Muennighoff/xP3x-sample huggingface dataset:
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1105/C:
Ayoub had an array $a$ of integers of size $n$ and this array had two interesting properties:   All the integers in the array were between $l$ and $r$ (inclusive).  The sum of all the elements was divisible by $3$. 

Unfortunately, Ayoub has lost his array, but he remembers the size of the array $n$ and the numbers $l$ and $r$, so he asked you to find the number of ways to restore the array. 

Since the answer could be very large, print it modulo $10^9 + 7$ (i.e. the remainder when dividing by $10^9 + 7$). In case there are no satisfying arrays (Ayoub has a wrong memory), print $0$.


-----Input-----

The first and only line contains three integers $n$, $l$ and $r$ ($1 \le n \le 2 \cdot 10^5 , 1 \le l \le r \le 10^9$) — the size of the lost array and the range of numbers in the array.


-----Output-----

Print the remainder when dividing by $10^9 + 7$ the number of ways to restore the array.


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

Output
3

Input
3 2 2

Output
1

Input
9 9 99

Output
711426616



-----Note-----

In the first example, the possible arrays are : $[1,2], [2,1], [3, 3]$.

In the second example, the only possible array is $[2, 2, 2]$.

I tried it in Python, but could not do it. Can you solve it?
targets: from sys import stdin, stdout
prime = 10**9 + 7

def unitMatrix(size):
    out = [[0]*size for i in range(size)]
    for i in range(size):
        out[i][i] = 1
    return out

def matrixMult(pre, post):
    rows = len(pre)
    mid = len(post)
    columns = len(post[0])
    out = [[0]*columns for i in range(rows)]
    for i in range(rows):
        for j in range(columns):
            for k in range(mid):
                out[i][j] += pre[i][k]*post[k][j]
                out[i][j] = out[i][j]%prime
    return out

def matrixExp(base, power):
    if power == 0:
        return unitMatrix(len(base))
    else:
        if power%2 == 0:
            half = matrixExp(base, power/2)
            return matrixMult(half, half)
        else:
            half = matrixExp(base, (power-1)/2)
            return matrixMult(matrixMult(half, half), base)

def main():
    n, l, r = [int(i) for i in stdin.readline().split()]

    vals = [[1], [0], [0]]
    mods = [(r-l+1)//3 ,(r-l+1)//3 ,(r-l+1)//3]
    remainder = (r-l+1)%3
    if remainder >= 1:
        mods[l%3] += 1
    if remainder == 2:
        mods[(l+1)%3] += 1
        
    transforms = [[mods[0], mods[2], mods[1]], [mods[1], mods[0], mods[2]], [mods[2], mods[1], mods[0]]]
    power = matrixExp(transforms, n)
    out = matrixMult(power, vals)
    print(out[0][0])

main()
inputs: I found an interesting problem on https://atcoder.jp/contests/abc067/tasks/arc078_a:
Snuke and Raccoon have a heap of N cards. The i-th card from the top has the integer a_i written on it.
They will share these cards.
First, Snuke will take some number of cards from the top of the heap, then Raccoon will take all the remaining cards.
Here, both Snuke and Raccoon have to take at least one card.
Let the sum of the integers on Snuke's cards and Raccoon's cards be x and y, respectively.
They would like to minimize |x-y|.
Find the minimum possible value of |x-y|.

-----Constraints-----
 - 2 \leq N \leq 2 \times 10^5
 - -10^{9} \leq a_i \leq 10^{9}
 - a_i is an integer.

-----Input-----
Input is given from Standard Input in the following format:
N
a_1 a_2 ... a_{N}

-----Output-----
Print the answer.

-----Sample Input-----
6
1 2 3 4 5 6

-----Sample Output-----
1

If Snuke takes four cards from the top, and Raccoon takes the remaining two cards, x=10, y=11, and thus |x-y|=1. This is the minimum possible value.

I tried it in Python, but could not do it. Can you solve it?
targets: n=int(input())
a=list(map(int,input().split()))
ans=2*(10**14)+1
s=sum(a)
x=0
y=s
for i in range(n-1):
    x+=a[i]
    y-=a[i]
    ans=min(ans,abs(x-y))
print(ans)
inputs: Solve in Python:
As you know America’s Presidential Elections are about to take place and the most popular leader of the Republican party Donald Trump is famous for throwing allegations against anyone he meets.

He goes to a rally and meets n people which he wants to offend. For each person i he can choose an integer between 1 to max[i].

He wants to decide in how many ways he can offend all these persons (N) given the condition that all numbers chosen by him for each person are distinct.
So he needs your help to find out the number of ways in which he can do that. If no solution is possible print 0

-----Input-----
The first line of the input contains an integer T (1<=T<=100) denoting the number of test cases. The description of T test cases follows.

The first line of each test case contains a single integer N denoting the number of people Trump wants to offend. The second line contains N space-separated integers maxnumber[0], maxnumber[1], ..., maxnumber[n-1] denoting the maxnumber that trump can choose for each person. 

-----Output-----
For each test case, output a single line containing the number of ways Trump can assign numbers to the people, modulo 1,000,000,007. If it's impossible to assign distinct integers to the people, print 0

-----Constraints-----
- 1 ≤ T ≤ 100
- 1 ≤ N ≤ 50
- 1 ≤ Maxnumber[i] ≤ 3000

-----Example-----
Input:
3
1
4
2
10 5
4
2 3 1 3

Output:
4
45
0


-----Explanation-----
In case 1, He can choose any number from 1 to 4

In case 2,Out of the total 50 combination he can not take (1,1) ,(2,2) , (3,3) ,(4,4) or (5,5).
targets: for t in range(int(input())):
 n = int(input())
 a = sorted(map(int,input().split()))
 ans = 1
 for i in range(n):
  ans *= (a[i]-i)
  ans %= (10**9+7)
  if (ans == 0):
   break
 print(ans)
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/960/A:
A has a string consisting of some number of lowercase English letters 'a'. He gives it to his friend B who appends some number of letters 'b' to the end of this string. Since both A and B like the characters 'a' and 'b', they have made sure that at this point, at least one 'a' and one 'b' exist in the string.

B now gives this string to C and he appends some number of letters 'c' to the end of the string. However, since C is a good friend of A and B, the number of letters 'c' he appends is equal to the number of 'a' or to the number of 'b' in the string. It is also possible that the number of letters 'c' equals both to the number of letters 'a' and to the number of letters 'b' at the same time.

You have a string in your hands, and you want to check if it is possible to obtain the string in this way or not. If it is possible to obtain the string, print "YES", otherwise print "NO" (without the quotes).


-----Input-----

The first and only line consists of a string $S$ ($ 1 \le |S| \le 5\,000 $). It is guaranteed that the string will only consist of the lowercase English letters 'a', 'b', 'c'.


-----Output-----

Print "YES" or "NO", according to the condition.


-----Examples-----
Input
aaabccc

Output
YES

Input
bbacc

Output
NO

Input
aabc

Output
YES



-----Note-----

Consider first example: the number of 'c' is equal to the number of 'a'. 

Consider second example: although the number of 'c' is equal to the number of the 'b', the order is not correct.

Consider third example: the number of 'c' is equal to the number of 'b'.

I tried it in Python, but could not do it. Can you solve it?
targets: s = input()
flag = 0
ans = 1
aa = s.count("a")
bb = s.count("b")
cc = s.count("c")
if min(aa,bb) == 0 or (bb != cc and aa != cc):
    print("NO")
else:
    for i in s:
        if flag == 0:
            if i == "b":
                flag = 1
                continue
            if i != "a":
                ans = 0
                break
        elif flag == 1:
            if i == "c":
                flag = 2
                continue
            if i != "b":
                ans = 0
                break
        else:
            if i != "c":
                ans = 0
                break
    if ans == 1:
        print("YES")
    else:
        print("NO")
inputs: Solve in Python:
You are given strings S and T consisting of lowercase English letters.
You can perform the following operation on S any number of times:
Operation: Choose two distinct lowercase English letters c_1 and c_2, then replace every occurrence of c_1 with c_2, and every occurrence of c_2 with c_1.
Determine if S and T can be made equal by performing the operation zero or more times.

-----Constraints-----
 - 1 \leq |S| \leq 2 \times 10^5
 - |S| = |T|
 - S and T consists of lowercase English letters.

-----Input-----
Input is given from Standard Input in the following format:
S
T

-----Output-----
If S and T can be made equal, print Yes; otherwise, print No.

-----Sample Input-----
azzel
apple

-----Sample Output-----
Yes

azzel can be changed to apple, as follows:
 - Choose e as c_1 and l as c_2. azzel becomes azzle.
 - Choose z as c_1 and p as c_2. azzle becomes apple.
targets: import bisect,collections,copy,itertools,math,string
import sys
def I(): return int(sys.stdin.readline().rstrip())
def LI(): return list(map(int,sys.stdin.readline().rstrip().split()))
def S(): return sys.stdin.readline().rstrip()
def LS(): return list(sys.stdin.readline().rstrip().split())
def main():


    s = S()
    t = S()
    
    dic1 = collections.defaultdict(str)
    dic2 = collections.defaultdict(str)
    
    for i in range(len(s)):
        if dic1[t[i]] == "":
            dic1[t[i]] = s[i]
        else:
            if dic1[t[i]] != s[i]:
                print("No")
                return

        if dic2[s[i]] == "":
            dic2[s[i]] = t[i]
        else:
            if dic2[s[i]] != t[i]:
                print("No")
                return
        
    print("Yes")

main()
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/802/B:
Whereas humans nowadays read fewer and fewer books on paper, book readership among marmots has surged. Heidi has expanded the library and is now serving longer request sequences.


-----Input-----

Same as the easy version, but the limits have changed: 1 ≤ n, k ≤ 400 000.


-----Output-----

Same as the easy version.


-----Examples-----
Input
4 100
1 2 2 1

Output
2

Input
4 1
1 2 2 1

Output
3

Input
4 2
1 2 3 1

Output
3

I tried it in Python, but could not do it. Can you solve it?
targets: # https://codeforces.com/problemset/problem/802/B
import heapq

n, k  = map(int, input().split())
a     = list(map(int, input().split()))
d     = {}
pos   = {}
Q     = []
cnt   = 0

for i, x in enumerate(a):
    if x not in pos:
        pos[x] = []
    pos[x].append(i)
    
for i, x in enumerate(a):
   
    if x not in d:
        cnt += 1
        
        if len(d) == k:
            pos_, x_ = heapq.heappop(Q)
            del d[x_]
        d[x] = 1
        
    pos[x].pop(0)
    
    if len(pos[x]) > 0:
        heapq.heappush(Q, (-pos[x][0], x))
    else:
        heapq.heappush(Q, (-float('inf'), x))   
        
print(cnt)
inputs: I found an interesting problem on https://atcoder.jp/contests/abc107/tasks/abc107_a:
There is an N-car train.
You are given an integer i. Find the value of j such that the following statement is true: "the i-th car from the front of the train is the j-th car from the back."

-----Constraints-----
 - 1 \leq N \leq 100
 - 1 \leq i \leq N

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

-----Output-----
Print the answer.

-----Sample Input-----
4 2

-----Sample Output-----
3

The second car from the front of a 4-car train is the third car from the back.

I tried it in Python, but could not do it. Can you solve it?
targets: n, i = map(int, input().split())
print(n - i + 1)
inputs: I found an interesting problem on https://leetcode.com/problems/rearrange-spaces-between-words/:
You are given a string text of words that are placed among some number of spaces. Each word consists of one or more lowercase English letters and are separated by at least one space. It's guaranteed that text contains at least one word.
Rearrange the spaces so that there is an equal number of spaces between every pair of adjacent words and that number is maximized. If you cannot redistribute all the spaces equally, place the extra spaces at the end, meaning the returned string should be the same length as text.
Return the string after rearranging the spaces.
 
Example 1:
Input: text = "  this   is  a sentence "
Output: "this   is   a   sentence"
Explanation: There are a total of 9 spaces and 4 words. We can evenly divide the 9 spaces between the words: 9 / (4-1) = 3 spaces.

Example 2:
Input: text = " practice   makes   perfect"
Output: "practice   makes   perfect "
Explanation: There are a total of 7 spaces and 3 words. 7 / (3-1) = 3 spaces plus 1 extra space. We place this extra space at the end of the string.

Example 3:
Input: text = "hello   world"
Output: "hello   world"

Example 4:
Input: text = "  walks  udp package   into  bar a"
Output: "walks  udp  package  into  bar  a "

Example 5:
Input: text = "a"
Output: "a"

 
Constraints:

1 <= text.length <= 100
text consists of lowercase English letters and ' '.
text contains at least one word.

I tried it in Python, but could not do it. Can you solve it?
targets: class Solution:
    def reorderSpaces(self, text: str) -> str:

        # separate words and spaces
        words = text.split(' ')
        words = list(filter(lambda x: len(x) > 0, words))
        # print(words)
        
        # get their counts so we can do some math
        wordsLen = len(words)
        spacesLen = text.count(' ')
        
        # print(wordsLen, spacesLen)
        if(wordsLen == 1):
            return words[0] + ' ' * spacesLen
        evenDistSpaces = spacesLen // (wordsLen-1)
        endDistSpaces = spacesLen % (wordsLen-1)
        
        # print(evenDistSpaces, endDistSpaces)
        
        space = ' ' * evenDistSpaces
        end = ' ' *  endDistSpaces
        
        print(len(space), len(end))
        
        resultString = space.join(words)
        resultString += end
        
        return resultString
inputs: I found an interesting problem on https://www.codewars.com/kata/5a084a098ba9146690000969:
This kata requires you to convert minutes (`int`) to hours and minutes in the format `hh:mm` (`string`).

If the input is `0` or negative value, then you should return `"00:00"`

**Hint:** use the modulo operation to solve this challenge. The modulo operation simply returns the remainder after a division. For example the remainder of 5 / 2 is 1, so 5 modulo 2 is 1.


## Example

If the input is `78`, then you should return `"01:18"`, because 78 minutes converts to 1 hour and 18 minutes.


Good luck! :D

I tried it in Python, but could not do it. Can you solve it?
targets: def time_convert(num):
    return '{:02}:{:02}'.format(*divmod(max(0, num), 60))
inputs: I found an interesting problem on https://www.codechef.com/ICOD2016/problems/ICODE16G:
Abhishek is fond of playing cricket very much. One morning, he is playing cricket with his friends. Abhishek  is a right-hand batsman

.He has to face  all types of balls either good or  bad. There are total 26 balls in the game and each ball is represented

by one of the following two ways:-
1. "g" denotes a good ball.
2. "b" denotes a bad  ball.
All 26 balls are represented by lower case letters (a,b,.....z).
Balls faced by Abhishek are represented as a string s, all the characters of which are lower case  i.e, within  26 above mentioned balls.
A substring s[l...r] (1≤l≤r≤|s|) of string s=s1s2...s|s| (where |s| is the length of string s) is string slsl+1...sr.
The substring s[l...r] is good, if among the letters slsl+1...sr, there are at most k bad ones (refer sample explanation ).
Your task is to find out the number of distinct good substrings for the given string s. Two substrings s[x...y] and s[p...q] are considered distinct if their contents are different, i.e. s[x...y]≠s[p...q].

-----Input Format-----
First Line contains an integer T, number of test cases. For each test case, first line contains a string - a sequence of balls faced by Abhishek.

And, next line contains a string of characters "g" and "b" of length 26 characters. If the ith character of this string equals "1", then the i-th English letter is good, otherwise it's bad. That is, the first character of this string corresponds to letter "a", the second one corresponds to letter "b" and so on.

And, the third line of the test case consists of a single integer k (0≤k≤|s|) — the maximum acceptable number of bad characters in a good substring.

-----Output Format -----
For each test case, print a single integer — the number of distinct good substrings of string s.

-----Constraints-----
- 1<=T<=1000
- 1<=|s|<=2000
- 0<=k<=|s|

-----Subtasks-----
Subtask 1 : 20 Points
- 1<=T<=10
- 1<=|s|<=20
- 0<=k<=|s|
Subtask 2 : 80 Points 
Original Constraints
Sample Input
2
ababab
bgbbbbbbbbbbbbbbbbbbbbbbbb
1
acbacbacaa
bbbbbbbbbbbbbbbbbbbbbbbbbb
2
Sample...

I tried it in Python, but could not do it. Can you solve it?
targets: import sys
for _ in range(0,eval(input())):   
    d,c,inp,mp,n,q=[],0,list(map(ord,list(sys.stdin.readline().strip()))),sys.stdin.readline().strip(),eval(input()),ord('a')
    for i in range(0,len(inp)):
        nn,h=n,0
        for j in range(i,len(inp)):
                if ( (mp[inp[j]-q]=='g') or  nn>0 ):
                    if((mp[inp[j]-q]=='g') == False ):                
                        nn=nn-1
                    h=(h*256)^inp[j]
                    d+=[h]
                else:
                    break  
    print(len(set(d)))
inputs: I found an interesting problem on https://www.codewars.com/kata/58983deb128a54b530000be6:
Write a function that checks the braces status in a string, and return `True` if all braces are properly closed, or `False` otherwise. Available types of brackets: `()`, `[]`, `{}`.

**Please note, you need to write this function without using regex!**

## Examples
```python
'([[some](){text}here]...)'  =>  True
'{([])}'                     =>  True
'()[]{}()'                   =>  True
'(...[]...{(..())}[abc]())'  =>  True
'1239(df){'                  =>  False
'[()])'                      =>  False
')12[x]34('                  =>  False
```
Don't forget to rate this kata! Thanks :)

I tried it in Python, but could not do it. Can you solve it?
targets: def braces_status(string):
    s = "".join(list(filter(lambda ch: ch in "{[()]}",list(string))))
    while '{}' in s or '()' in s or '[]' in s:
        s=s.replace('{}','')
        s=s.replace('[]','')
        s=s.replace('()','')
    return s==''
inputs: Solve in Python:
You are given a bracket sequence $s$ (not necessarily a regular one). A bracket sequence is a string containing only characters '(' and ')'.

A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"), and ")(", "(" and ")" are not.

Your problem is to calculate the number of regular bracket sequences of length $2n$ containing the given bracket sequence $s$ as a substring (consecutive sequence of characters) modulo $10^9+7$ ($1000000007$).


-----Input-----

The first line of the input contains one integer $n$ ($1 \le n \le 100$) — the half-length of the resulting regular bracket sequences (the resulting sequences must have length equal to $2n$).

The second line of the input contains one string $s$ ($1 \le |s| \le 200$) — the string $s$ that should be a substring in each of the resulting regular bracket sequences ($|s|$ is the length of $s$).


-----Output-----

Print only one integer — the number of regular bracket sequences containing the given bracket sequence $s$ as a substring. Since this number can be huge, print it modulo $10^9+7$ ($1000000007$).


-----Examples-----
Input
5
()))()

Output
5

Input
3
(()

Output
4

Input
2
(((

Output
0



-----Note-----

All regular bracket sequences satisfying the conditions above for the first example:   "(((()))())";  "((()()))()";  "((()))()()";  "(()(()))()";  "()((()))()". 

All regular bracket sequences satisfying the conditions above for the second example:   "((()))";  "(()())";  "(())()";  "()(())". 

And there is no regular bracket sequences of length $4$ containing "(((" as a substring in the third example.
targets: MOD=10**9+7
n=int(input())
s=[c=='(' for c in input()]
m=len(s)
z=[[0,0]]
for v in s:
 a=z[-1][v]
 z[-1][v]=len(z)
 z.append(z[a][:])
z[m][0]=z[m][1]=m
dp=[[0 for _ in range(m+1)] for _ in range(n+1)]
dp[0][0]=1
for _ in range(2*n):
 ndp=[[0 for _ in range(m+1)] for _ in range(n+1)]
 for i in range(n+1):
  for j in range(m+1):
   if i>0:ndp[i-1][z[j][0]]=(ndp[i-1][z[j][0]]+dp[i][j])%MOD
   if i<n:ndp[i+1][z[j][1]]=(ndp[i+1][z[j][1]]+dp[i][j])%MOD
 dp=ndp
print(dp[0][m])
inputs: I found an interesting problem on https://www.codewars.com/kata/5886faac54a7111c21000072:
# Task
 A media access control address (MAC address) is a unique identifier assigned to network interfaces for communications on the physical network segment.

 The standard (IEEE 802) format for printing MAC-48 addresses in human-friendly form is six groups of two hexadecimal digits (0 to 9 or A to F), separated by hyphens (e.g. 01-23-45-67-89-AB).

# Example

 For `inputString = "00-1B-63-84-45-E6"`, the output should be `true`;
 
 For `inputString = "Z1-1B-63-84-45-E6"`, the output should be `false`;
 
 For `inputString = "not a MAC-48 address"`, the output should be `false`.
 
# Input/Output

 - `[input]` string `inputString`
 
 - `[output]` a boolean value

    `true` if inputString corresponds to MAC-48 address naming rules, `false` otherwise.

I tried it in Python, but could not do it. Can you solve it?
targets: is_mac_48_address = lambda address: bool( __import__("re").match('-'.join(['[0-9A-F]{2}']*6) + '$', address) )
inputs: I found an interesting problem on https://www.codewars.com/kata/52ec24228a515e620b0005ef:
# How many ways can you make the sum of a number?

From wikipedia: https://en.wikipedia.org/wiki/Partition_(number_theory)#

>In number theory and combinatorics, a partition of a positive integer *n*, also called an *integer partition*, is a way of writing n as a sum of positive integers. Two sums that differ only in the order of their summands are considered the same partition. If order matters, the sum becomes a composition. For example, 4 can be partitioned in five distinct ways:
```
4
3 + 1
2 + 2
2 + 1 + 1
1 + 1 + 1 + 1
```

## Examples

### Basic

```python
exp_sum(1) # 1
exp_sum(2) # 2  -> 1+1 , 2
exp_sum(3) # 3 -> 1+1+1, 1+2, 3
exp_sum(4) # 5 -> 1+1+1+1, 1+1+2, 1+3, 2+2, 4
exp_sum(5) # 7 -> 1+1+1+1+1, 1+1+1+2, 1+1+3, 1+2+2, 1+4, 5, 2+3

exp_sum(10) # 42
```

### Explosive

```python
exp_sum(50) # 204226
exp_sum(80) # 15796476
exp_sum(100) # 190569292
```

See [here](http://www.numericana.com/data/partition.htm) for more examples.

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

@functools.lru_cache(1 << 16)
def f(n, m):
    if n == 0:
        return 1
    return sum(f(n - i, i) for i in range(1, min(n, m)+1))

def exp_sum(n):
    return f(n, n)
inputs: I found an interesting problem on https://atcoder.jp/contests/abc046/tasks/abc046_a:
AtCoDeer the deer recently bought three paint cans.
The color of the one he bought two days ago is a, the color of the one he bought yesterday is b, and the color of the one he bought today is c.
Here, the color of each paint can is represented by an integer between 1 and 100, inclusive.
Since he is forgetful, he might have bought more than one paint can in the same color.
Count the number of different kinds of colors of these paint cans and tell him.

-----Constraints-----
 - 1≦a,b,c≦100

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

-----Output-----
Print the number of different kinds of colors of the paint cans.

-----Sample Input-----
3 1 4

-----Sample Output-----
3

Three different colors: 1, 3, and 4.

I tried it in Python, but could not do it. Can you solve it?
targets: penki=list(map(int,input().split()))
ans=len(set(penki))
print(ans)
inputs: Solve in Python:
Bob watches TV every day. He always sets the volume of his TV to $b$. However, today he is angry to find out someone has changed the volume to $a$. Of course, Bob has a remote control that can change the volume.

There are six buttons ($-5, -2, -1, +1, +2, +5$) on the control, which in one press can either increase or decrease the current volume by $1$, $2$, or $5$. The volume can be arbitrarily large, but can never be negative. In other words, Bob cannot press the button if it causes the volume to be lower than $0$.

As Bob is so angry, he wants to change the volume to $b$ using as few button presses as possible. However, he forgets how to do such simple calculations, so he asks you for help. Write a program that given $a$ and $b$, finds the minimum number of presses to change the TV volume from $a$ to $b$.


-----Input-----

Each test contains multiple test cases. The first line contains the number of test cases $T$ ($1 \le T \le 1\,000$). Then the descriptions of the test cases follow.

Each test case consists of one line containing two integers $a$ and $b$ ($0 \le a, b \le 10^{9}$) — the current volume and Bob's desired volume, respectively.


-----Output-----

For each test case, output a single integer — the minimum number of presses to change the TV volume from $a$ to $b$. If Bob does not need to change the volume (i.e. $a=b$), then print $0$.


-----Example-----
Input
3
4 0
5 14
3 9

Output
2
3
2



-----Note-----

In the first example, Bob can press the $-2$ button twice to reach $0$. Note that Bob can not press $-5$ when the volume is $4$ since it will make the volume negative. 

In the second example, one of the optimal ways for Bob is to press the $+5$ twice, then press $-1$ once.

In the last example, Bob can press the $+5$ once, then press $+1$.
targets: T = int(input())

for _ in range(T):
	a, b = map(int, input().split())

	d = abs(a - b)

	ans = (d // 5)

	d = d % 5

	if d == 1 or d == 2:
		ans += 1

	if d == 3 or d == 4:
		ans += 2

	print(ans)
inputs: I found an interesting problem on https://www.codewars.com/kata/5993e6f701726f0998000030:
# Disclaimer

This Kata is an insane step-up from [GiacomoSorbi's Kata](https://www.codewars.com/kata/total-increasing-or-decreasing-numbers-up-to-a-power-of-10/python),
so I recommend to solve it first before trying this one.

# Problem Description

A positive integer `n` is called an *increasing number* if its digits are in increasing order
(e.g. 123, 144568, 56799).

Similarly, `n` is called a *decreasing number* if its digits are in decreasing order
(e.g. 210, 76642, 998500).

Note that the numbers whose digits are all the same (e.g. 1111, 22222) are both increasing and decreasing.

Given the maximum number of digits (`max_digits`), how many positive integers in the range are either increasing or decreasing (or both)?
Since your answer will be very large, please give your answer **modulo 12345787**.

Also note that, unlike Giacomo's version, the number zero is excluded from the counts
(because it's not positive).

# Constraints

`1 <= max_digits <= 10 ** 9`

**Note the input size!**

The input will be always a valid integer.

# Examples

```python
# Up to two digits, all numbers are either increasing or decreasing
insane_inc_or_dec(1) == 9
insane_inc_or_dec(2) == 99
insane_inc_or_dec(3) == 474
insane_inc_or_dec(4) == 1674
insane_inc_or_dec(5) == 4953
insane_inc_or_dec(6) == 12951
```

# Acknowledgement

This problem was inspired by [Project Euler #113: Non-bouncy Numbers](https://projecteuler.net/problem=113).

If you enjoyed this Kata, please also have a look at [my other Katas](https://www.codewars.com/users/Bubbler/authored)!

I tried it in Python, but could not do it. Can you solve it?
targets: from functools import reduce
from operator import mul

def insane_inc_or_dec(x):   
    return (reduce(mul,[x + i + i * (i == 10) for i in range(1, 11)]) // 3628800 - 10 * x - 2) % 12345787
inputs: I found an interesting problem on https://leetcode.com/problems/sequential-digits/:
An integer has sequential digits if and only if each digit in the number is one more than the previous digit.
Return a sorted list of all the integers in the range [low, high] inclusive that have sequential digits.
 
Example 1:
Input: low = 100, high = 300
Output: [123,234]
Example 2:
Input: low = 1000, high = 13000
Output: [1234,2345,3456,4567,5678,6789,12345]

 
Constraints:

10 <= low <= high <= 10^9

I tried it in Python, but could not do it. Can you solve it?
targets: class Solution:
    def sequentialDigits(self, low: int, high: int) -> List[int]:
        nums = []
        max_digits = 9
        for depth in range(2, max_digits + 1):
            nums += self.genNum(max_digits, depth)
        return [n for n in nums if n >= low and n <= high]
    
    def genNum(self, max_digits, depth):
        result = []
        for startDigit in range(1, max_digits-depth+2):
            num = startDigit
            for lvl in range(depth-1):
                num = num * 10 + num % 10 + 1
            result.append(num)
        return result
inputs: I found an interesting problem on https://atcoder.jp/contests/abc116/tasks/abc116_d:
There are N pieces of sushi. Each piece has two parameters: "kind of topping" t_i and "deliciousness" d_i.
You are choosing K among these N pieces to eat.
Your "satisfaction" here will be calculated as follows:
 - The satisfaction is the sum of the "base total deliciousness" and the "variety bonus".
 - The base total deliciousness is the sum of the deliciousness of the pieces you eat.
 - The variety bonus is x*x, where x is the number of different kinds of toppings of the pieces you eat.
You want to have as much satisfaction as possible.
Find this maximum satisfaction.

-----Constraints-----
 - 1 \leq K \leq N  \leq 10^5
 - 1 \leq t_i \leq N
 - 1 \leq d_i \leq 10^9
 - All values in input are integers.

-----Input-----
Input is given from Standard Input in the following format:
N K
t_1 d_1
t_2 d_2
.
.
.
t_N d_N

-----Output-----
Print the maximum satisfaction that you can obtain.

-----Sample Input-----
5 3
1 9
1 7
2 6
2 5
3 1

-----Sample Output-----
26

If you eat Sushi 1,2 and 3:
 - The base total deliciousness is 9+7+6=22.
 - The variety bonus is 2*2=4.
Thus, your satisfaction will be 26, which is optimal.

I tried it in Python, but could not do it. Can you solve it?
targets: n,k=list(map(int,input().split()))
td=sorted([list(map(int,input().split())) for i in range(n)],reverse=True,key=lambda x:x[1])

ans=0
kl=dict()
for i in range(k):
    ans+=td[i][1]
    if td[i][0] in kl:
        kl[td[i][0]]+=1
    else:
        kl[td[i][0]]=1
l=len(kl)
ans+=(l**2)
ans_=ans
now=k-1
for i in range(k,n):
    if td[i][0] not in kl:
        #最小を求めるのをもっと早く
        while now>=0:
            if kl[td[now][0]]>1:
                mi=td[now]
                kl[mi[0]]-=1
                now-=1
                break
            now-=1
        if now==-1:
            break
        else:
            ans=ans+2*l+1-mi[1]+td[i][1]
            kl[td[i][0]]=1
            ans_=max(ans,ans_)
            l+=1

print(ans_)
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1191/B:
Tokitsukaze is playing a game derivated from Japanese mahjong. In this game, she has three tiles in her hand. Each tile she owns is a suited tile, which means it has a suit (manzu, pinzu or souzu) and a number (a digit ranged from $1$ to $9$). In this problem, we use one digit and one lowercase letter, which is the first character of the suit, to represent a suited tile. All possible suited tiles are represented as 1m, 2m, $\ldots$, 9m, 1p, 2p, $\ldots$, 9p, 1s, 2s, $\ldots$, 9s.

In order to win the game, she must have at least one mentsu (described below) in her hand, so sometimes she should draw extra suited tiles. After drawing a tile, the number of her tiles increases by one. She can draw any tiles she wants, including those already in her hand.

Do you know the minimum number of extra suited tiles she needs to draw so that she can win?

Here are some useful definitions in this game:  A mentsu, also known as meld, is formed by a koutsu or a shuntsu;  A koutsu, also known as triplet, is made of three identical tiles, such as [1m, 1m, 1m], however, [1m, 1p, 1s] or [1m, 4m, 7m] is NOT a koutsu;  A shuntsu, also known as sequence, is made of three sequential numbered tiles in the same suit, such as [1m, 2m, 3m] and [5s, 7s, 6s], however, [9m, 1m, 2m] or [1m, 2p, 3s] is NOT a shuntsu. 

Some examples:   [2m, 3p, 2s, 4m, 1s, 2s, 4s] — it contains no koutsu or shuntsu, so it includes no mentsu;  [4s, 3m, 3p, 4s, 5p, 4s, 5p] — it contains a koutsu, [4s, 4s, 4s], but no shuntsu, so it includes a mentsu;  [5p, 5s, 9m, 4p, 1s, 7p, 7m, 6p] — it contains no koutsu but a shuntsu, [5p, 4p, 6p] or [5p, 7p, 6p], so it includes a mentsu. 

Note that the order of tiles is unnecessary and you can assume the number of each type of suited tiles she can draw is infinite.


-----Input-----

The only line contains three strings — the tiles in Tokitsukaze's hand. For each string, the first character is a digit ranged from $1$ to $9$ and the second character is m, p or s.


-----Output-----

Print a single integer — the minimum...

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

t.sort()

if t.count(t[0]) == 3:
  print('0')
elif t.count(t[0]) == 2 or t.count(t[1]) == 2:
  print('1')
else:
  num = list(map(int, [t[0][0], t[1][0], t[2][0]]))
  suit = [t[0][1], t[1][1], t[2][1]]
  if len(set(suit)) == 3:
    print('2')
  elif len(set(suit)) == 1:
    if num[1] == num[0] + 1 or num[2] == num[1] + 1:
      if num[2] == num[0] + 2:
        print('0')
      else:
        print('1')
    elif num[1] == num[0] + 2 or num[2] == num[1] + 2:
        print('1')
    else:
      print('2')
  else:
    if suit[0] == suit[1]:
      if num[1] - num[0] in [1, 2]:
        print('1')
      else:
        print('2')
    elif suit[1] == suit[2]:
      if num[2] - num[1] in [1, 2]:
        print('1')
      else:
        print('2')
    else:
      if num[2] - num[0] in [1, 2]:
        print('1')
      else:
        print('2')
inputs: Solve in Python:
#### Task:

Your job here is to implement a method, `approx_root` in Ruby/Python/Crystal and `approxRoot` in JavaScript/CoffeeScript, that takes one argument, `n`, and returns the approximate square root of that number, rounded to the nearest hundredth and computed in the following manner.

1.      Start with `n = 213` (as an example). 
2.      To approximate the square root of n, we will first find the greatest perfect square that is below or equal to `n`. (In this example, that would be 196, or 14 squared.) We will call the square root of this number (which means sqrt 196, or 14) `base`.
3.      Then, we will take the lowest perfect square that is greater than or equal to `n`. (In this example, that would be 225, or 15 squared.)
4.      Next, subtract 196 (greatest perfect square less than or equal to `n`) from `n`. (213 - 196 = **17**) We will call this value `diff_gn`.
5.      Find the difference between the lowest perfect square greater than or equal to `n` and the greatest perfect square less than or equal to `n`. (225 – 196 = **29**) We will call this value `diff_lg`.
6.      Your final answer is `base` + (`diff_gn` / `diff_lg`). In this example: 14 + (17 / 29) which is 14.59, rounded to the nearest hundredth.

Just to clarify, if the input is a perfect square itself, you should return the exact square of the input.

In case you are curious, the approximation (computed like above) for 213 rounded to four decimal places is 14.5862. The actual square root of 213 is 14.5945. 

Inputs will always be positive whole numbers. If you are having trouble understanding it, let me know with a comment, or take a look at the second group of the example cases.

#### Some examples:

```python
approx_root(400) #=> 20
approx_root(401) #=> 
  # smallest perfect square above 401 is 441 or 21 squared
  # greatest perfect square below 401 is 400 or 20 squared
  # difference between 441 and 400 is 41
  # difference between 401 and 400 is 1
  # answer is 20 + (1 / 41) which becomes 20.02, rounded to the nearest hundredth
  #...
targets: def approx_root(n):
    base = int(n ** 0.5)
    diff_gn = n - base ** 2
    diff_lg = base * 2 + 1
    return round(base + diff_gn / diff_lg, 2)
inputs: Solve in Python:
Sereja has got an array, consisting of n integers, a_1, a_2, ..., a_{n}. Sereja is an active boy, so he is now going to complete m operations. Each operation will have one of the three forms:  Make v_{i}-th array element equal to x_{i}. In other words, perform the assignment a_{v}_{i} = x_{i}.  Increase each array element by y_{i}. In other words, perform n assignments a_{i} = a_{i} + y_{i} (1 ≤ i ≤ n).  Take a piece of paper and write out the q_{i}-th array element. That is, the element a_{q}_{i}. 

Help Sereja, complete all his operations.


-----Input-----

The first line contains integers n, m (1 ≤ n, m ≤ 10^5). The second line contains n space-separated integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ 10^9) — the original array.

Next m lines describe operations, the i-th line describes the i-th operation. The first number in the i-th line is integer t_{i} (1 ≤ t_{i} ≤ 3) that represents the operation type. If t_{i} = 1, then it is followed by two integers v_{i} and x_{i}, (1 ≤ v_{i} ≤ n, 1 ≤ x_{i} ≤ 10^9). If t_{i} = 2, then it is followed by integer y_{i} (1 ≤ y_{i} ≤ 10^4). And if t_{i} = 3, then it is followed by integer q_{i} (1 ≤ q_{i} ≤ n).


-----Output-----

For each third type operation print value a_{q}_{i}. Print the values in the order, in which the corresponding queries follow in the input.


-----Examples-----
Input
10 11
1 2 3 4 5 6 7 8 9 10
3 2
3 9
2 10
3 1
3 10
1 1 10
2 10
2 10
3 1
3 10
3 9

Output
2
9
11
20
30
40
39
targets: import sys
n,m=list(map(int,sys.stdin.readline().split()))

L=list(map(int,sys.stdin.readline().split()))
c=0
Ans=""
for i in range(m):
    x=list(map(int,sys.stdin.readline().split()))
    if(x[0]==1):
        L[x[1]-1]=x[2]-c
    elif(x[0]==2):
        c+=x[1]
    else:
        Ans+=str(L[x[1]-1]+c)+"\n"
sys.stdout.write(Ans)
inputs: Solve in Python:
Alena has successfully passed the entrance exams to the university and is now looking forward to start studying.

One two-hour lesson at the Russian university is traditionally called a pair, it lasts for two academic hours (an academic hour is equal to 45 minutes).

The University works in such a way that every day it holds exactly n lessons. Depending on the schedule of a particular group of students, on a given day, some pairs may actually contain classes, but some may be empty (such pairs are called breaks).

The official website of the university has already published the schedule for tomorrow for Alena's group. Thus, for each of the n pairs she knows if there will be a class at that time or not.

Alena's House is far from the university, so if there are breaks, she doesn't always go home. Alena has time to go home only if the break consists of at least two free pairs in a row, otherwise she waits for the next pair at the university.

Of course, Alena does not want to be sleepy during pairs, so she will sleep as long as possible, and will only come to the first pair that is presented in her schedule. Similarly, if there are no more pairs, then Alena immediately goes home.

Alena appreciates the time spent at home, so she always goes home when it is possible, and returns to the university only at the beginning of the next pair. Help Alena determine for how many pairs she will stay at the university. Note that during some pairs Alena may be at the university waiting for the upcoming pair.


-----Input-----

The first line of the input contains a positive integer n (1 ≤ n ≤ 100) — the number of lessons at the university. 

The second line contains n numbers a_{i} (0 ≤ a_{i} ≤ 1). Number a_{i} equals 0, if Alena doesn't have the i-th pairs, otherwise it is equal to 1. Numbers a_1, a_2, ..., a_{n} are separated by spaces.


-----Output-----

Print a single number — the number of pairs during which Alena stays at the university.


-----Examples-----
Input
5
0 1 0 1 1

Output
4

Input
7
1 0 1 0 0 1...
targets: #!/usr/bin/env python3
# -*- coding: utf-8 -*-

import re

N = int(input())
S = input()
S = S.replace(" ", "")
S = S.strip("0")

ret = 0
for seg in re.split("00+", S):
    ret += len(seg)

print(ret)
inputs: Solve in Python:
Chef has a number N, Cheffina challenges the chef to check the divisibility of all the permutation of N by 2. If any of the permutations is divisible by 2 then print 1 else print 0.

-----Input:-----
- First-line will contain $T$, the number of test cases. Then the test cases follow. 
- Each test case contains a single line of input, $N$. 

-----Output:-----
For each test case, output in a single line answer 1 or 0.

-----Constraints-----
- $1 \leq T \leq 10^6$
- $1 \leq N \leq 10^6$

-----Sample Input:-----
2
19
385

-----Sample Output:-----
0
1
targets: for _ in range(int(input())):
    r=int(input())
    while(r!=0):
        k=r%10
        if(k%2==0):
            print("1")
            break
        r=r//10
    else:
        print("0")
inputs: I found an interesting problem on https://atcoder.jp/contests/abc059/tasks/arc072_b:
Alice and Brown loves games. Today, they will play the following game.
In this game, there are two piles initially consisting of X and Y stones, respectively.
Alice and Bob alternately perform the following operation, starting from Alice:
 - Take 2i stones from one of the piles. Then, throw away i of them, and put the remaining i in the other pile. Here, the integer i (1≤i) can be freely chosen as long as there is a sufficient number of stones in the pile.
The player who becomes unable to perform the operation, loses the game.
Given X and Y, determine the winner of the game, assuming that both players play optimally.

-----Constraints-----
 - 0 ≤ X, Y ≤ 10^{18}

-----Input-----
Input is given from Standard Input in the following format:
X Y

-----Output-----
Print the winner: either Alice or Brown.

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

-----Sample Output-----
Brown

Alice can do nothing but taking two stones from the pile containing two stones. As a result, the piles consist of zero and two stones, respectively. Then, Brown will take the two stones, and the piles will consist of one and zero stones, respectively. Alice will be unable to perform the operation anymore, which means Brown's victory.

I tried it in Python, but could not do it. Can you solve it?
targets: # 解説AC
X,Y = map(int, input().split())
print("Alice" if abs(X - Y) > 1 else "Brown")
inputs: Solve in Python:
For an integer N, we will choose a permutation \{P_1, P_2, ..., P_N\} of \{1, 2, ..., N\}.
Then, for each i=1,2,...,N, let M_i be the remainder when i is divided by P_i.
Find the maximum possible value of M_1 + M_2 + \cdots + M_N.

-----Constraints-----
 - N is an integer satisfying 1 \leq N \leq 10^9.

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

-----Output-----
Print the maximum possible value of M_1 + M_2 + \cdots + M_N.

-----Sample Input-----
2

-----Sample Output-----
1

When the permutation \{P_1, P_2\} = \{2, 1\} is chosen, M_1 + M_2 = 1 + 0 = 1.
targets: n=int(input())
print((n*(n-1)//2))
inputs: I found an interesting problem on https://www.hackerrank.com/challenges/py-if-else/problem:
=====Problem Statement=====
Given an integer, n, perform the following conditional actions:
If n is odd, print Weird
If n is even and in the inclusive range of 2 to 5, print Not Weird
If n is even and in the inclusive range of 6 to 20, print Weird
If n is even and greater than 20, print Not Weird

=====Input Format=====
A single line containing a positive integer, n.

=====Constraints=====
1≤n≤100

=====Output Format=====
Print Weird if the number is weird. Otherwise, print Not Weird.

I tried it in Python, but could not do it. Can you solve it?
targets: def __starting_point():
    n = int(input())
    if n % 2 == 1 or 6 <= n <= 20:
        print("Weird")
    else:
        print("Not Weird")

__starting_point()
inputs: I found an interesting problem on https://www.codewars.com/kata/5550d638a99ddb113e0000a2:
This problem takes its name by arguably the most important event in the life of the ancient historian Josephus: according to his tale, he and his 40 soldiers were trapped in a cave by the Romans during a siege.

Refusing to surrender to the enemy, they instead opted for mass suicide, with a twist: **they formed a circle and proceeded to kill one man every three, until one last man was left (and that it was supposed to kill himself to end the act)**.

Well, Josephus and another man were the last two and, as we now know every detail of the story, you may have correctly guessed that they didn't exactly follow through the original idea.

You are now to create a function that returns a Josephus permutation, taking as parameters the initial *array/list of items* to be permuted as if they were in a circle and counted out every *k* places until none remained.

**Tips and notes:** it helps to start counting from 1 up to n, instead of the usual range 0..n-1; k will always be >=1.

For example, with n=7 and k=3 `josephus(7,3)` should act this way.
```
[1,2,3,4,5,6,7] - initial sequence
[1,2,4,5,6,7] => 3 is counted out and goes into the result [3]
[1,2,4,5,7] => 6 is counted out and goes into the result [3,6]
[1,4,5,7] => 2 is counted out and goes into the result [3,6,2]
[1,4,5] => 7 is counted out and goes into the result [3,6,2,7]
[1,4] => 5 is counted out and goes into the result [3,6,2,7,5]
[4] => 1 is counted out and goes into the result [3,6,2,7,5,1]
[] => 4 is counted out and goes into the result [3,6,2,7,5,1,4]
```
So our final result is:
```
josephus([1,2,3,4,5,6,7],3)==[3,6,2,7,5,1,4]
```
For more info, browse the Josephus Permutation page on wikipedia; related kata: Josephus Survivor.

Also, [live game demo](https://iguacel.github.io/josephus/) by [OmniZoetrope](https://www.codewars.com/users/OmniZoetrope).

I tried it in Python, but could not do it. Can you solve it?
targets: def josephus(itemlist, interval):
    length = len(itemlist)

    if interval == 1:
        return itemlist
    
    count = 1
    disposal = [] #list which stores order in which items get deleted

    while len(disposal) != length: #loop runs unti; every item has been deleted
        index = 0 #gives an index. Useful if there are duplicates of items in a list
        for item in itemlist:
            if count == interval: 
                
                disposal.append(item)
                itemlist.pop(index)
                count = 1  
                
                try:
                    print(itemlist[index]) #sees if there is another number after the deleted one. Catches numbers at the end of the list which get's deleted
                except:
                    break

            count += 1
            index += 1
    return disposal