 - Power of Cryptography = 高精度整数

http://poj.org/problem?id=2109

给一个大概100位的高精度整数p，找他的n次方根，貌似题目有问题，不一定会有开根开得尽的情况，这个时候要找的就是最大的根。

那这样有什么意思呢？

这种题按道理要用Java去写的，可以先从p和n的关系找到根的位数，然后在这个范围里面二分会更快。

具体来说，比如22位的p，7次方根，就是22/7的上整，取4位数。

还有就是因为题目的原因，double的精度对100位的高精度整数开根是足够的，也就是不会的话可以试试莽一发long double？

https://blog.csdn.net/code_pang/article/details/8263971

但是long double反而是过不了，不知道是为什么，有人用泰勒公式分析到double的误差不超过9.0e-7，说明double还是比较给力的。

果然是歪门邪道，这种还是乖乖写Java的大数。

原文地址：https://www.cnblogs.com/Inko/p/11723245.html

时间： 2024-10-08 01:29:49

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