Muennighoff/xP3x-sample huggingface dataset:
inputs: I found an interesting problem on https://www.codewars.com/kata/59ca8e8e1a68b7de740001f4:
Given two arrays of strings, return the number of times each string of the second array appears in the first array.

#### Example

```python
array1 = ['abc', 'abc', 'xyz', 'cde', 'uvw']
array2 = ['abc', 'cde', 'uap']
```

How many times do the elements in `array2` appear in `array1`? 

* `'abc'` appears twice in the first array (2)
* `'cde'` appears only once (1)
* `'uap'` does not appear in the first array (0)

Therefore, `solve(array1, array2) = [2, 1, 0]`

Good luck!

If you like this Kata, please try:

[Word values](https://www.codewars.com/kata/598d91785d4ce3ec4f000018)

[Non-even substrings](https://www.codewars.com/kata/59da47fa27ee00a8b90000b4)

I tried it in Python, but could not do it. Can you solve it?
targets: def solve(first,second):
    return [first.count(word) for word in second]
inputs: Solve in Python:
Given a rational number n

``` n >= 0, with denominator strictly positive```  

- as a string (example: "2/3" in Ruby, Python, Clojure, JS, CS, Go) 
- or as two strings (example: "2" "3" in Haskell, Java, CSharp, C++, Swift) 
- or as a rational or decimal number (example: 3/4, 0.67 in R) 
- or two integers (Fortran)

decompose 
this number as a sum of rationals with numerators equal to one and without repetitions
(2/3 = 1/2 + 1/6 is correct but not 2/3 = 1/3 + 1/3, 1/3 is repeated).
    
The algorithm must be "greedy", so at each stage the new rational obtained in the decomposition must have a denominator as small as possible. 
In this manner the sum of a few fractions in the decomposition gives a rather good approximation of the rational to decompose.
    
2/3 = 1/3 + 1/3 doesn't fit because of the repetition but also because the first 1/3 has a denominator bigger than the one in 1/2 
in the decomposition 2/3 = 1/2 + 1/6.
    
### Example: 
(You can see other examples in "Sample Tests:")
```
decompose("21/23") or "21" "23" or 21/23 should return 

["1/2", "1/3", "1/13", "1/359", "1/644046"] in Ruby, Python, Clojure, JS, CS, Haskell, Go

"[1/2, 1/3, 1/13, 1/359, 1/644046]" in Java, CSharp, C++

"1/2,1/3,1/13,1/359,1/644046" in C, Swift, R
```  

### Notes
1) The decomposition of 21/23 as
```
21/23 = 1/2 + 1/3 + 1/13 + 1/598 + 1/897
```
is exact but don't fulfill our requirement because 598 is bigger than 359.
Same for 
```
21/23 = 1/2 + 1/3 + 1/23 + 1/46 + 1/69 (23 is bigger than 13)
or 
21/23 = 1/2 + 1/3 + 1/15 + 1/110 + 1/253 (15 is bigger than 13).
```
    
2) The rational given to decompose could be greater than one or equal to one, in which case the first "fraction" will be an integer 
(with an implicit denominator of 1). 

3) If the numerator parses to zero the function "decompose" returns [] (or "".

4) The number could also be a decimal which can be expressed as a rational. 

examples:

`0.6` in Ruby, Python, Clojure,JS, CS, Julia, Go
  
`"66" "100"` in Haskell, Java, CSharp, C++, C, Swift, Scala,...
targets: from fractions import Fraction
from math import ceil

def decompose(n):
    n = Fraction(n)
    lst, n = [int(n)], n - int(n)
    while n > 0:
        next_denom = int(ceil(1/n))
        lst.append(next_denom)
        n -= Fraction(1, next_denom)
    return ([] if lst[0] == 0 else [str(lst[0])]) + ["1/{}".format(d) for d in lst[1:]]
inputs: Solve in Python:
## Task

Given a positive integer as input, return the output as a string in the following format: 

Each element, corresponding to a digit of the number, multiplied by a power of 10 in such a way that with the sum of these elements you can obtain the original number.

## Examples

Input | Output
---   | ---
0     | ""
56    | "5\*10+6"
60    | "6\*10"
999   | "9\*100+9\*10+9"
10004 | "1\*10000+4"

Note: `input >= 0`
targets: def simplify(n):
    return "+".join([f"{x}*{10**i}" if i else x for i, x in enumerate(str(n)[::-1]) if x != "0"][::-1])
inputs: Solve in Python:
10^{10^{10}} participants, including Takahashi, competed in two programming contests.
In each contest, all participants had distinct ranks from first through 10^{10^{10}}-th.
The score of a participant is the product of his/her ranks in the two contests.
Process the following Q queries:
 - In the i-th query, you are given two positive integers A_i and B_i. Assuming that Takahashi was ranked A_i-th in the first contest and B_i-th in the second contest, find the maximum possible number of participants whose scores are smaller than Takahashi's.

-----Constraints-----
 - 1 \leq Q \leq 100
 - 1\leq A_i,B_i\leq 10^9(1\leq i\leq Q)
 - All values in input are integers.

-----Input-----
Input is given from Standard Input in the following format:
Q
A_1 B_1
:
A_Q B_Q

-----Output-----
For each query, print the maximum possible number of participants whose scores are smaller than Takahashi's.

-----Sample Input-----
8
1 4
10 5
3 3
4 11
8 9
22 40
8 36
314159265 358979323

-----Sample Output-----
1
12
4
11
14
57
31
671644785

Let us denote a participant who was ranked x-th in the first contest and y-th in the second contest as (x,y).
In the first query, (2,1) is a possible candidate of a participant whose score is smaller than Takahashi's. There are never two or more participants whose scores are smaller than Takahashi's, so we should print 1.
targets: n = int(input())
for i in range(n):
    a,b= list(map(int, input().split( )))

    #解説参照
    #何度見ても解けない
    #どちらかは高橋君より良い
    #a<=bとする
    #a=b (1~a-1)*2
    #a<b c**2<abとする
    #c(c+1)>=ab？2c-2 / 2c-1

    if a==b:
        print((2*a-2))
    else:
        c = int((a*b)**(1/2))
        if c*c==a*b:
            c -= 1
        if c*c+c>=a*b:
            print((2*c-2))
        else:
            print((2*c-1))
inputs: Solve in Python:
You are given a sequence of N$N$ powers of an integer k$k$; let's denote the i$i$-th of these powers by kAi$k^{A_i}$. You should partition this sequence into two non-empty contiguous subsequences; each element of the original sequence should appear in exactly one of these subsequences. In addition, the product of (the sum of elements of the left subsequence) and (the sum of elements of the right subsequence) should be maximum possible.
Find the smallest position at which you should split this sequence in such a way that this product is maximized.

-----Input-----
- The first line of the input contains a single integer T$T$ denoting the number of test cases. The description of T$T$ test cases follows.
- The first line of each test case contains two space-separated integers N$N$ and k$k$.
- The second line contains N$N$ space-separated integers A1,A2,…,AN$A_1, A_2, \dots, A_N$.

-----Output-----
For each test case, print a single line containing one integer — the size of the left subsequence. If there is more than one possible answer, print the smallest possible one.

-----Constraints-----
- 1≤T≤10$1 \le T \le 10$
- 2≤N≤105$2 \le N \le 10^5$
- 2≤k≤109$2 \le k \le 10^9$
- 0≤Ai≤105$0 \le A_i \le 10^5$ for each valid i$i$

-----Subtasks-----
Subtask #1 (30 points):
- 2≤N≤1,000$2 \le N \le 1,000$
- 0≤Ai≤1,000$0 \le A_i \le 1,000$ for each valid i$i$
Subtask #2 (70 points): original constraints

-----Example Input-----
1
5 2
1 1 3 3 5

-----Example Output-----
4

-----Explanation-----
Example case 1: The actual sequence of powers is [21,21,23,23,25]=[2,2,8,8,32]$[2^1, 2^1, 2^3, 2^3, 2^5] = [2, 2, 8, 8, 32]$. The maximum product is 20⋅32=640$20 \cdot 32 = 640$. In the optimal solution, the sequence is partitioned into [2,2,8,8]$[2, 2, 8, 8]$ and [32]$[32]$.
targets: # cook your dish here
for i in range(int(input())):
    n,k=[int(i) for i in input().split()]
    arr=[(k**int(i)) for i in input().split()]
    v1=sum(arr)
    v1=v1//2
    v2=0
    for i in range(len(arr)):
     v2+=arr[i]
     if(v2>=v1):
      break
    if(i==0):
     print(1)
    else:
     if(abs(v2-v1)<abs(v2-arr[i]-v1)):
      print(i+1)
     else:
      print(i)