FallingFactorialPower | Topcoder Archive

Problem Statement

A number n taken to the falling factorial power k is defined as n*(n-1)*...*(n-k+1). We will denote it by n^^k. For example, 7^^3=7*6*5=210. By definition, n^^1=n.

We will now continue this definition to the non-positive values of k using the following fact: (n-k)*(n^^k)=n^^(k+1), or, in other words, n^^k=(n^^(k+1))/(n-k). It is directly derived from the above definition.

By using it, we find:

n^^0=n^^1/(n-0)=1,
n^^(-1)=n^^0/(n+1)=1/(n+1),
n^^(-2)=1/(n+1)/(n+2),
and, in general, n^^(-k)=1/(n+1)/(n+2)/.../(n+k).

For example, 3^^(-1)=1/4=0.25, 2^^(-3)=1/3/4/5=1/60=0.016666...

Given a positive int n and an int k, return a double containing the value of n taken to the falling factorial power of k.

Definition

Class:: FallingFactorialPower
Method:: compute
Parameters:: int, int
Returns:: double
Method signature:: double compute(int n, int k)
(be sure your method is public)

Notes

Your return must have relative or absolute error less than 1E-9.

Constraints

n will be between 1 and 10, inclusive.
k will be between -5 and 5, inclusive.

Examples

```
7
```
```
3
```
```
Returns: 210.0
```
7^^3=7*6*5=210.
```
10
```
```
1
```
```
Returns: 10.0
```
```
5
```
```
0
```
```
Returns: 1.0
```
```
3
```
```
-1
```
```
Returns: 0.25
```
```
2
```
```
-3
```
```
Returns: 0.016666666666666666
```

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