meetforever/GRPO_CL huggingface dataset:
user: Challenge:
Given two null-terminated strings in the arguments "string" and "prefix", determine if "string" starts with the "prefix" string. Return 1 (or any other "truthy" value) if true, 0 if false.

Example:
```
startsWith("hello world!", "hello"); // should return 1.
startsWith("hello world!", "HELLO"); // should return 0.
startsWith("nowai", "nowaisir"); // should return 0.
```

Addendum:
For this problem, an empty "prefix" string should always return 1 (true) for any value of "string".

If the length of the "prefix" string is greater than the length of the "string", return 0.

The check should be case-sensitive, i.e. startsWith("hello", "HE") should return 0, whereas startsWith("hello", "he") should return 1.

The length of the "string" as well as the "prefix" can be defined by the formula: 0 <= length < +Infinity

No characters should be ignored and/or omitted during the test, e.g. whitespace characters should not be ignored.

Answer: {"fn_name": "starts_with", "inputs": [["hello world!", "hello"], ["hello world!", "HELLO"], ["nowai", "nowaisir"], ["", ""], ["abc", ""], ["", "abc"]], "outputs": [[true], [false], [false], [true], [true], [false]]}
user: In this kata you will be given an **integer n**, which is the number of times that is thown a coin. You will have to return an array of string for all the possibilities (heads[H] and tails[T]). Examples:
```coin(1) should return {"H", "T"}```
```coin(2) should return {"HH", "HT", "TH", "TT"}```
```coin(3) should return {"HHH", "HHT", "HTH", "HTT", "THH", "THT", "TTH", "TTT"}```
When finished sort them alphabetically.

In C and C++ just return a ```char*``` with all elements separated by```,``` (without space):
```coin(2) should return "HH,HT,TH,TT"```
INPUT:
```0 < n < 18```
Careful with performance!! You'll have to pass 3 basic test (n = 1, n = 2, n = 3), many medium tests (3 < n <= 10) and many large tests (10 < n < 18)

Answer: {"fn_name": "coin", "inputs": [[1], [2], [3]], "outputs": [[["H", "T"]], [["HH", "HT", "TH", "TT"]], [["HHH", "HHT", "HTH", "HTT", "THH", "THT", "TTH", "TTT"]]]}
user: In this kata you have to correctly return who is the "survivor", ie: the last element of a Josephus permutation.

Basically you have to assume that n people are put into a circle and that they are eliminated in steps of k elements, like this:

```
josephus_survivor(7,3) => means 7 people in a circle;
one every 3 is eliminated until one remains
[1,2,3,4,5,6,7] - initial sequence
[1,2,4,5,6,7] => 3 is counted out
[1,2,4,5,7] => 6 is counted out
[1,4,5,7] => 2 is counted out
[1,4,5] => 7 is counted out
[1,4] => 5 is counted out
[4] => 1 counted out, 4 is the last element - the survivor!
```

The above link about the "base" kata description will give you a more thorough insight about the origin of this kind of permutation, but basically that's all that there is to know to solve this kata.

**Notes and tips:** using the solution to the other kata to check your function may be helpful, but as much larger numbers will be used, using an array/list to compute the number of the survivor may be too slow; you may assume that both n and k will always be >=1.

Answer: {"fn_name": "josephus_survivor", "inputs": [[7, 3], [11, 19], [40, 3], [14, 2], [100, 1], [1, 300], [2, 300], [5, 300], [7, 300], [300, 300]], "outputs": [[4], [10], [28], [13], [100], [1], [1], [1], [7], [265]]}
user: JavaScript provides a built-in parseInt method.

It can be used like this:

- `parseInt("10")` returns `10`
- `parseInt("10 apples")` also returns `10`

We would like it to return `"NaN"` (as a string) for the second case because the input string is not a valid number.

You are asked to write a `myParseInt` method with the following rules:

- It should make the conversion if the given string only contains a single integer value (and possibly spaces - including tabs, line feeds... - at both ends)
- For all other strings (including the ones representing float values), it should return NaN
- It should assume that all numbers are not signed and written in base 10

Answer: {"fn_name": "my_parse_int", "inputs": [["9"], ["09"], ["009"], ["  9"], ["  9  "], ["\t9\n"], ["5 friends"], ["5friends"], ["I <3 u"], ["17.42"], ["0x10"], ["123~~"], ["1 1"], ["1 2 3"], ["1.0"]], "outputs": [[9], [9], [9], [9], [9], [9], ["NaN"], ["NaN"], ["NaN"], ["NaN"], ["NaN"], ["NaN"], ["NaN"], ["NaN"], ["NaN"]]}
user: You are standing on top of an amazing Himalayan mountain. The view is absolutely breathtaking! you want to take a picture on your phone, but... your memory is full again! ok, time to sort through your shuffled photos and make some space...

Given a gallery of photos, write a function to sort through your pictures.
You get a random hard disk drive full of pics, you must return an array with the 5 most recent ones PLUS the next one (same year and number following the one of the last).

You will always get at least a photo and all pics will be in the format `YYYY.imgN`

Examples:
```python
sort_photos["2016.img1","2016.img2","2015.img3","2016.img4","2013.img5"]) ==["2013.img5","2015.img3","2016.img1","2016.img2","2016.img4","2016.img5"]
sort_photos["2016.img1"]) ==["2016.img1","2016.img2"]
```

Answer: {"fn_name": "sort_photos", "inputs": [[["2016.img1", "2016.img2", "2016.img3", "2016.img4", "2016.img5"]], [["2016.img4", "2016.img5", "2016.img1", "2016.img3", "2016.img2"]], [["2012.img2", "2016.img1", "2016.img3", "2016.img4", "2016.img5"]], [["2016.img1", "2013.img3", "2016.img2", "2015.img3", "2012.img7", "2016.img4", "2013.img5"]], [["2016.img7", "2016.img2", "2016.img3", "2015.img3", "2012.img8", "2016.img4", "2016.img5"]]], "outputs": [[["2016.img1", "2016.img2", "2016.img3", "2016.img4", "2016.img5", "2016.img6"]], [["2016.img1", "2016.img2", "2016.img3", "2016.img4", "2016.img5", "2016.img6"]], [["2012.img2", "2016.img1", "2016.img3", "2016.img4", "2016.img5", "2016.img6"]], [["2013.img5", "2015.img3", "2016.img1", "2016.img2", "2016.img4", "2016.img5"]], [["2016.img2", "2016.img3", "2016.img4", "2016.img5", "2016.img7", "2016.img8"]]]}
user: *** No Loops Allowed ***

You will be given an array (a) and a value (x). All you need to do is check whether the provided array contains the value, without using a loop.

Array can contain numbers or strings. X can be either. Return true if the array contains the value, false if not. With strings you will need to account for case.

Looking for more, loop-restrained fun? Check out the other kata in the series:

 https://www.codewars.com/kata/no-loops-1-small-enough
 https://www.codewars.com/kata/no-loops-3-copy-within

Answer: {"fn_name": "check", "inputs": [[[66, 101], 66], [[80, 117, 115, 104, 45, 85, 112, 115], 45], [["t", "e", "s", "t"], "e"], [["what", "a", "great", "kata"], "kat"], [[66, "codewars", 11, "alex loves pushups"], "alex loves pushups"], [["come", "on", 110, "2500", 10, "!", 7, 15], "Come"], [["when's", "the", "next", "Katathon?", 9, 7], "Katathon?"], [[8, 7, 5, "bored", "of", "writing", "tests", 115], 45], [["anyone", "want", "to", "hire", "me?"], "me?"]], "outputs": [[true], [true], [true], [false], [true], [false], [true], [false], [true]]}
user: Given an array of integers `a` and integers `t` and `x`, count how many elements in the array you can make equal to `t` by **increasing** / **decreasing** it by `x` (or doing nothing).
*EASY!*

```python
# ex 1

a = [11, 5, 3]
t = 7
x = 2

count(a, t, x) # => 3
```
- you can make 11 equal to 7 by subtracting 2 twice
- you can make 5 equal to 7 by adding 2
- you can make 3 equal to 7 by adding 2 twice

```python
# ex 2

a = [-4,6,8]
t = -7
x = -3

count(a, t, x) # => 2
```

## Constraints
**-10^(18) <= a[i],t,x <= 10^(18)**

**3 <= |a| <= 10^(4)**

Answer: {"fn_name": "count", "inputs": [[[11, 5, 3], 7, 2], [[-4, 6, 8], -7, -3]], "outputs": [[3], [2]]}
user: Given a string `S` and a character `C`, return an array of integers representing the shortest distance from the current character in `S` to `C`.

### Notes

* All letters will be lowercase.
* If the string is empty, return an empty array.
* If the character is not present, return an empty array.

## Examples

```python
shortest_to_char("lovecodewars", "e") == [3, 2, 1, 0, 1, 2, 1, 0, 1, 2, 3, 4]
shortest_to_char("aaaabbbb", "b") == [4, 3, 2, 1, 0, 0, 0, 0]
shortest_to_char("", "b") == []
shortest_to_char("abcde", "") == []
```

___

If you liked it, please rate  :D

Answer: {"fn_name": "shortest_to_char", "inputs": [["lovecodewars", "e"], ["aaaaa", "a"], ["aabbaabb", "a"], ["aaaabbbb", "b"], ["aaaaa", "b"], ["lovecoding", ""], ["", ""]], "outputs": [[[3, 2, 1, 0, 1, 2, 1, 0, 1, 2, 3, 4]], [[0, 0, 0, 0, 0]], [[0, 0, 1, 1, 0, 0, 1, 2]], [[4, 3, 2, 1, 0, 0, 0, 0]], [[]], [[]], [[]]]}
user: The objective is to disambiguate two given names: the original with another

This kata is slightly more evolved than the previous one: [Author Disambiguation: to the point!](https://www.codewars.com/kata/580a429e1cb4028481000019).

The function ```could_be``` is still given the original name and another one to test
against. 

```python
# should return True even with 'light' variations (more details in section below)
> could_be("Chuck Norris", u"chück!")
True

# should False otherwise (whatever you may personnaly think)
> could_be("Chuck Norris", "superman")
False
``` 

**Watch out**: When accents comes into the game, they will enter through **UTF-8 unicodes. **

The function should be tolerant with regards to:

 * upper and lower cases: ```could_be(A, a) : True```
 * accents: ```could_be(E, é) : True```
 * dots: ```could_be(E., E) : True```
 * same for other ending punctuations in [!,;:?]: ```could_be(A, A!) : True```

On the other hand, more consideration needs to be given to *composed names*...
Let's be bold about it: if you have any, they will be considered as a whole :

```python
# We still have:
> could_be("Carlos Ray Norris", "Carlos Ray Norris")
True
> could_be("Carlos-Ray Norris", "Carlos-Ray Norris")
True

# But:
> could_be("Carlos Ray Norris", "Carlos-Ray Norris")
False
> could_be("Carlos-Ray Norris", "Carlos Ray Norris")
False
> could_be("Carlos-Ray Norris", "Carlos Ray-Norris")
False
```
 
Among the valid combinaisons of the fullname "Carlos Ray Norris", you will find

```python
could_be("Carlos Ray Norris", "carlos ray") : True
could_be("Carlos Ray Norris", "Carlos. Ray, Norris;") : True
could_be("Carlos Ray Norris", u"Carlòs! Norris") : True
```

Too easy ? Try the next step: [Author Disambiguation: Signatures worth it](https://www.codewars.com/kata/author-disambiguation-signatures-worth-it)

Answer: {"fn_name": "could_be", "inputs": [["Carlos Ray Norris", "Carlos Ray Norris"], ["Carlos Ray Norris", "Carlos Ray"], ["Carlos Ray Norris", "Ray Norris"], ["Carlos Ray Norris", "Carlos Norris"], ["Carlos Ray Norris", "Norris"], ["Carlos Ray Norris", "Carlos"], ["Carlos Ray Norris", "Norris Carlos"], ["Carlos Ray Norris", "carlos ray norris"], ["Carlos Ray Norris", "Norris! ?ray"], ["Carlos Ray Norris", "Carlos. Ray; Norris,"], ["Carlos Ray Norris", "Carlos:Ray Norris"], ["Carlos-Ray Norris", "Carlos-Ray Norris:"], ["Carlos Ray-Norris", "Carlos? Ray-Norris"], ["Carlos Ray Norris", "Carlos Ray Norr"], ["Carlos Ray Norris", "Ra Norris"], ["", "C"], ["", ""], ["Carlos Ray Norris", " "], ["Carlos-Ray Norris", "Carlos Ray-Norris"], ["Carlos Ray Norris", "Carlos-Ray Norris"], ["Carlos Ray Norris", "Carlos Ray-Norris"], ["Carlos Ray", "Carlos Ray Norris"], ["Carlos", "Carlos Ray Norris"]], "outputs": [[true], [true], [true], [true], [true], [true], [true], [true], [true], [true], [true], [true], [true], [false], [false], [false], [false], [false], [false], [false], [false], [false], [false]]}
user: A number m of the form 10x + y is divisible by 7 if and only if x − 2y is divisible by 7. In other words, subtract twice the last digit 
from the number formed by the remaining digits. Continue to do this until a number known to be divisible or not by 7 is obtained; 
you can stop when this number has *at most* 2 digits because you are supposed to know if a number of at most 2 digits is divisible by 7 or not.

The original number is divisible by 7 if and only if the last number obtained using this procedure is divisible by 7. 

Examples:

1 - `m = 371 -> 37 − (2×1) -> 37 − 2 = 35` ; thus, since 35 is divisible by 7, 371 is divisible by 7. 

The number of steps to get the result is `1`.

2 - `m = 1603 -> 160 - (2 x 3) -> 154 -> 15 - 8 = 7` and 7 is divisible by 7. 

3 - `m = 372 -> 37 − (2×2) -> 37 − 4 = 33` ; thus, since 33 is not divisible by 7, 372 is not divisible by 7. 

The number of steps to get the result is `1`.

4 - `m = 477557101->47755708->4775554->477547->47740->4774->469->28` and 28 is divisible by 7, so is 477557101. 

The number of steps is 7.

# Task:
Your task is to return to the function ```seven(m)``` (m integer >= 0) an array (or a pair, depending on the language) of numbers, 
the first being the *last* number `m` with at most 2 digits obtained by your function (this last `m` will be divisible or not by 7), the second one being the number of steps to get the result.

## Forth Note:
Return on the stack `number-of-steps, last-number-m-with-at-most-2-digits `

## Examples:
```
seven(371) should return [35, 1]
seven(1603) should return [7, 2]
seven(477557101) should return [28, 7]
```

Answer: {"fn_name": "seven", "inputs": [[1021], [477557101], [477557102], [2340029794923400297949], [1603], [371], [1369851], [483], [483595], [0], [170232833617690725]], "outputs": [[[10, 2]], [[28, 7]], [[47, 7]], [[14, 20]], [[7, 2]], [[35, 1]], [[0, 5]], [[42, 1]], [[28, 4]], [[0, 0]], [[0, 16]]]}
user: In this Kata, you will implement the [Luhn Algorithm](http://en.wikipedia.org/wiki/Luhn_algorithm), which is used to help validate credit card numbers.

Given a positive integer of up to 16 digits, return ```true``` if it is a valid credit card number, and ```false``` if it is not.

Here is the algorithm:
  
* Double every other digit, scanning **from right to left**, starting from the second digit (from the right).

  Another way to think about it is: if there are an **even** number of digits, double every other digit starting with the **first**; if there are an **odd** number of digits, double every other digit starting with the **second**:

  ```
  1714 ==> [1*, 7, 1*, 4] ==> [2, 7, 2, 4]
  
  12345 ==> [1, 2*, 3, 4*, 5] ==> [1, 4, 3, 8, 5]
  
  891 ==> [8, 9*, 1] ==> [8, 18, 1]
  ```

*  If a resulting number is greater than `9`, replace it with the sum of its own digits (which is the same as subtracting `9` from it):

  ```
  [8, 18*, 1] ==> [8, (1+8), 1] ==> [8, 9, 1]
  
  or:
  
  [8, 18*, 1] ==> [8, (18-9), 1] ==> [8, 9, 1]
  ```
  
* Sum all of the final digits:

  ```
  [8, 9, 1] ==> 8 + 9 + 1 = 18
  ```

* Finally, take that sum and divide it by `10`.  If the remainder equals zero, the original credit card number is valid.
    ```
    18 (modulus) 10 ==> 8 , which is not equal to 0, so this is not a valid credit card number
    ```

```if:fsharp,csharp
For F# and C# users:

The input will be a string of spaces and digits `0..9`

```

Answer: {"fn_name": "validate", "inputs": [[123], [1], [2121], [1230], [8675309], [4111111111111111], [26], [2626262626262626], [91], [92], [912030], [922030]], "outputs": [[false], [false], [true], [true], [false], [true], [true], [true], [true], [false], [true], [false]]}
user: Imagine a triangle of numbers which follows this pattern:

 * Starting with the number "1", "1" is positioned at the top of the triangle. As this is the 1st row, it can only support a single number.
 * The 2nd row can support the next 2 numbers: "2" and "3"
 * Likewise, the 3rd row, can only support the next 3 numbers: "4", "5", "6"
 * And so on; this pattern continues.

```
    1
   2 3
  4 5 6
 7 8 9 10
...
```

Given N, return the sum of all numbers on the Nth Row:

1 <= N <= 10,000

Answer: {"fn_name": "cumulative_triangle", "inputs": [[1], [2], [3], [4], [15], [150], [100], [500], [1000], [10000]], "outputs": [[1], [5], [15], [34], [1695], [1687575], [500050], [62500250], [500000500], [500000005000]]}
user: ##Overview

Write a helper function that takes in a Time object and converts it to a more human-readable format. You need only go up to '_ weeks ago'.
```python
to_pretty(0) => "just now"

to_pretty(40000) => "11 hours ago"
```
##Specifics
 - The output will be an amount of time, t, included in one of the following phrases: "just now", "[t] seconds ago", "[t] minutes ago", "[t] hours ago", "[t] days ago", "[t] weeks ago".
 - You will have to handle the singular cases. That is, when t = 1, the phrasing will be one of "a second ago", "a minute ago", "an hour ago", "a day ago", "a week ago".
 - The amount of time is always rounded down to the nearest integer. For example, if the amount of time is actually 11.73 hours ago, the return value will be "11 hours ago".
 - Only times in the past will be given, with the range "just now" to "52 weeks ago"

Answer: {"fn_name": "to_pretty", "inputs": [[0], [1], [30], [90], [180], [4000], [40000], [140000], [400000], [1000000], [10000000]], "outputs": [["just now"], ["a second ago"], ["30 seconds ago"], ["a minute ago"], ["3 minutes ago"], ["an hour ago"], ["11 hours ago"], ["a day ago"], ["4 days ago"], ["a week ago"], ["16 weeks ago"]]}
user: Ho ho! So you think you know integers, do you? Well then, young wizard, tell us what the Nth digit of the [Champernowne constant](https://en.wikipedia.org/wiki/Champernowne_constant) is!

The constant proceeds like this: `0.12345678910111213141516...`

I hope you see the pattern!

Conjure a function that will accept an integer, `n`, and return the (one-indexed) `n`th digit of Champernowne's constant. Can you get it to run in _constant_ time?

For example:

`n = 1` should return `0` (the very first digit)

`n = 2` should return `1` (we ignore the period character since it's not a digit!)

`n = 20` should return `4` (that's the `4` in the number `14`, 20th in sequence)

For any invalid values, such as `0` and below, or non-integers, return... `NaN`!

I hope (for your sake) that you've been practicing your mathemagical spells, because a naïve solution will _not_ be fast enough to compete in this championship!

Invoke with _precision_, and be wary of rounding errors in the realms of enormity!

May the best integer win!

Answer: {"fn_name": "champernowneDigit", "inputs": [[1], [2], [3], [4], [5], [6], [7], [8], [9], [10], [11], [12], [20], [35], [36], [37], [100], [101], [102], [103], [104], [105], [190], [2890], [38890], [488890], [5888890], [3678608], [3678609], [3678610], [3678611], [3678612], [3678613], [3678614], [3897249], [3897189], [3897309], [3897369], [3898749], [3898809], [3898869], [3898929], [3898989], [3899049], [3899109], [999999999], [1000599563], [1000599564], [1000599565], [1000599566], [1000599567], [1000599568], [101800813569], [77199254740991], [501337501337101]], "outputs": [[0], [1], [2], [3], [4], [5], [6], [7], [8], [9], [1], [0], [4], [2], [2], [2], [4], [5], [5], [5], [6], [5], [9], [9], [9], [9], [9], [6], [1], [9], [6], [3], [1], [6], [5], [4], [6], [7], [0], [1], [2], [3], [4], [5], [6], [8], [1], [2], [3], [5], [2], [3], [6], [7], [3]]}
user: # Description:

 Find the longest successive exclamation marks and question marks combination in the string. A successive exclamation marks and question marks combination must contains two part: a substring of "!" and a substring "?", they are adjacent. 
 
 If more than one result are found, return the one which at left side; If no such a combination found, return `""`.

# Examples

```
find("!!") === ""
find("!??") === "!??"
find("!?!!") === "?!!"
find("!!???!????") === "!!???"
find("!!???!?????") === "!?????"
find("!????!!!?") === "????!!!" 
find("!?!!??!!!?") === "??!!!"
```

# Note
Please don't post issue about difficulty or duplicate. Because:
>[That's unfair on the kata creator. This is a valid kata and introduces new people to javascript some regex or loops, depending on how they tackle this problem.  --matt c](https://www.codewars.com/kata/remove-exclamation-marks/discuss#57fabb625c9910c73000024e)

Answer: {"fn_name": "find", "inputs": [["!!"], ["!??"], ["!?!!"], ["!!???!????"], ["!!???!?????"], ["!????!!!?"], ["!?!!??!!!?"]], "outputs": [[""], ["!??"], ["?!!"], ["!!???"], ["!?????"], ["????!!!"], ["??!!!"]]}
user: Everyone Knows AdFly and their Sister Sites.

If we see the Page Source of an ad.fly site we can see this perticular line:
Believe it or not This is actually the Encoded url which you would skip to.

The Algorithm is as Follows:
```
1) The ysmm value is broken like this

ysmm   = 0 1 2 3 4 5 6 7 8 9  = "0123456789"
code1  = 0   2   4   6   8    = "02468"
code2  =   9   7   5   3   1  = "97531"

2) code1+code2 is base64 Decoded

3) The result will be of this form :
https://adf.ly/go.php?u=

4)  has to be further decoded and the result is to be returned
```

Your Task:

```
Make a function to Encode text to ysmm value.

and a function to Decode ysmm value.
```

Note: Take random values for The first 2 int values.
I personally hate trivial error checking like giving integer input in place of string. 

```
Input For Decoder & Encoder: Strings
Return "Invalid" for invalid Strings
```

Answer: {"fn_name": "adFly_decoder", "inputs": [["O=T0ZToPdRHJRmwdcOz1oGvTL22lFzkRZhih5GsbezSw9kndbvyR50wYawHIAF/SdhT1"], ["N=z0dDoMdyHIRmwac1zMolvWLz2RFmkMZiiZ5HsZeySw9kndbvyR50wYawHIAF/SdhT1"], ["lololol"]], "outputs": [["http://yahoo.com"], ["http://google.com"], ["Invalid"]]}
user: ## Description:

 Remove all exclamation marks from the end of words. Words are separated by spaces in the sentence.

### Examples

```
remove("Hi!") === "Hi"
remove("Hi!!!") === "Hi"
remove("!Hi") === "!Hi"
remove("!Hi!") === "!Hi"
remove("Hi! Hi!") === "Hi Hi"
remove("!!!Hi !!hi!!! !hi") === "!!!Hi !!hi !hi"
```

Answer: {"fn_name": "remove", "inputs": [["Hi!"], ["Hi!!!"], ["!Hi"], ["!Hi!"], ["Hi! Hi!"]], "outputs": [["Hi"], ["Hi"], ["!Hi"], ["!Hi"], ["Hi Hi"]]}
user: Bob has ladder. He wants to climb this ladder, but being a precocious child, he wonders about exactly how many ways he could to climb this `n` size ladder using jumps of up to distance `k`.

Consider this example...

n = 5\
k = 3

Here, Bob has ladder of length 5, and with each jump, he can ascend up to 3 steps (he can either jump step 1 or 2 or 3). This gives the below possibilities

```
1 1 1 1 1
1 1 1 2
1 1 2 1 
1 2 1 1
2 1 1 1
1 2 2
2 2 1
2 1 2
1 1 3
1 3 1
3 1 1
2 3
3 2
```

Your task to calculate number of ways to climb ladder of length `n` with upto `k` steps for Bob. (13 in above case)

Constraints:

```python
1<=n<=50
1<=k<=15
```

_Tip: try fibonacci._

Answer: {"fn_name": "count_ways", "inputs": [[1, 3], [3, 3], [2, 3], [5, 3], [4, 3], [10, 6], [14, 7]], "outputs": [[1], [4], [2], [13], [7], [492], [7936]]}
user: Implement a function, `multiples(m, n)`, which returns an array of the first `m` multiples of the real number `n`. Assume that `m` is a positive integer.

Ex.
```
multiples(3, 5.0)
```
should return
```
[5.0, 10.0, 15.0]
```

Answer: {"fn_name": "multiples", "inputs": [[3, 5], [1, 3.14], [5, -1]], "outputs": [[[5, 10, 15]], [[3.14]], [[-1, -2, -3, -4, -5]]]}