反向析因 [英] Reverse factorial

问题描述

好了，我们都知道，如果N给出很容易计算出n个！但是，我们的反？

N！给予，你要找到N - 这可能吗？我很好奇。

解决方案

1. 设置 X = 1 。
2. 生成 F = X！
3. 为：F =输入？如果是，则 X 是N
4. 如果没有，则设置 X = X + 1 ，然后在＃2重新开始。

您可以通过使用 F 的previous结果来计算新 F （优化新F =新的X *老˚F）。

它一样快会相反的方向，如果不是更快，因为除法一般需要长于乘法。一个给定的阶乘 A！保证具有小于 A 所有整数的因素除了A，所以你' ð花同样多的时间分解出来的那些，你会只计算运行阶乘。

Well, we all know that if N is given it's easy to calculate N!. But what about the inverse?

N! is given and you are about to find N - Is that possible ? I'm curious.

解决方案

1. Set X=1.
2. Generate F=X!
3. Is F = the input? If yes, then X is N.
4. If not, then set X=X+1, then start again at #2.

You can optimize by using the previous result of F to compute the new F (new F = new X * old F).

It's just as fast as going the opposite direction, if not faster, given that division generally takes longer than multiplication. A given factorial A! is guaranteed to have all integers less than A as factors in addition to A, so you'd spend just as much time factoring those out as you would just computing a running factorial.

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