Muennighoff/xP3x-sample huggingface dataset:
inputs: I found an interesting problem on https://leetcode.com/problems/maximum-width-of-binary-tree/:
Given a binary tree, write a function to get the maximum width of the given tree. The width of a tree is the maximum width among all levels. The binary tree has the same structure as a full binary tree, but some nodes are null. 

The width of one level is defined as the length between the end-nodes (the leftmost and right most non-null nodes in the level, where the null nodes between the end-nodes are also counted into the length calculation.

Example 1:

Input: 

           1
         /   \
        3     2
       / \     \  
      5   3     9 

Output: 4
Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).



Example 2:

Input: 

          1
         /  
        3    
       / \       
      5   3     

Output: 2
Explanation: The maximum width existing in the third level with the length 2 (5,3).



Example 3:

Input: 

          1
         / \
        3   2 
       /        
      5      

Output: 2
Explanation: The maximum width existing in the second level with the length 2 (3,2).


Example 4:

Input: 

          1
         / \
        3   2
       /     \  
      5       9 
     /         \
    6           7
Output: 8
Explanation:The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).




Note:
Answer will in the range of 32-bit signed integer.

I tried it in Python, but could not do it. Can you solve it?
targets: class Solution:
     
     
     def widthOfBinaryTree(self, root):
         """
         :type root: TreeNode
         :rtype: int
         """
         if not root: return 0
 
         maxlen=1
         start=1
         end=1
         tlen=1
         l=[[root,1]]
         
         while tlen:
             llen=tlen
             tlen=0
             while llen:
                 cur=l.pop(0)
                 llen-=1
                 if cur[0]:
                     l.append([cur[0].left,2*cur[1]-1])
                     l.append([cur[0].right,2*cur[1]])
                     tlen+=2
 
             if len(l)>0:
                 for item in reversed(l):
                     if item[0]:
                         end=item[1]
                         break
                 if end != item[1]: return maxlen
                 
                 for item in l:
                     if item[0]:
                         start=item[1]
                         break;
                 
                 end = end - start + 1
                 if(end > maxlen):  maxlen=end
                 
                 start=end=-1;
                 
         return maxlen
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/621/B:
Today, Wet Shark is given n bishops on a 1000 by 1000 grid. Both rows and columns of the grid are numbered from 1 to 1000. Rows are numbered from top to bottom, while columns are numbered from left to right.

Wet Shark thinks that two bishops attack each other if they share the same diagonal. Note, that this is the only criteria, so two bishops may attack each other (according to Wet Shark) even if there is another bishop located between them. Now Wet Shark wants to count the number of pairs of bishops that attack each other.


-----Input-----

The first line of the input contains n (1 ≤ n ≤ 200 000) — the number of bishops.

Each of next n lines contains two space separated integers x_{i} and y_{i} (1 ≤ x_{i}, y_{i} ≤ 1000) — the number of row and the number of column where i-th bishop is positioned. It's guaranteed that no two bishops share the same position.


-----Output-----

Output one integer — the number of pairs of bishops which attack each other. 


-----Examples-----
Input
5
1 1
1 5
3 3
5 1
5 5

Output
6

Input
3
1 1
2 3
3 5

Output
0



-----Note-----

In the first sample following pairs of bishops attack each other: (1, 3), (1, 5), (2, 3), (2, 4), (3, 4) and (3, 5). Pairs (1, 2), (1, 4), (2, 5) and (4, 5) do not attack each other because they do not share the same diagonal.

I tried it in Python, but could not do it. Can you solve it?
targets: def nC2(n):
    return n * (n - 1) // 2

SIZE = 1000

N = int(input())
a = [[0] * SIZE for i in range(SIZE)]
for i in range(N):
    x, y = map(int, input().split())
    a[x-1][y-1] = 1

ans = 0

#1
for sx in range(SIZE):
    t_cnt = 0
    
    x = sx; y = 0;
    while x >= 0:
        t_cnt += a[x][y]
        x -= 1; y += 1
        
    ans += nC2(t_cnt)

#2
for sy in range(1, SIZE):
    t_cnt = 0
    
    x = SIZE - 1; y = sy;
    while y < SIZE:
        t_cnt += a[x][y]
        x -= 1; y += 1
        
    ans += nC2(t_cnt)

#3
for sx in range(SIZE):
    t_cnt = 0
    
    x = sx; y = 0;
    while x < SIZE:
        t_cnt += a[x][y]
        x += 1; y += 1
        
    ans += nC2(t_cnt)

#4
for sy in range(1, SIZE):
    t_cnt = 0
    
    x = 0; y = sy;
    while y < SIZE:
        t_cnt += a[x][y]
        x += 1; y += 1
        
    ans += nC2(t_cnt)

print(ans)
inputs: Solve in Python:
Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm's runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

Example 1:


Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:


Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]
targets: class Solution:
     def searchRange(self, nums, target):
         """
         :type nums: List[int]
         :type target: int
         :rtype: List[int]
         """
         #m = int(len(nums)/2)
         #upper, lower = nums[:m], nums[m:]
         s, e = -1, -1
         l, u = 0, len(nums)-1
         if not nums or target > nums[u] or target < nums[l]:
             return [s, e]
         m = int((l+u)/2)
         while u >= l:
             if nums[m] > target:
                 if m == u:
                     break
                 u = m
                 #if int((l+u)/2) == u:
                 #    break
                 m = int((l+u)/2)
             elif nums[m] < target:
                 l = m
                 if int((l+u)/2) == l:
                     m = l+1
                 else:
                     m = int((l+u)/2)
             else:
                 s = e = m
                 while 0 < s and nums[s-1] == nums[s]:
                     s-=1
                 while e < len(nums)-1 and nums[e+1] == nums[e]:
                     e+=1
                 break
         return [s, e]
inputs: I found an interesting problem on https://www.codewars.com/kata/5ac69d572f317bdfc3000124:
In mathematics, a **pandigital number** is a number that in a given base has among its significant digits each digit used in the base at least once. For example, 1234567890 is a pandigital number in base 10.

For simplification, in this kata, we will consider pandigital numbers in *base 10* and with all digits used *exactly once*. The challenge is to calculate a sorted sequence of pandigital numbers, starting at a certain `offset` and with a specified `size`.

Example:
```python
  > get_sequence(0, 5)
  [1023456789, 1023456798, 1023456879, 1023456897, 1023456978]
```

Rules:
- We are looking for positive pandigital numbers in base 10.
- Each digit should occur `exactly once`.
- A pandigital number can't start with digit zero.
- The offset is an integer (negative, zero or positive number) (long in Java)
- The size is a positive integer number (int in Java)
- Return the `size` pandigital numbers which are not smaller than the `offset`. If there is not enough `size` pandigital numbers, just return all of them.
- Return an empty array if nothing is found.

I tried it in Python, but could not do it. Can you solve it?
targets: from itertools import permutations as perms

def is_pand(n):
    return len(set(list(str(n)))) == len(str(n))

def next_pan(n):
    while True:
        if is_pand(n):yield n
        n += 1

def get_sequence(n, k):
    if n < 1023456789: n = 1023456789
    elif n >= 9999999999: return []
    if not is_pand(n): n = next(next_pan(n))
    res = []
    for i in next_pan(n):
        res.append(i)
        if len(res)== k: break
    return res
inputs: Solve in Python:
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

Example 1:


Input: 1->2->3->3->4->4->5
Output: 1->2->5


Example 2:


Input: 1->1->1->2->3
Output: 2->3
targets: # Definition for singly-linked list.
 # class ListNode:
 #     def __init__(self, x):
 #         self.val = x
 #         self.next = None
 from collections import OrderedDict
 
 class Solution:
     def deleteDuplicates(self, head):
         """
         :type head: ListNode
         :rtype: ListNode
         """
         lookup = OrderedDict()
         
         while head:
             if head.val in lookup:
                 lookup[head.val] += 1
             else:
                 lookup[head.val] = 1
             head = head.next
         
         previous = None
         head = None
         for key in lookup:
             
             if (lookup[key] != 1):
                 continue
                 
             node = ListNode(key)
             if previous:
                 previous.next = node
             else:
                 head = node
             previous = node
 
         return(head)
inputs: Solve in Python:
Given an array A of positive lengths, return the largest perimeter of a triangle with non-zero area, formed from 3 of these lengths.
If it is impossible to form any triangle of non-zero area, return 0.
 



Example 1:
Input: [2,1,2]
Output: 5


Example 2:
Input: [1,2,1]
Output: 0


Example 3:
Input: [3,2,3,4]
Output: 10


Example 4:
Input: [3,6,2,3]
Output: 8

 
Note:

3 <= A.length <= 10000
1 <= A[i] <= 10^6
targets: class Solution:
    def largestPerimeter(self, A: List[int]) -> int:
        if len(A)<3:
            return 0
        
        A.sort(reverse=True)
        
        while len(A)>2:
            max_num = A.pop(0)
            if max_num >= A[0] + A[1]:
                continue
            else:
                return max_num+A[0]+A[1]
        return 0
inputs: I found an interesting problem on https://atcoder.jp/contests/abc086/tasks/abc086_a:
AtCoDeer the deer found two positive integers, a and b.
Determine whether the product of a and b is even or odd.

-----Constraints-----
 - 1 ≤ a,b ≤ 10000
 - a and b are integers.

-----Input-----
Input is given from Standard Input in the following format:
a b

-----Output-----
If the product is odd, print Odd; if it is even, print Even.

-----Sample Input-----
3 4

-----Sample Output-----
Even

As 3 × 4 = 12 is even, print Even.

I tried it in Python, but could not do it. Can you solve it?
targets: a,b=list(map(int,input().split()))

if(a*b)%2==0:
  print('Even')
else:
  print('Odd')
inputs: Solve in Python:
An array is called `centered-N` if some `consecutive sequence` of elements of the array sum to `N` and this sequence is preceded and followed by the same number of elements. 

Example:
```
[3,2,10,4,1,6,9] is centered-15
because the sequence 10,4,1 sums to 15 and the sequence 
is preceded by two elements [3,2] and followed by two elements [6,9]

```

Write a method called `isCenteredN` that returns :

- `true` if its array argument is `not empty` and `centered-N` or empty and centered-0
- otherwise returns `false`.
targets: def is_centered(arr,n):
    for i in range(len(arr)):
        for j in range(len(arr)+1):
            if j>i:
                if sum(arr[i:j]) == n and i == len(arr)-j:
                    return True
    return True if n==0 and arr!=arr[::-1] else False
inputs: I found an interesting problem on https://www.codewars.com/kata/5897cdc26551af891c000124:
Hofstadter sequences are a family of related integer sequences, among which the first ones were described by an American professor Douglas Hofstadter in his book Gödel, Escher, Bach. 

### Task
Today we will be implementing the rather chaotic recursive sequence of integers called Hofstadter Q.
The Hofstadter Q is defined as:

As the author states in the aforementioned book:It is reminiscent of the Fibonacci definition in that each new value is a sum of two
previous values-but not of the immediately previous two values. Instead, the two
immediately previous values tell how far to count back to obtain the numbers to be added
to make the new value.
The function produces the starting sequence:  
`1, 1, 2, 3, 3, 4, 5, 5, 6 . . .`  
Test info: 100 random tests, n is always positive
Good luck!

I tried it in Python, but could not do it. Can you solve it?
targets: import sys; sys.setrecursionlimit(10000)
from functools import lru_cache

@lru_cache(maxsize=None)
def Q(n):
    if n <= 2:
        return 1
    return Q(n - Q(n-1)) + Q(n - Q(n-2))
    
def hofstadter_Q(n):
    return Q(n)
inputs: I found an interesting problem on https://www.codechef.com/problems/PRFYIT:
We say that a binary string (a string containing only characters '0' and '1') is pure if it does not contain either of the strings "0101" or "1010" as a subsequence.
Recall that string T is a subsequence of string S if we can delete some of the letters of S (possibly none) such that the resulting string will become T.
You are given a binary string $S$ with length $N$. We want to make this string pure by deleting some (possibly zero) characters from it. What is the minimum number of characters we have to delete?

-----Input-----
- The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
- The first and only line of each test case contains a single string $S$ with length $N$.

-----Output-----
For each test case, print a single line containing one integer — the minimum number of characters we have to delete from $S$.

-----Constraints-----
- $1 \le T \le 40$
- $1 \le N \le 1,000$
- $S$ contains only characters '0' and '1'

-----Example Input-----
4
010111101
1011100001011101
0110
111111

-----Example Output-----
2
3
0
0

-----Explanation-----
Example case 1: We can delete the first and third character of our string. There is no way to make the string pure by deleting only one character.
Example case 3: The given string is already pure, so the answer is zero.

I tried it in Python, but could not do it. Can you solve it?
targets: for _ in range(int(input())):
 bi = input().strip()
 dp = [0 if i < 2 else len(bi) for i in range(6)]
 for c in bi:
  if c == '1':
   dp[3] = min(dp[3], dp[0])
   dp[0] += 1
   dp[5] = min(dp[5], dp[2])
   dp[2] += 1
   dp[4] += 1
  else:
   dp[2] = min(dp[2], dp[1])
   dp[1] += 1
   dp[4] = min(dp[4], dp[3])
   dp[3] += 1
   dp[5] += 1
 print(min(dp))
inputs: Solve in Python:
We have A cards, each of which has an integer 1 written on it. Similarly, we also have B cards with 0s and C cards with -1s.
We will pick up K among these cards. What is the maximum possible sum of the numbers written on the cards chosen?

-----Constraints-----
 - All values in input are integers.
 - 0 \leq A, B, C
 - 1 \leq K \leq A + B + C \leq 2 \times 10^9

-----Input-----
Input is given from Standard Input in the following format:
A B C K

-----Output-----
Print the maximum possible sum of the numbers written on the cards chosen.

-----Sample Input-----
2 1 1 3

-----Sample Output-----
2

Consider picking up two cards with 1s and one card with a 0.
In this case, the sum of the numbers written on the cards is 2, which is the maximum possible value.
targets: A, B, C, K = list(map(int, input().split()))

a, c = min(A, K), max(0, min(C, K - A - B))
print((a - c))
inputs: Solve in Python:
n passengers board an airplane with exactly n seats. The first passenger has lost the ticket and picks a seat randomly. But after that, the rest of passengers will:

Take their own seat if it is still available, 
Pick other seats randomly when they find their seat occupied 

What is the probability that the n-th person can get his own seat?
 
Example 1:
Input: n = 1
Output: 1.00000
Explanation: The first person can only get the first seat.
Example 2:
Input: n = 2
Output: 0.50000
Explanation: The second person has a probability of 0.5 to get the second seat (when first person gets the first seat).

 
Constraints:

1 <= n <= 10^5
targets: class Solution:
    def nthPersonGetsNthSeat(self, n: int) -> float:
        # if n == 1:
        #     return 1.0
        # else:
        #     return 0.5
        
        if n == 1:
            return 1.0
        sum_results = 0.0
        for i in range(2, n+1):
            p = 1/i * (1 + sum_results)
            sum_results += p
        return p
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/276/C:
The little girl loves the problems on array queries very much.

One day she came across a rather well-known problem: you've got an array of n elements (the elements of the array are indexed starting from 1); also, there are q queries, each one is defined by a pair of integers l_{i}, r_{i} (1 ≤ l_{i} ≤ r_{i} ≤ n). You need to find for each query the sum of elements of the array with indexes from l_{i} to r_{i}, inclusive.

The little girl found the problem rather boring. She decided to reorder the array elements before replying to the queries in a way that makes the sum of query replies maximum possible. Your task is to find the value of this maximum sum.


-----Input-----

The first line contains two space-separated integers n (1 ≤ n ≤ 2·10^5) and q (1 ≤ q ≤ 2·10^5) — the number of elements in the array and the number of queries, correspondingly.

The next line contains n space-separated integers a_{i} (1 ≤ a_{i} ≤ 2·10^5) — the array elements.

Each of the following q lines contains two space-separated integers l_{i} and r_{i} (1 ≤ l_{i} ≤ r_{i} ≤ n) — the i-th query.


-----Output-----

In a single line print a single integer — the maximum sum of query replies after the array elements are reordered.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.


-----Examples-----
Input
3 3
5 3 2
1 2
2 3
1 3

Output
25

Input
5 3
5 2 4 1 3
1 5
2 3
2 3

Output
33

I tried it in Python, but could not do it. Can you solve it?
targets: n, q = map(int, input().split())
t = [0] + list(map(int, input().split()))
p = [0] * (n + 1)

for i in range(q):
    l, r = map(int, input().split())
    p[l - 1] += 1
    p[r] -= 1

for i in range(1, n + 1):
    p[i] += p[i - 1]

t.sort()
p.sort()

j = 0
while p[j] == 0: j += 1

print(sum(p[i] * t[i] for i in range(j, n + 1)))
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/448/D:
Bizon the Champion isn't just charming, he also is very smart.

While some of us were learning the multiplication table, Bizon the Champion had fun in his own manner. Bizon the Champion painted an n × m multiplication table, where the element on the intersection of the i-th row and j-th column equals i·j (the rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table is the k-th largest number? Bizon the Champion always answered correctly and immediately. Can you repeat his success?

Consider the given multiplication table. If you write out all n·m numbers from the table in the non-decreasing order, then the k-th number you write out is called the k-th largest number.


-----Input-----

The single line contains integers n, m and k (1 ≤ n, m ≤ 5·10^5; 1 ≤ k ≤ n·m).


-----Output-----

Print the k-th largest number in a n × m multiplication table.


-----Examples-----
Input
2 2 2

Output
2

Input
2 3 4

Output
3

Input
1 10 5

Output
5



-----Note-----

A 2 × 3 multiplication table looks like this:

1 2 3

2 4 6

I tried it in Python, but could not do it. Can you solve it?
targets: from sys import stdin

n, m, k = [int(x) for x in stdin.readline().split()]
be, en = 1, k + 1

while be < en:
    mid = (be + en + 1) >> 1
    be1, cur = (mid + m - 1) // m, 0
    for i in range(1, be1):
        cur += m

    for i in range(be1, n + 1):
        cur += (mid - 1) // i

    if cur <= k - 1:
        be = mid
    else:
        en = mid - 1

print(be)
inputs: Solve in Python:
Binod and his family live in Codingland. They have a festival called N-Halloween.
The festival is celebrated for N  consecutive days. On each day Binod gets some candies from his mother. He may or may not take them. On a given day ,  Binod will be sad if he collected candies on that day and he does not have at least X   candies remaining  from the candies he collected that day at the end of that day . His friend Pelu who is very fond of candies asks for X  candies from Binod on a day if Binod collects any candy on that day. He does so immediately after Binod collects them. Binod being a good friend never denies him.  

Given a list a   where ai   denotes the number of candies Binod may take on the ith   day and  Q  queries having a number X , find the maximum number of candies Binod can collect so that he won't be sad after the festival ends.
Input format: 

First-line contains two integers N and Q denoting the number of festival days and number of queries respectively.

The second line is the array whose ith   element is the maximum number of candies Binod may
collect on that day.

Each of the next Q lines contains the number X i.e the minimum number of candies Binod wants to have at the end of every day on which he collects any candy.
Output format: 

For every query output, a single integer denoting the max number of candies Binod may
collect (for the given X ) to be happy after the festival ends.
Constraints:  

1<=N,Q<=105  

1<=Q<=105  

0<=ai<=109  

0<=X<=109  

Sample Input : 

5 2

4 6 5 8 7

1

2

Sample Output : 

30

30

Sample Input:  

6 3 

20 10 12 3 30 5 

2 

6 

13 

Sample Output  

77 

62 

30 

Explanation: 

In the first query of sample input 1, Binod can collect the maximum number of given chocolates for a given day each day as after the end of every day Binod will have a number of collected
chocolates on that day greater than equal to 1 even after giving 1 chocolate to Pelu. So the
the answer is 4+6+5+8+7=30.
targets: # cook your dish here
import copy
import bisect
n,q=list(map(int,input().split()))

a=list(map(int,input().split()))

a.sort()
b=copy.copy(a)
for i in range(1,len(b)):
 b[i]+=b[i-1]
##print(b)
for i in range(q):
 x=int(input())
 ans=bisect.bisect_left(a,x*2)
 if ans==0:
  ans1=b[n-1]
 else:
  ans1=b[n-1]-b[ans-1]
 print(ans1)
inputs: I found an interesting problem on https://www.codewars.com/kata/56445cc2e5747d513c000033:
In Russia, there is an army-purposed station named UVB-76 or "Buzzer" (see also https://en.wikipedia.org/wiki/UVB-76). Most of time specific "buzz" noise is being broadcasted, but on very rare occasions, the buzzer signal is interrupted and a voice transmission in Russian takes place. Transmitted messages have always the same format like this:

 MDZHB 01 213 SKIF 38 87 23 95 

or: 

 MDZHB 80 516 GANOMATIT 21 23 86 25 

Message format consists of following parts:

 Initial keyword "MDZHB"; 
 Two groups of digits, 2 digits in first and 3 in second ones; 
 Some keyword of arbitrary length consisting only of uppercase letters; 
 Final 4 groups of digits with 2 digits in each group. 

  
Your task is to write a function that can validate the correct UVB-76 message. Function should return "True" if message is in correct format and "False" otherwise.

I tried it in Python, but could not do it. Can you solve it?
targets: import re
def validate(message):
  return bool(re.fullmatch(r'MDZHB \d{2} \d{3} [A-Z]+ \d{2} \d{2} \d{2} \d{2}', message))
inputs: I found an interesting problem on https://www.codewars.com/kata/5899f1df27926b7d000000eb:
The Tower of Hanoi problem involves 3 towers.  A number of rings decreasing in size are placed on one tower.  All rings must then be moved to another tower, but at no point can a larger ring be placed on a smaller ring.

Your task:  Given a number of rings, return the MINIMUM number of moves needed to move all the rings from one tower to another.

Reference: Tower of Hanoi, Courtesy of Coolmath Games

NB:  This problem may seem very complex, but in reality there is an amazingly simple formula to calculate the minimum number.  Just Learn how to solve the problem via the above link (if you are not familiar with it), and then think hard.  Your solution should be in no way extraordinarily long and complex.  The Kata ranking is for figuring out the solution, but the coding skills required are minimal.

I tried it in Python, but could not do it. Can you solve it?
targets: def tower_of_hanoi(rings):
    moves = 0
    for r in range(rings):
        moves *= 2
        moves += 1
    return moves
inputs: I found an interesting problem on https://www.codewars.com/kata/596c26187bd547f3a6000050:
# Task
A newspaper is published in Walrusland. Its heading is `s1` , it consists of lowercase Latin letters. 

Fangy the little walrus wants to buy several such newspapers, cut out their headings, glue them one to another in order to get one big string. 

After that walrus erase several letters from this string in order to get a new word `s2`. 

It is considered that when Fangy erases some letter, there's no whitespace formed instead of the letter. That is, the string remains unbroken and it still only consists of lowercase Latin letters.

For example, the heading is `"abc"`. If we take two such headings and glue them one to the other one, we get `"abcabc"`. If we erase the 1st letter("a") and 5th letter("b"), we get a word `"bcac"`.

Given two string `s1` and `s2`, return the least number of newspaper headings `s1`, which Fangy will need to receive the word `s2`. If it is impossible to get the word `s2` in the above-described manner, return `-1`.

# Example

For `s1="abc", s2="bcac"`, the output should be `2`.

```
"abcabc" --> "bcac"
 x   x
```

For `s1="abc", s2="xyz"`, the output should be `-1`.

It's impossible to get the word `s2`.

I tried it in Python, but could not do it. Can you solve it?
targets: import re
def buy_newspaper(s1, s2):
    if not set(s2) <= set(s1):
        return -1
    regex = ''.join(c + '?' for c in s1)
    return len([match for match in re.findall(regex, s2) if match])
inputs: Solve in Python:
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._
targets: def count_ways(n, k):
    steps = [1] * (n + 1)
    for i in range(1, n + 1):
        if i <= k:
            steps[i] = sum(steps[j] for j in range(0, i))
        else:
            steps[i] = sum(steps[i - j] for j in range(1, k + 1))
    return steps[-1]
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/485/A:
One industrial factory is reforming working plan. The director suggested to set a mythical detail production norm. If at the beginning of the day there were x details in the factory storage, then by the end of the day the factory has to produce $x \operatorname{mod} m$ (remainder after dividing x by m) more details. Unfortunately, no customer has ever bought any mythical detail, so all the details produced stay on the factory. 

The board of directors are worried that the production by the given plan may eventually stop (that means that there will be а moment when the current number of details on the factory is divisible by m). 

Given the number of details a on the first day and number m check if the production stops at some moment.


-----Input-----

The first line contains two integers a and m (1 ≤ a, m ≤ 10^5).


-----Output-----

Print "Yes" (without quotes) if the production will eventually stop, otherwise print "No".


-----Examples-----
Input
1 5

Output
No

Input
3 6

Output
Yes

I tried it in Python, but could not do it. Can you solve it?
targets: a,m = map(int,input().split())

for i in range(100000):
    if a%m==0:
        print("Yes")
        quit()
    else:
        a+=a%m
print("No")
inputs: Solve in Python:
We have a board with a 2 \times N grid.
Snuke covered the board with N dominoes without overlaps.
Here, a domino can cover a 1 \times 2 or 2 \times 1 square.
Then, Snuke decided to paint these dominoes using three colors: red, cyan and green.
Two dominoes that are adjacent by side should be painted by different colors.
Here, it is not always necessary to use all three colors.
Find the number of such ways to paint the dominoes, modulo 1000000007.
The arrangement of the dominoes is given to you as two strings S_1 and S_2 in the following manner:
 - Each domino is represented by a different English letter (lowercase or uppercase).
 - The j-th character in S_i represents the domino that occupies the square at the i-th row from the top and j-th column from the left.

-----Constraints-----
 - 1 \leq N \leq 52
 - |S_1| = |S_2| = N
 - S_1 and S_2 consist of lowercase and uppercase English letters.
 - S_1 and S_2 represent a valid arrangement of dominoes.

-----Input-----
Input is given from Standard Input in the following format:
N
S_1
S_2

-----Output-----
Print the number of such ways to paint the dominoes, modulo 1000000007.

-----Sample Input-----
3
aab
ccb

-----Sample Output-----
6

There are six ways as shown below:
targets: #template
def inputlist(): return [int(j) for j in input().split()]
#template

mod = 10**9+7

N = int(input())

if N == 1:
    print((3))
    return

S1 = list(input())
S2 = list(input())

blocks = []
s = S1[0]
for i in range(1,N):
    if S1[i] == S1[i-1]:
        s += S1[i]
    else:
        blocks.append(s)
        s = S1[i]
    if i == N-1:
        blocks.append(s)

n = len(blocks)

last_blocks = [0]*n

for i in range(1,n):
    last_blocks[i] = len(blocks[i-1])

ans = 1
for i in range(n):
    tmp = blocks[i]
    last_block = last_blocks[i]
    if last_block == 0:
        if len(tmp) == 1:
            ans *= 3
        else:
            ans*=6
    if last_block == 1:
        ans *= 2
    if last_block == 2:
        if len(tmp) == 1:
            ans*=1
        if len(tmp) == 2:
            ans*=3
    ans %= mod

print((ans%mod))
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/509/B:
There are n piles of pebbles on the table, the i-th pile contains a_{i} pebbles. Your task is to paint each pebble using one of the k given colors so that for each color c and any two piles i and j the difference between the number of pebbles of color c in pile i and number of pebbles of color c in pile j is at most one.

In other words, let's say that b_{i}, c is the number of pebbles of color c in the i-th pile. Then for any 1 ≤ c ≤ k, 1 ≤ i, j ≤ n the following condition must be satisfied |b_{i}, c - b_{j}, c| ≤ 1. It isn't necessary to use all k colors: if color c hasn't been used in pile i, then b_{i}, c is considered to be zero.


-----Input-----

The first line of the input contains positive integers n and k (1 ≤ n, k ≤ 100), separated by a space — the number of piles and the number of colors respectively.

The second line contains n positive integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ 100) denoting number of pebbles in each of the piles.


-----Output-----

If there is no way to paint the pebbles satisfying the given condition, output "NO" (without quotes) .

Otherwise in the first line output "YES" (without quotes). Then n lines should follow, the i-th of them should contain a_{i} space-separated integers. j-th (1 ≤ j ≤ a_{i}) of these integers should be equal to the color of the j-th pebble in the i-th pile. If there are several possible answers, you may output any of them.


-----Examples-----
Input
4 4
1 2 3 4

Output
YES
1
1 4
1 2 4
1 2 3 4

Input
5 2
3 2 4 1 3

Output
NO

Input
5 4
3 2 4 3 5

Output
YES
1 2 3
1 3
1 2 3 4
1 3 4
1 1 2 3 4

I tried it in Python, but could not do it. Can you solve it?
targets: n, colors = map(int, input().split())
lens = list(map(int, input().split()))
q = [[] for i in range(n)]

def end():
    for i in range(n):
        if lens[i] != len(q[i]):
            return False
    return True

mi = min(lens)
for i in range(n):
    while len(q[i]) < min(lens[i], 1 + mi):
        q[i].append(1)

k = 1
while not end():
    k += 1
    for i in range(n):
        if len(q[i]) < lens[i]:
            q[i].append(k)

ma = max(map(max, q))
if ma <= colors:
    print('YES')
    for w in q:
        for i in w:
            print(i, end=' ')
        print()
else:
    print('NO')
inputs: Solve in Python:
For every string, after every occurrence of `'and'` and `'but'`, insert the substring `'apparently'` directly after the occurrence.

If input does not contain 'and' or 'but', return the original string. If a blank string, return `''`.

If substring `'apparently'` is already directly after an `'and'` and/or `'but'`, do not add another. (Do not add duplicates).


# Examples:

Input 1 

    'It was great and I've never been on live television before but sometimes I don't watch this.'

Output 1

    'It was great and apparently I've never been on live television before but apparently sometimes I don't watch this.'

Input 2

    'but apparently'

Output 2

    'but apparently' 
(no changes because `'apparently'` is already directly after `'but'` and there should not be a duplicate.)

An occurrence of `'and'` and/or `'but'` only counts when it is at least one space separated. For example `'andd'` and `'bbut'` do not count as occurrences, whereas `'b but'` and `'and d'` does count.

reference that may help:
https://www.youtube.com/watch?v=rz5TGN7eUcM
targets: def apparently(string):
    a, r = string.split(), []
    for i in range(len(a)):
        if a[i] in ('and', 'but') and i == len(a)-1: r.append(a[i]+' apparently')
        elif a[i] in ('and', 'but') and a[i+1] != 'apparently': r.append(a[i]+' apparently')
        else: r.append(a[i])
    return ' '.join(r)
inputs: I found an interesting problem on https://www.codewars.com/kata/5ae43ed6252e666a6b0000a4:
Welcome

This kata is inspired by This Kata


We have a string s

We have a number n

Here is a function that takes your string, concatenates the even-indexed chars to the front, odd-indexed chars to the back.

Examples

    s = "Wow Example!"
    result = "WwEapeo xml!"
    s = "I'm JomoPipi"
    result = "ImJm ii' ooPp"
    
The Task:

return the result of the string after applying the function to it n times.

example where s = "qwertyuio" and n = 2:

    after 1 iteration  s = "qetuowryi"
    after 2 iterations s = "qtorieuwy"
    return "qtorieuwy"

 Note 

there's a lot of tests, big strings,
and n is greater than a billion

so be ready to optimize.

after doing this: do it's best friend!

# Check out my other kata!

        
Matrix Diagonal Sort OMG
String -> N iterations -> String
String -> X iterations -> String
ANTISTRING
Array - squareUp b!
Matrix - squareUp b!
Infinitely Nested Radical Expressions
pipi Numbers!

I tried it in Python, but could not do it. Can you solve it?
targets: def jumbled_string(s, n):
    idx  = list(range(0,len(s),2)) + list(range(1,len(s),2))
    lst = []
    while not lst or s != lst[0]:
        lst.append(s)
        s = ''.join(s[i] for i in idx)
        if len(lst) == n: return s
    return lst[ n%len(lst) ]
inputs: Solve in Python:
=====Function Descriptions=====
Basic mathematical functions operate element-wise on arrays. They are available both as operator overloads and as functions in the NumPy module.

import numpy

a = numpy.array([1,2,3,4], float)
b = numpy.array([5,6,7,8], float)

print a + b                     #[  6.   8.  10.  12.]
print numpy.add(a, b)           #[  6.   8.  10.  12.]

print a - b                     #[-4. -4. -4. -4.]
print numpy.subtract(a, b)      #[-4. -4. -4. -4.]

print a * b                     #[  5.  12.  21.  32.]
print numpy.multiply(a, b)      #[  5.  12.  21.  32.]

print a / b                     #[ 0.2         0.33333333  0.42857143  0.5       ]
print numpy.divide(a, b)        #[ 0.2         0.33333333  0.42857143  0.5       ]

print a % b                     #[ 1.  2.  3.  4.]
print numpy.mod(a, b)           #[ 1.  2.  3.  4.]

print a**b                      #[  1.00000000e+00   6.40000000e+01   2.18700000e+03   6.55360000e+04]
print numpy.power(a, b)         #[  1.00000000e+00   6.40000000e+01   2.18700000e+03   6.55360000e+04]

=====Problem Statement=====
You are given two integer arrays, A and B of dimensions NXM.
Your task is to perform the following operations:
1. Add (A + B)
2. Subtract (A - B)
3. Multiply (A * B)
4. Integer Division (A / B)
5. Mod (A % B)
6. Power (A ** B)

=====Input Format=====
The first line contains two space separated integers, N and M.
The next N lines contains M space separated integers of array A.
The following N lines contains M space separated integers of array B.

=====Output Format=====
Print the result of each operation in the given order under Task.
targets: import numpy
n,m = list(map(int,input().split()))
ar1 = []
ar2 = []
for i in range(n):
    tmp = list(map(int,input().split()))
    ar1.append(tmp)
for i in range(n):
    tmp = list(map(int,input().split()))
    ar2.append(tmp)
np_ar1 = numpy.array(ar1)
np_ar2 = numpy.array(ar2)
print((np_ar1 + np_ar2))
print((np_ar1 - np_ar2))
print((np_ar1 * np_ar2))
print((np_ar1 // np_ar2))
print((np_ar1 % np_ar2))
print((np_ar1 ** np_ar2))
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/691/C:
You are given a positive decimal number x.

Your task is to convert it to the "simple exponential notation".

Let x = a·10^{b}, where 1 ≤ a < 10, then in general case the "simple exponential notation" looks like "aEb". If b equals to zero, the part "Eb" should be skipped. If a is an integer, it should be written without decimal point. Also there should not be extra zeroes in a and b.


-----Input-----

The only line contains the positive decimal number x. The length of the line will not exceed 10^6. Note that you are given too large number, so you can't use standard built-in data types "float", "double" and other.


-----Output-----

Print the only line — the "simple exponential notation" of the given number x.


-----Examples-----
Input
16

Output
1.6E1

Input
01.23400

Output
1.234

Input
.100

Output
1E-1

Input
100.

Output
1E2

I tried it in Python, but could not do it. Can you solve it?
targets: string_list = input().split(sep='.')
integer = string_list[0].lstrip('0')
if len(string_list) == 1:
    fractional = ''
else:
    fractional = string_list[1].rstrip('0')
if integer:
    if fractional:
        a = "{}.{}{}".format(integer[0], integer[1:], fractional)
    else:
        a = "{}.{}".format(integer[0], integer[1:].rstrip('0'))
    b = len(integer) - 1
    if a[-1] == '.':
        a = a[:-1]
    if b > 0:
        print("E".join((a, str(b))))
    else:
        print(a)
else:
    old_len = len(fractional)
    fractional = fractional.lstrip('0')
    zero_count = old_len - len(fractional)
    a = "{}.{}".format(fractional[0], fractional[1:])
    b = -zero_count - 1
    if a[-1] == '.':
        a = a[:-1]
    print("E".join((a, str(b))))
inputs: I found an interesting problem on https://www.codechef.com/LTIME40/problems/LTM40CD:
Chef likes problems on geometry a lot. Please help him to solve one such problem.
Find all possible triangles with integer sides which has the radius of inscribed circle (also known as incircle) equal to R.
Two triangles are said to be different if they have at least one different side lengths. Formally, let there be two triangles T1, T2. Let a, b, c denote the sides of triangle T1, such that a ≤ b ≤ c. Similarly, Let d, e, f denote the sides of triangle T2, such that d ≤ e ≤ f. Then T1 will said to be different from T2 if either a ≠ d, or b ≠ e or c ≠ f.

-----Input-----
There is a single test case per test file.
The only line of input contains an integer R.

-----Output-----
Output in first line single number - number of triangles satisfying statement.
Order the sides of triangles in non-decreasing order. Output all triangles in non-decreasing order, i.e. order first by smallest sides, otherwise by second smallest sides, if first and second sides equal, then by third.

-----Constraints-----
- 1 ≤ R ≤ 100 

-----Subtasks-----
- Subtask #1: (20 points)  1 ≤ R ≤ 3
- Subtask #2: (30 points)  1 ≤ R ≤ 20
- Subtask #3: (50 points)  Original constraints

-----Example-----
Input:2

Output:5
5 12 13
6 8 10
6 25 29
7 15 20
9 10 17

I tried it in Python, but could not do it. Can you solve it?
targets: from math import sqrt
r = int(input())
cnt=0
rt=[]
for i in range(1,16*r):
 for j in range(i, 460):
  for k in range(j+1, j+i):
   #print i,j,k
   s = float((i+j+k))/2
   #print s,i,j,k,s*(s-i)*(s-j)*(s-k)
   area = sqrt(abs(s*(s-i)*(s-j)*(s-k)))
   #print(float(2*area))/(i+j+k)
   if (r==(float(2*area))/(i+j+k)):
    cnt+=1
    #print i,j,k,area
    rt.append([i,j,k])
    
print(cnt)

for i in rt:
 for j in i:
  print(j, end=' ')
 print()
inputs: Solve in Python:
Write a function that when given a number >= 0, returns an Array of ascending length subarrays.

```
pyramid(0) => [ ]
pyramid(1) => [ [1] ]
pyramid(2) => [ [1], [1, 1] ]
pyramid(3) => [ [1], [1, 1], [1, 1, 1] ]
```

**Note:** the subarrays should be filled with `1`s
targets: def pyramid(n):
    result = []
    for i in range (1, n + 1):
        result.append(i * [1])
    return result
inputs: Solve in Python:
Inna and Dima decided to surprise Sereja. They brought a really huge candy matrix, it's big even for Sereja! Let's number the rows of the giant matrix from 1 to n from top to bottom and the columns — from 1 to m, from left to right. We'll represent the cell on the intersection of the i-th row and j-th column as (i, j). Just as is expected, some cells of the giant candy matrix contain candies. Overall the matrix has p candies: the k-th candy is at cell (x_{k}, y_{k}).

The time moved closer to dinner and Inna was already going to eat p of her favourite sweets from the matrix, when suddenly Sereja (for the reason he didn't share with anyone) rotated the matrix x times clockwise by 90 degrees. Then he performed the horizontal rotate of the matrix y times. And then he rotated the matrix z times counterclockwise by 90 degrees. The figure below shows how the rotates of the matrix looks like. [Image] 

Inna got really upset, but Duma suddenly understood two things: the candies didn't get damaged and he remembered which cells contained Inna's favourite sweets before Sereja's strange actions. Help guys to find the new coordinates in the candy matrix after the transformation Sereja made!


-----Input-----

The first line of the input contains fix integers n, m, x, y, z, p (1 ≤ n, m ≤ 10^9; 0 ≤ x, y, z ≤ 10^9; 1 ≤ p ≤ 10^5).

Each of the following p lines contains two integers x_{k}, y_{k} (1 ≤ x_{k} ≤ n; 1 ≤ y_{k} ≤ m) — the initial coordinates of the k-th candy. Two candies can lie on the same cell.


-----Output-----

For each of the p candies, print on a single line its space-separated new coordinates.


-----Examples-----
Input
3 3 3 1 1 9
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
3 3

Output
1 3
1 2
1 1
2 3
2 2
2 1
3 3
3 2
3 1



-----Note-----

Just for clarity. Horizontal rotating is like a mirroring of the matrix. For matrix:

QWER      REWQ 

ASDF  ->  FDSA

ZXCV      VCXZ
targets: n, m, x, y, z, p = list(map(int, input().split()))
n, m, x, y, z = n + 1, m + 1, x % 4, y % 2, (4 - z) % 4

def a(i, j, n, m, k):
    if k == 0: return i, j, n, m
    if k == 1: return j, n - i, m, n
    if k == 2: return n - i, m - j, n, m
    return m - j, i, m, n

def b(i, j, m, k):
    if k == 0: return i, j
    if k == 1: return i, m - j

for i in range(p):
    u, v = list(map(int, input().split()))
    u, v, q, p = a(u, v, n, m, x)
    u, v = b(u, v, p, y)
    u, v, q, p = a(u, v, q, p, z)
    print(u, v)
inputs: Solve in Python:
Leonardo Fibonacci in a rare portrait of his younger days

I assume you are all familiar with the famous Fibonacci sequence, having to get each number as the sum of the previous two (and typically starting with either `[0,1]` or `[1,1]` as the first numbers).

While there are plenty of variation on it ([including](https://www.codewars.com/kata/tribonacci-sequence) [a few](https://www.codewars.com/kata/fibonacci-tribonacci-and-friends) [I wrote](https://www.codewars.com/kata/triple-shiftian-numbers/)), usually the catch is all the same: get a starting (signature) list of numbers, then proceed creating more with the given rules.

What if you were to get to get two parameters, one with the signature (starting value) and the other with the number you need to sum at each iteration to obtain the next one?

And there you have it, getting 3 parameters:

* a signature of length `length`
* a second parameter is a list/array of indexes of the last `length` elements you need to use to obtain the next item in the sequence (consider you can end up not using a few or summing the same number multiple times)' in other words, if you get a signature of length `5` and `[1,4,2]` as indexes, at each iteration you generate the next number by summing the 2nd, 5th and 3rd element (the ones with indexes `[1,4,2]`) of the last 5 numbers
* a third and final parameter is of course which sequence element you need to return (starting from zero, I don't want to bother you with adding/removing 1s or just coping with the fact that after years on CodeWars we all count as computers do):

```python
custom_fib([1,1],[0,1],2) == 2 #classical fibonacci!
custom_fib([1,1],[0,1],3) == 3 #classical fibonacci!
custom_fib([1,1],[0,1],4) == 5 #classical fibonacci!
custom_fib([3,5,2],[0,1,2],4) == 17 #similar to my Tribonacci
custom_fib([7,3,4,1],[1,1],6) == 2 #can you figure out how it worked ;)?
```
targets: from collections import deque

def custom_fib(signature,indexes,n):
    s = deque(signature, maxlen=len(signature))
    for i in range(n):
        s.append(sum(s[i] for i in indexes))
    return s[0]
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/342/A:
Xenia the mathematician has a sequence consisting of n (n is divisible by 3) positive integers, each of them is at most 7. She wants to split the sequence into groups of three so that for each group of three a, b, c the following conditions held:  a < b < c;  a divides b, b divides c. 

Naturally, Xenia wants each element of the sequence to belong to exactly one group of three. Thus, if the required partition exists, then it has $\frac{n}{3}$ groups of three.

Help Xenia, find the required partition or else say that it doesn't exist.


-----Input-----

The first line contains integer n (3 ≤ n ≤ 99999) — the number of elements in the sequence. The next line contains n positive integers, each of them is at most 7.

It is guaranteed that n is divisible by 3.


-----Output-----

If the required partition exists, print $\frac{n}{3}$ groups of three. Print each group as values of the elements it contains. You should print values in increasing order. Separate the groups and integers in groups by whitespaces. If there are multiple solutions, you can print any of them.

If there is no solution, print -1.


-----Examples-----
Input
6
1 1 1 2 2 2

Output
-1

Input
6
2 2 1 1 4 6

Output
1 2 4
1 2 6

I tried it in Python, but could not do it. Can you solve it?
targets: a=[0 for i in range(8)]
N=int(input())
b=list(map(int, input().split()))
for k in range(N):
    a[b[k-1]]=a[b[k-1]]+1;
if a[5] or a[7]:
    print(-1)
    return
A,C=a[4],a[3]
B=a[6]-C
if A>=0 and B>=0 and C>=0 and A+B==a[2] and A+B+C==a[1]:
    for i in range(A):
        print("1 2 4")
    for i in range(B):
        print("1 2 6")
    for i in range(C):
        print("1 3 6")
else: print(-1)