HardIneq | Topcoder Archive

Problem Statement

A very famous algorithms text presents its readers with a seemingly simple math problem in its opening chapter:
Find the smallest positive integer value of n such that a*(n^k) < k^n, where a and k are given integral values and ^ denotes exponentiation.
Solving this problem algebraically involves using the overly complicated Lambert W function. Instead you are to solve the problem by creating a class HardIneq with a method biggestVal which, given a and k, returns the smallest positive integer value of n such that a*(n^k) < k^n.

Definition

Class:: HardIneq
Method:: biggestVal
Parameters:: int, int
Returns:: int
Method signature:: int biggestVal(int a, int k)
(be sure your method is public)

Notes

Remember that log(x^y) = y * log(x) and that log(x*y) = log(x) + log(y)

Constraints

a will be between 1 and 10000, inclusive
k will be between 2 and 1000, inclusive

Examples

```
10
```
```
5
```
```
Returns: 8
```
if n = 1, we have a*(n^k) = 10*(1^5) = 10, and k^n = 5^1 = 5. 10 is not less than 5. n=2: 10*2^5 is not less than 5^2 n=2: 10*3^5 is not less than 5^3 and so on. Not until we get to n = 8, and have 10*8^5 = 327680, and 5^8 = 390625 is the inequality satisfied.
```
1
```
```
2
```
```
Returns: 1
```
1*(1^2) is less than 2^1, so we return 1.
```
2000
```
```
200
```
```
Returns: 202
```
```
27
```
```
3
```
```
Returns: 10
```
Note that 27*9^3 is equal to 3^9.
```
729
```
```
6
```
```
Returns: 13
```

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