# codeforces #341 div 2 D. Rat Kwesh and Cheese

## http://codeforces.com/contest/621/problem/D 题意：给出12个式子，问哪个最大。 思路：主要记住两个。一个是比较指数形式的数一个常用办法是取对数，同时要考虑是否能取对数，分情况讨论对于不能取对数的情况经过变换去取对数。第二个是取了两次对数后比较时候的最大值可能是小于0的。所以初始时置于0不够小。官方题解说得很清楚。

The tricky Rat Kwesh has finally made an appearance; it is time to prepare for some tricks. But truly, we didn’t expect it to be so hard for competitors though. Especially the part about taking log of a negative number.

We need a way to deal with xyz and xyz. We cannot directly compare them, 200200200 is way too big. So what we do? Take log! is an increasing function on positive numbers (we can see this by taking , then , which is positive when we are dealing with positive numbers). So if , then x ≥ y.

When we take log, But yz can still be 200200, which is still far too big. So now what can we do? Another log! But is it legal? When x = 0.1 for example, , so we cannot take another log. When can we take another log, however? We need to be a positive number. yz will always be positive, so all we need is for to be positive. This happens when x > 1. So ifx, y, z > 1, everything will be ok.

There is another good observation to make. If one of x, y, z is greater than 1, then we can always achieve some expression (out of those 12) whose value is greater than 1. But if x < 1, then xa will never be greater than 1. So if at least one of x, y, z is greater than 1, then we can discard those bases that are less than or equal to 1. In this case, . Remember that , so . Similarly, .

The last case is when x ≤ 1, y ≤ 1, z ≤ 1. Then, notice that for example, . But the denominator of this fraction is something we recognize, because 10 / 3 > 1. So if all x, y, z < 1, then it is the same as the original problem, except we are looking for the minimum this time.