Muennighoff/xP3x-sample huggingface dataset:
inputs: Solve in Python:
According to an old legeng, a long time ago Ankh-Morpork residents did something wrong to miss Fortune, and she cursed them. She said that at some time n snacks of distinct sizes will fall on the city, and the residents should build a Snacktower of them by placing snacks one on another. Of course, big snacks should be at the bottom of the tower, while small snacks should be at the top.

Years passed, and once different snacks started to fall onto the city, and the residents began to build the Snacktower. [Image] 

However, they faced some troubles. Each day exactly one snack fell onto the city, but their order was strange. So, at some days the residents weren't able to put the new stack on the top of the Snacktower: they had to wait until all the bigger snacks fell. Of course, in order to not to anger miss Fortune again, the residents placed each snack on the top of the tower immediately as they could do it.

Write a program that models the behavior of Ankh-Morpork residents.


-----Input-----

The first line contains single integer n (1 ≤ n ≤ 100 000) — the total number of snacks.

The second line contains n integers, the i-th of them equals the size of the snack which fell on the i-th day. Sizes are distinct integers from 1 to n. 


-----Output-----

Print n lines. On the i-th of them print the sizes of the snacks which the residents placed on the top of the Snacktower on the i-th day in the order they will do that. If no snack is placed on some day, leave the corresponding line empty.


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

Output
3
 
2 1
Input
5
4 5 1 2 3

Output
 
5 4
 
 
3 2 1



-----Note-----

In the example a snack of size 3 fell on the first day, and the residents immediately placed it. On the second day a snack of size 1 fell, and the residents weren't able to place it because they were missing the snack of size 2. On the third day a snack of size 2 fell, and the residents immediately placed it. Right after that they placed the snack of size 1 which had fallen before.
targets: n = int(input())
maxsn = n-1
a = list(map(int, input().split()))
snacks = [0] * 100001
for i in range(n):
    snacks[a[i]-1] = 1
    while snacks[maxsn] == 1:
        snacks[maxsn] = 0
        print(maxsn + 1, end=' ')
        maxsn -= 1
    print()
inputs: Solve in Python:
Let's say you have a bunch of points, and you want to round them all up and calculate the area of the smallest polygon containing all of the points (nevermind why, you just want a challenge). What you're looking for is the area of the *convex hull* of these points. Here is an example, delimited in blue :



## Your task

Implement a function that will compute the area covered by the convex hull that can be formed from an array of points, the area being rounded to two decimal places. The points are given as `(x,y)`, like in an orthonormal coordinates system.

```python
points = [(0, 0), (0, 3), (4, 0)]
convex_hull_area(points) == 6.00
```

*Note* : In Python, the scipy module has a [ready made solution](https://docs.scipy.org/doc/scipy/reference/generated/scipy.spatial.ConvexHull.html) for this. Of course, if you use it here, you are lame.

*P. S.* : If you enjoy this kata, you may also like this one, which asks you to compute a convex hull, without finding its area.
targets: def cross(p1,p2,p):
    return (p[1] - p1[1]) * (p2[0] - p1[0]) - (p2[1] - p1[1]) * (p[0] - p1[0])
    
def dist_to(p1):
    def f(p):
        return (p1[0]-p[0]) * (p1[0]-p[0]) + (p1[1]-p[1]) * (p1[1]-p[1])
    return f

def xy_yx(p1,p2):
    return p1[0]*p2[1] - p1[1]*p2[0]
    
def convex_hull_area(pts):
    if len(pts) == 0:
        return 0
    pts = set(pts)
    new = min(pts)
    head = new
    area = 0
    while 1 :
        tail = new
        d_fn = dist_to(tail)
        new = next(iter(pts))
        for test in pts:
            if test != new and test != tail:
                c = cross(tail,new,test)
                if tail == new or c > 0: 
                    new = test
                elif c == 0 :
                    new = max(new,test,key=d_fn)
        area += xy_yx(tail, new)
        if new is head:
            return round(abs(area)/2.0,2)
inputs: Solve in Python:
Takahashi loves gold coins. He gains 1000 happiness points for each 500-yen coin he has and gains 5 happiness points for each 5-yen coin he has. (Yen is the currency of Japan.)
Takahashi has X yen. If he exchanges his money so that he will gain the most happiness points, how many happiness points will he earn?
(We assume that there are six kinds of coins available: 500-yen, 100-yen, 50-yen, 10-yen, 5-yen, and 1-yen coins.)

-----Constraints-----
 - 0 \leq X \leq 10^9
 - X is an integer.

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

-----Output-----
Print the maximum number of happiness points that can be earned.

-----Sample Input-----
1024

-----Sample Output-----
2020

By exchanging his money so that he gets two 500-yen coins and four 5-yen coins, he gains 2020 happiness points, which is the maximum number of happiness points that can be earned.
targets: def __starting_point():

    x = int(input())
    
    ans = (x // 500)
    if ans > 0:
        x = x - (500 * ans)
        ans *= 1000
    
    ans += (x // 5) * 5
    print(ans)
__starting_point()
inputs: I found an interesting problem on https://www.codewars.com/kata/5b713d7187c59b53e60000b0:
If we multiply the integer      `717 (n)` by `7 (k)`, the result will be equal to `5019`.

Consider all the possible ways that this last number may be split as a string and calculate their corresponding sum obtained by adding the substrings as integers. When we add all of them up,... surprise, we got the original number `717`:

```
Partitions as string                Total Sums
['5', '019']                        5 + 19 = 24
['50', '19']                       50 + 19 = 69
['501', '9']                      501 + 9 = 510
['5', '0', '19']                5 + 0 + 19 = 24
['5', '01', '9']                 5 + 1 + 9 = 15
['50', '1', '9']                50 + 1 + 9 = 60
['5', '0', '1', '9']         5 + 0 + 1 + 9 = 15
                            ____________________
                              Big Total:    717
                            ____________________
```
In fact, `717` is one of the few integers that has such property with a factor `k = 7`.

Changing the factor `k`, for example to `k = 3`, we may see that the integer `40104` fulfills this property.

Given an integer `start_value` and an integer `k`, output the smallest integer `n`, but higher than `start_value`, that fulfills the above explained properties.

If by chance, `start_value`, fulfills the property, do not return `start_value` as a result, only the next integer. Perhaps you may find this assertion redundant if you understood well the requirement of the kata: "output the smallest integer `n`, but higher than `start_value`"

The values for `k` in the input may be one of these: `3, 4, 5, 7`

### Features of the random tests

If you want to understand the style and features of the random tests, see the *Notes* at the end of these instructions.

The random tests are classified in three parts.

- Random tests each with one of the possible values of `k` and a random `start_value` in the interval `[100, 1300]`
- Random tests each with a `start_value` in a larger interval for each value of `k`, as follows:
 - for `k = 3`, a random `start value` in the range...

I tried it in Python, but could not do it. Can you solve it?
targets: def partitions(s):
    if s:
        for i in range(1, len(s)+1):
            for p in partitions(s[i:]):
                yield [s[:i]] + p
    else:
        yield []
def add_values (product_string): 
  product_list = partitions(product_string)
  return sum([sum (map(int,items)) for items in list(product_list)[:-1]])
  
def next_higher(start_value,k):
    product = (start_value+1) * k
    while add_values(str(product)) != start_value:
        start_value+=1
        product = start_value*k
    return start_value
inputs: I found an interesting problem on https://atcoder.jp/contests/abc143/tasks/abc143_c:
There are N slimes lining up from left to right. The colors of these slimes will be given as a string S of length N consisting of lowercase English letters. The i-th slime from the left has the color that corresponds to the i-th character of S.
Adjacent slimes with the same color will fuse into one larger slime without changing the color. If there were a slime adjacent to this group of slimes before fusion, that slime is now adjacent to the new larger slime.
Ultimately, how many slimes will be there?

-----Constraints-----
 - 1 \leq N \leq 10^5
 - |S| = N
 - S consists of lowercase English letters.

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

-----Output-----
Print the final number of slimes.

-----Sample Input-----
10
aabbbbaaca

-----Sample Output-----
5

Ultimately, these slimes will fuse into abaca.

I tried it in Python, but could not do it. Can you solve it?
targets: N = int(input())
S = str(input())
cnt = 1

for i in range(1,N):
    if S[i] == S[i-1]:
        pass
    else:
        cnt += 1

print(cnt)
inputs: Solve in Python:
The new £5 notes have been recently released in the UK and they've certainly became a sensation! Even those of us who haven't been carrying any cash around for a while, having given in to the convenience of cards, suddenly like to have some of these in their purses and pockets. But how many of them could you get with what's left from your salary after paying all bills? The programme that you're about to write will count this for you!

Given a salary and the array of bills, calculate your disposable income for a month and return it as a number of new £5 notes you can get with that amount. If the money you've got (or do not!) doesn't allow you to get any £5 notes return 0.

£££ GOOD LUCK! £££
targets: def get_new_notes(salary,bills):
    return max((salary-sum(bills))//5, 0)
inputs: Solve in Python:
A magic island Geraldion, where Gerald lives, has its own currency system. It uses banknotes of several values. But the problem is, the system is not perfect and sometimes it happens that Geraldionians cannot express a certain sum of money with any set of banknotes. Of course, they can use any number of banknotes of each value. Such sum is called unfortunate. Gerald wondered: what is the minimum unfortunate sum?


-----Input-----

The first line contains number n (1 ≤ n ≤ 1000) — the number of values of the banknotes that used in Geraldion. 

The second line contains n distinct space-separated numbers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ 10^6) — the values of the banknotes.


-----Output-----

Print a single line — the minimum unfortunate sum. If there are no unfortunate sums, print  - 1.


-----Examples-----
Input
5
1 2 3 4 5

Output
-1
targets: def __starting_point():

    n = int(input())
    a = [int(x) for x in input().split()]
    a.sort()

    if a[0] == 1:
        print(-1)
    else:
        print(1)
        

__starting_point()
inputs: Solve in Python:
Phoenix loves beautiful arrays. An array is beautiful if all its subarrays of length $k$ have the same sum. A subarray of an array is any sequence of consecutive elements.

Phoenix currently has an array $a$ of length $n$. He wants to insert some number of integers, possibly zero, into his array such that it becomes beautiful. The inserted integers must be between $1$ and $n$ inclusive. Integers may be inserted anywhere (even before the first or after the last element), and he is not trying to minimize the number of inserted integers.


-----Input-----

The input consists of multiple test cases. The first line contains an integer $t$ ($1 \le t \le 50$) — the number of test cases.

The first line of each test case contains two integers $n$ and $k$ ($1 \le k \le n \le 100$).

The second line of each test case contains $n$ space-separated integers ($1 \le a_i \le n$) — the array that Phoenix currently has. This array may or may not be already beautiful.


-----Output-----

For each test case, if it is impossible to create a beautiful array, print -1. Otherwise, print two lines.

The first line should contain the length of the beautiful array $m$ ($n \le m \le 10^4$). You don't need to minimize $m$.

The second line should contain $m$ space-separated integers ($1 \le b_i \le n$) — a beautiful array that Phoenix can obtain after inserting some, possibly zero, integers into his array $a$. You may print integers that weren't originally in array $a$.

If there are multiple solutions, print any. It's guaranteed that if we can make array $a$ beautiful, we can always make it with resulting length no more than $10^4$.


-----Example-----
Input
4
4 2
1 2 2 1
4 3
1 2 2 1
3 2
1 2 3
4 4
4 3 4 2

Output
5
1 2 1 2 1
4
1 2 2 1
-1
7
4 3 2 1 4 3 2


-----Note-----

In the first test case, we can make array $a$ beautiful by inserting the integer $1$ at index $3$ (in between the two existing $2$s). Now, all subarrays of length $k=2$ have the same sum $3$. There exists many other possible solutions, for example:   $2, 1, 2, 1, 2, 1$...
targets: for _ in range(int(input())):
    n, k = list(map(int, input().split()))
    arr = list(map(int, input().split()))
    if len(set(arr)) > k:
        print(-1)
    else:
        result = []
        temp = list(set(arr))
        for i in range(1, n + 1):
            if len(temp) == k:
                break
            if i not in temp:
                temp.append(i)
                
        for i in range(len(arr)):
            result.extend(temp)
        print(len(result))
        print(*result)
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/707/C:
Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such that you can construct a right triangle with segments of lengths corresponding to triple. Such triples are called Pythagorean triples.

For example, triples (3, 4, 5), (5, 12, 13) and (6, 8, 10) are Pythagorean triples.

Here Katya wondered if she can specify the length of some side of right triangle and find any Pythagorean triple corresponding to such length? Note that the side which length is specified can be a cathetus as well as hypotenuse.

Katya had no problems with completing this task. Will you do the same?


-----Input-----

The only line of the input contains single integer n (1 ≤ n ≤ 10^9) — the length of some side of a right triangle.


-----Output-----

Print two integers m and k (1 ≤ m, k ≤ 10^18), such that n, m and k form a Pythagorean triple, in the only line.

In case if there is no any Pythagorean triple containing integer n, print  - 1 in the only line. If there are many answers, print any of them.


-----Examples-----
Input
3

Output
4 5
Input
6

Output
8 10
Input
1

Output
-1
Input
17

Output
144 145
Input
67

Output
2244 2245


-----Note-----[Image]

Illustration for the first sample.

I tried it in Python, but could not do it. Can you solve it?
targets: n = int(input())
if n % 4 == 0:
    temp = n // 4
    m = temp * 3
    k = temp * 5
elif n % 2 == 0:
    n //=  2
    m = n**2 // 2
    k = m + 1
    m *= 2
    k *= 2
else:
    m = n**2 // 2
    k = m + 1
if 3 > n:
    print("-1")
else:
    print(m,k)
inputs: Solve in Python:
----- Statement -----

You need to find a string which has exactly K positions in it such that the character at that position comes alphabetically later than the character immediately after it. If there are many such strings, print the one which has the shortest length. If there is still a tie, print the string which comes the lexicographically earliest (would occur earlier in a dictionary).

-----Input-----
The first line contains the number of test cases T. Each test case contains an integer K (≤ 100).

-----Output-----
Output T lines, one for each test case, containing the required string.  Use only lower-case letters a-z.

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

-----Sample Output-----
ba
cba
targets: x = 'abcdefghijklmnopqrstuvwxyz'
for i in range(int(input())):
 num = int(input())
 r = ''
 while 1:
  r = x[num::-1]+r
  if num<26:
   break
  num-=25
 print(r)
inputs: Solve in Python:
Given a pattern and a string str, find if str follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str.

Example 1:


Input: pattern = "abba", str = "dog cat cat dog"
Output: true

Example 2:


Input:pattern = "abba", str = "dog cat cat fish"
Output: false

Example 3:


Input: pattern = "aaaa", str = "dog cat cat dog"
Output: false

Example 4:


Input: pattern = "abba", str = "dog dog dog dog"
Output: false

Notes:
You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.
targets: class Solution:
     def wordPattern(self, pattern, str):
         """
         :type pattern: str
         :type str: str
         :rtype: bool
         """
         return list(map(pattern.find, pattern)) == list(map(str.split().index, str.split()))
inputs: Solve in Python:
You are given two positive integers a and b.
Let x be the average of a and b.
Print x rounded up to the nearest integer.

-----Constraints-----
 - a and b are integers.
 - 1 \leq a, b \leq 100

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

-----Output-----
Print x rounded up to the nearest integer.

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

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

The average of 1 and 3 is 2.0, and it will be rounded up to the nearest integer, 2.
targets: #ABC082A
a,b = list(map(int,input().split()))
a+=b
print(((a+2-1)//2))
inputs: I found an interesting problem on https://www.codechef.com/ALGT2013/problems/TNMALG07:
In a bizarre game of chess ,knight was so drunk, that instead of his usual move he started walking straight. In every move Knight jumps on 2n steps forward (n is number of block that he had travelled so far from starting) but after that he has to take either 1 step forward or backward.
Now the Knight needs to get to position X so King (i.e. You) needs to decide the order of his backward or forward step in such a way that he can reach its destination in minimum number of steps. Remember he always travels in a straight line and the length of the board is infinite.

-----Input-----
The first line of the input contains an integer T denoting the number of test cases, for each test case enter value X ( i.e. destination)
Note : initially knight is at n = 1.

-----Output-----
For each test case the output should be string of numbers 1 & 2 where 1 denotes backward step and 2 denote the forward step 
Note : for no solution print 0.

-----Constraints-----
- 1 ≤ T ≤ 100
- 1 ≤ X ≤ 10^10

-----Example-----
Input:
2
17
10
Output:
2111
0

-----Explanation-----
Case 1 : starting from  n = 1 , knight moves to n = 3 ('2') , 5 ('1') , 9 ('1') , 17 ('1') i.e. string printed is 2 1 1 1
Case 2 : no solution is possible

I tried it in Python, but could not do it. Can you solve it?
targets: t=int(input())
getBin = lambda x: x >= 0 and str(bin(x))[2:] or "-" + str(bin(x))[3:]
while(t):
 n=int(input())
 if(n%2==0):
  print(0)
 else:
  lst=[]
  arr=getBin(n)
  for i in range(0,len(arr)-1):
   if(arr[i]=='1'):
    lst.append("2")
   if(arr[i]=='0'):
    lst.append("1")
  print("".join(lst))
 t=t-1
inputs: Solve in Python:
DropCaps means that the first letter of the starting word of the paragraph should be in caps and the remaining lowercase, just like you see in the newspaper. 

But for a change, let's do that for each and every word of the given String. Your task is to capitalize every word that has length greater than 2, leaving smaller words as they are.

*should work also on Leading and Trailing Spaces and caps.

```python
drop_cap('apple') => "Apple"
drop_cap('apple of banana'); => "Apple of Banana"
drop_cap('one   space'); => "One   Space" 
drop_cap('   space WALK   '); => "   Space Walk   " 
```

**Note:** you will be provided atleast one word and should take string as input and return string as output.
targets: import re

def drop_cap(s):
    return re.sub(r'\S{3,}', lambda m: m.group(0).title(), s)
inputs: Solve in Python:
Stepan has n pens. Every day he uses them, and on the i-th day he uses the pen number i. On the (n + 1)-th day again he uses the pen number 1, on the (n + 2)-th — he uses the pen number 2 and so on.

On every working day (from Monday to Saturday, inclusive) Stepan spends exactly 1 milliliter of ink of the pen he uses that day. On Sunday Stepan has a day of rest, he does not stend the ink of the pen he uses that day. 

Stepan knows the current volume of ink in each of his pens. Now it's the Monday morning and Stepan is going to use the pen number 1 today. Your task is to determine which pen will run out of ink before all the rest (that is, there will be no ink left in it), if Stepan will use the pens according to the conditions described above.


-----Input-----

The first line contains the integer n (1 ≤ n ≤ 50 000) — the number of pens Stepan has.

The second line contains the sequence of integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ 10^9), where a_{i} is equal to the number of milliliters of ink which the pen number i currently has.


-----Output-----

Print the index of the pen which will run out of ink before all (it means that there will be no ink left in it), if Stepan will use pens according to the conditions described above. 

Pens are numbered in the order they are given in input data. The numeration begins from one. 

Note that the answer is always unambiguous, since several pens can not end at the same time.


-----Examples-----
Input
3
3 3 3

Output
2

Input
5
5 4 5 4 4

Output
5



-----Note-----

In the first test Stepan uses ink of pens as follows:   on the day number 1 (Monday) Stepan will use the pen number 1, after that there will be 2 milliliters of ink in it;  on the day number 2 (Tuesday) Stepan will use the pen number 2, after that there will be 2 milliliters of ink in it;  on the day number 3 (Wednesday) Stepan will use the pen number 3, after that there will be 2 milliliters of ink in it;  on the day number 4 (Thursday) Stepan will use the pen number 1, after that there will be 1...
targets: import sys

def Min(x, y):
    if x > y:
        return y
    else:
        return x

def Gcd(x, y):
    if x == 0:
        return y
    else:
        return Gcd(y % x, x)

def Lcm(x, y):
    return x * y // Gcd(x, y)

n = int(input())
a = [int(i) for i in input().split()]
d = [int(0) for i in range(0, n)]

ok = 0

cur = 0

len = Lcm(7, n)

for i in range(0, 7 * n):
    if a[i % n] == 0 :
        print(i % n + 1)
        ok = 1
        break
    if cur != 6:
        a[i % n] -= 1
        d[i % n] += 1
    cur = (cur + 1) % 7

if ok == 0:
    k = 10**20

    for i in range(0, n):
        a[i] += d[i]
        if d[i] == 0: continue
        if a[i] % d[i] > 0:
            k = Min(k, a[i] // d[i])
        else:
            k = Min(k, a[i] // d[i] - 1)

    if k == 10**20:
        k = 0

    for i in range(0, n):
        a[i] -= k * d[i]

    iter = 0
    cur = 0

    while True:
        if a[iter] == 0:
            print(iter % n + 1)
            break
        else:
            if cur != 6:
                a[iter] -= 1
            cur = (cur + 1) % 7
            iter = (iter + 1) % n
inputs: Solve in Python:
## The story you are about to hear is true
Our cat, Balor, sadly died of cancer in 2015.

While he was alive, the three neighborhood cats Lou, Mustache Cat, and Raoul all recognized our house and yard as Balor's territory, and would behave respectfully towards him and each other when they would visit. 

But after Balor died, gradually each of these three neighborhood cats began trying to claim his territory as their own, trying to drive the others away by growling, yowling, snarling, chasing, and even fighting, when one came too close to another, and no human was right there to distract or extract one of them before the situation could escalate. 

It is sad that these otherwise-affectionate animals, who had spent many afternoons peacefully sitting and/or lying near Balor and each other on our deck or around our yard, would turn on each other like that. However, sometimes, if they are far enough away from each other, especially on a warm day when all they really want to do is pick a spot in the sun and lie in it, they will ignore each other, and once again there will be a Peaceable Kingdom.

## Your Mission
In this, the first and simplest of a planned trilogy of cat katas :-), all you have to do is determine whether the distances between any visiting cats are large enough to make for a peaceful afternoon, or whether there is about to be an altercation someone will need to deal with by carrying one of them into the house or squirting them with water or what have you.

As input your function will receive a list of strings representing the yard as a grid, and an integer representing the minimum distance needed to prevent problems (considering the cats' current states of sleepiness). A point with no cat in it will be represented by a "-" dash. Lou, Mustache Cat, and Raoul will be represented by an upper case L, M, and R respectively. At any particular time all three cats may be in the yard, or maybe two, one, or even none. 

If the number of cats in the yard is one or none, or if the distances between all cats are...
targets: from itertools import combinations
from math import hypot


def peaceful_yard(yard, min_distance):
    l, yard = len(yard[0]), "".join(yard)
    cats = (divmod(yard.index(c), l) for c in "LMR" if c in yard)
    distances = (hypot(x2-x1, y2-y1) for (x1, y1), (x2, y2) in combinations(cats, 2))
    return all(min_distance <= d for d in distances)
    

# one-liner
#peaceful_yard = lambda y, m: all(m <= d for d in (hypot(x2-x1, y2-y1) for (x1, y1), (x2, y2) in combinations((divmod(i, len(y[0])) for i, c in enumerate("".join(y)) if c != "-"), 2)))

# alternatives:
# more intuitive but much slower:
#    cats = [divmod(i, l) for i, c in enumerate(yard) if c != "-"]
# less readable but avoid imports:
#    distances = [((x2-x1)**2 + (y2-y1)**2)**0.5 for i, (x1, y1) in enumerate(cats, 1) for (x2, y2) in cats[i:]]
inputs: Solve in Python:
Allen has a LOT of money. He has $n$ dollars in the bank. For security reasons, he wants to withdraw it in cash (we will not disclose the reasons here). The denominations for dollar bills are $1$, $5$, $10$, $20$, $100$. What is the minimum number of bills Allen could receive after withdrawing his entire balance?


-----Input-----

The first and only line of input contains a single integer $n$ ($1 \le n \le 10^9$).


-----Output-----

Output the minimum number of bills that Allen could receive.


-----Examples-----
Input
125

Output
3

Input
43

Output
5

Input
1000000000

Output
10000000



-----Note-----

In the first sample case, Allen can withdraw this with a $100$ dollar bill, a $20$ dollar bill, and a $5$ dollar bill. There is no way for Allen to receive $125$ dollars in one or two bills.

In the second sample case, Allen can withdraw two $20$ dollar bills and three $1$ dollar bills.

In the third sample case, Allen can withdraw $100000000$ (ten million!) $100$ dollar bills.
targets: n=int(input())
c=0
c+=n//100
n=n%100
c+=n//20
n=n%20
c+=n//10
n=n%10
c+=n//5
n=n%5
c+=n//1
n=n%1
print(c)
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/975/C:
Ivar the Boneless is a great leader. He is trying to capture Kattegat from Lagertha. The war has begun and wave after wave Ivar's warriors are falling in battle.

Ivar has $n$ warriors, he places them on a straight line in front of the main gate, in a way that the $i$-th warrior stands right after $(i-1)$-th warrior. The first warrior leads the attack.

Each attacker can take up to $a_i$ arrows before he falls to the ground, where $a_i$ is the $i$-th warrior's strength.

Lagertha orders her warriors to shoot $k_i$ arrows during the $i$-th minute, the arrows one by one hit the first still standing warrior. After all Ivar's warriors fall and all the currently flying arrows fly by, Thor smashes his hammer and all Ivar's warriors get their previous strengths back and stand up to fight again. In other words, if all warriors die in minute $t$, they will all be standing to fight at the end of minute $t$.

The battle will last for $q$ minutes, after each minute you should tell Ivar what is the number of his standing warriors.


-----Input-----

The first line contains two integers $n$ and $q$ ($1 \le n, q \leq 200\,000$) — the number of warriors and the number of minutes in the battle.

The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) that represent the warriors' strengths.

The third line contains $q$ integers $k_1, k_2, \ldots, k_q$ ($1 \leq k_i \leq 10^{14}$), the $i$-th of them represents Lagertha's order at the $i$-th minute: $k_i$ arrows will attack the warriors.


-----Output-----

Output $q$ lines, the $i$-th of them is the number of standing warriors after the $i$-th minute.


-----Examples-----
Input
5 5
1 2 1 2 1
3 10 1 1 1

Output
3
5
4
4
3

Input
4 4
1 2 3 4
9 1 10 6

Output
1
4
4
1



-----Note-----

In the first example:   after the 1-st minute, the 1-st and 2-nd warriors die.  after the 2-nd minute all warriors die (and all arrows left over are wasted), then they will be revived thus answer is 5 — all warriors are alive.  after the 3-rd minute, the 1-st warrior...

I tried it in Python, but could not do it. Can you solve it?
targets: #!/usr/bin/env python3

import bisect

[n, q] = list(map(int, input().strip().split()))
ais = list(map(int, input().strip().split()))
kis = list(map(int, input().strip().split()))

iais = [0 for _ in range(n + 1)]
for i in range(n):
	iais[i + 1] = iais[i] + ais[i]


s = 0
tot = iais[-1]
r = 0
for k in kis:
	s += k
	if s >= tot:
		print (n)
		s = 0
		r = 0
	else:
		r = bisect.bisect_right(iais, s, r) - 1
		print(n - r)
inputs: Solve in Python:
There are N children in AtCoder Kindergarten. Mr. Evi will arrange the children in a line, then give 1 candy to the first child in the line, 2 candies to the second child, ..., N candies to the N-th child. How many candies will be necessary in total?

-----Constraints-----
 - 1≦N≦100

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

-----Output-----
Print the necessary number of candies in total.

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

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

The answer is 1+2+3=6.
targets: n = int(input())
print((n*(n+1)//2))
inputs: Solve in Python:
Little Robber Girl likes to scare animals in her zoo for fun. She decided to arrange the animals in a row in the order of non-decreasing height. However, the animals were so scared that they couldn't stay in the right places.

The robber girl was angry at first, but then she decided to arrange the animals herself. She repeatedly names numbers l and r such that r - l + 1 is even. After that animals that occupy positions between l and r inclusively are rearranged as follows: the animal at position l swaps places with the animal at position l + 1, the animal l + 2 swaps with the animal l + 3, ..., finally, the animal at position r - 1 swaps with the animal r.

Help the robber girl to arrange the animals in the order of non-decreasing height. You should name at most 20 000 segments, since otherwise the robber girl will become bored and will start scaring the animals again.


-----Input-----

The first line contains a single integer n (1 ≤ n ≤ 100) — number of animals in the robber girl's zoo.

The second line contains n space-separated integers a_1, a_2, ..., a_{n} (1 ≤ a_{i} ≤ 10^9), where a_{i} is the height of the animal occupying the i-th place.


-----Output-----

Print the sequence of operations that will rearrange the animals by non-decreasing height.

The output should contain several lines, i-th of the lines should contain two space-separated integers l_{i} and r_{i} (1 ≤ l_{i} < r_{i} ≤ n) — descriptions of segments the robber girl should name. The segments should be described in the order the operations are performed.

The number of operations should not exceed 20 000.

If the animals are arranged correctly from the start, you are allowed to output nothing.


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

Output
1 4

Input
7
36 28 57 39 66 69 68

Output
1 4
6 7

Input
5
1 2 1 2 1

Output
2 5
3 4
1 4
1 4



-----Note-----

Note that you don't have to minimize the number of operations. Any solution that performs at most 20 000 operations is allowed.
targets: n = int(input())
a = list(map(int, input().split()))

swaps = []
for i in range(n):
    # find the value at element i
    smallest = a[i]
    best_pos = i
    for j in range(i + 1, n):
        if a[j] < smallest:
            smallest = a[j]
            best_pos = j
    # swap from i to j
    while best_pos > i:
        a[best_pos - 1], a[best_pos] = a[best_pos], a[best_pos - 1]
        swaps.append((best_pos, best_pos + 1))
        best_pos -= 1

for a, b in swaps:
    print(a, b)
inputs: I found an interesting problem on https://www.codewars.com/kata/580a0347430590220e000091:
Find the area of a rectangle when provided with one diagonal and one side of the rectangle. If the input diagonal is less than or equal to the length of the side, return "Not a rectangle". If the resultant area has decimals round it to two places.

`This kata is meant for beginners. Rank and upvote to bring it out of beta!`

I tried it in Python, but could not do it. Can you solve it?
targets: def area(d, l):
    if d <= l: return 'Not a rectangle'
    a = l * (d ** 2 - l ** 2) ** 0.5
    return round(a, 2) if a % 1 else int(a)
inputs: I found an interesting problem on https://www.codechef.com/problems/URCALC:
Write a program to obtain 2 numbers $($ $A$ $and$ $B$ $)$ and an arithmetic operator $(C)$ and then design a $calculator$ depending upon the operator entered by the user.
So for example if C="+", you have to sum the two numbers.
If C="-", you have to subtract the two numbers.
If C=" * ", you have to print the product.
If C=" / ", you have to divide the two numbers.

-----Input:-----
- First line will contain the first number $A$.
- Second line will contain the second number $B$.
- Third line will contain the operator $C$, that is to be performed on A and B.

-----Output:-----
Output a single line containing the answer, obtained by, performing the operator on the numbers. Your output will be considered to be correct if the difference between your output and the actual answer is not more than $10^{-6}$.

-----Constraints-----
- $-1000 \leq A \leq 1000$
- $-1000 \leq B \leq 1000$ $and$ $B \neq 0$
- $C$ $can$ $only$ $be$ $one$ $of$ $these$ $4$ $operators$ {" + ", " - ", " * ", " / "}

-----Sample Input:-----
8
2
/

-----Sample Output:-----
4.0

-----Sample Input:-----
5
3
+

-----Sample Output:-----
8

I tried it in Python, but could not do it. Can you solve it?
targets: # cook your dish here
a=int(input())
b=int(input())
c=input()
if c=="+":
 print(a+b)
elif c=="-":
 print(a-b)
elif c=="*":
 print(a*b)
elif c=="/":
 print(a/b)
inputs: I found an interesting problem on https://atcoder.jp/contests/abc173/tasks/abc173_e:
Given are N integers A_1,\ldots,A_N.
We will choose exactly K of these elements. Find the maximum possible product of the chosen elements.
Then, print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

-----Constraints-----
 - 1 \leq K \leq N \leq 2\times 10^5
 - |A_i| \leq 10^9

-----Input-----
Input is given from Standard Input in the following format:
N K
A_1 \ldots A_N

-----Output-----
Print the maximum product modulo (10^9+7), using an integer between 0 and 10^9+6 (inclusive).

-----Sample Input-----
4 2
1 2 -3 -4

-----Sample Output-----
12

The possible products of the two chosen elements are 2, -3, -4, -6, -8, and 12, so the maximum product is 12.

I tried it in Python, but could not do it. Can you solve it?
targets: # -*- coding: utf-8 -*-
import sys

def main():
    N,K = list(map(int, sys.stdin.readline().split()))
    A_list = list(map(int, sys.stdin.readline().split()))


    A_list.sort(key=lambda x: -abs(x))
    mod = 10**9 + 7

    L_plus  = R_plus  = -1 # the Index of plus value
    L_minus = R_minus = -1 # the Index of minus value
    # L : between index 0 to (K-1)
    # R : between index K to end


    for i in range(K-1, -1, -1):
        if (L_plus == -1) and (A_list[i] >= 0):
            L_plus = i
        if (L_minus == -1) and (A_list[i] < 0):
            L_minus = i
        if (L_plus != -1) and (L_minus != -1):
            break
    
    for i in range(K,N):
        if (R_plus == -1) and (A_list[i] >= 0):
            R_plus = i
        if (R_minus == -1) and (A_list[i] < 0):
            R_minus = i
        if (R_plus != -1) and (R_minus != -1):
            break


    cnt_minus = 0

    for i in range(K):
        if A_list[i] < 0:
            cnt_minus += 1


    if cnt_minus % 2 == 0:
        target_idx = [0, K-1]
    
    else: # cnt_minus % 2 != 0
        if (R_plus == -1) and \
           ( (L_plus == -1) or (R_minus == -1) ) :
        
            calc1 = calc2 = 0
        
        elif (R_plus != -1) and \
           ( (L_plus == -1) or (R_minus == -1) ) :
        
            calc1 = 1
            calc2 = 0
        
        elif (R_plus == -1) and \
           ( (L_plus != -1) and (R_minus != -1) ) :
        
            calc1 = 0
            calc2 = 1
        
        elif (R_plus != -1) and \
           ( (L_plus != -1) and (R_minus != -1) ) :
            
            calc1 = A_list[L_plus] * A_list[R_plus]
            calc2 = A_list[L_minus] * A_list[R_minus]


        if calc1 == calc2:
            target_idx = [N-K, N-1]
        
        elif calc1 > calc2:
            A_list[L_minus], A_list[R_plus] = A_list[R_plus], A_list[L_minus]
            target_idx = [0, K-1]

        elif calc1 < calc2:
            A_list[L_plus], A_list[R_minus] = A_list[R_minus], A_list[L_plus]
           ...
inputs: I found an interesting problem on https://leetcode.com/problems/maximum-length-of-pair-chain/:
You are given n pairs of numbers. In every pair, the first number is always smaller than the second number.



Now, we define a pair (c, d) can follow another pair (a, b) if and only if b < c. Chain of pairs can be formed in this fashion. 



Given a set of pairs, find the length longest chain which can be formed. You needn't use up all the given pairs. You can select pairs in any order.



Example 1:

Input: [[1,2], [2,3], [3,4]]
Output: 2
Explanation: The longest chain is [1,2] -> [3,4]



Note:

The number of given pairs will be in the range [1, 1000].

I tried it in Python, but could not do it. Can you solve it?
targets: import operator
 class Solution:
     def findLongestChain(self, pairs):
         """
         :type pairs: List[List[int]]
         :rtype: int
         """
         pairs.sort(key=operator.itemgetter(1, 0))
         maxNum = -float('inf')
         ret = 0
         for pair in pairs:
             if pair[0] > maxNum:
                 maxNum = pair[1]
                 ret += 1
         return ret
inputs: I found an interesting problem on https://www.codechef.com/problems/GUESSNUM:
Chef is playing a game with his brother Chefu. He asked Chefu to choose a positive integer $N$, multiply it by a given integer $A$, then choose a divisor of $N$ (possibly $N$ itself) and add it to the product. Let's denote the resulting integer by $M$; more formally, $M = A \cdot N + d$, where $d$ is some divisor of $N$.
Chefu told Chef the value of $M$ and now, Chef should guess $N$. Help him find all values of $N$ which Chefu could have chosen.

-----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 two space-separated integers $A$ and $M$.

-----Output-----
For each test case, print two lines. The first line should contain a single integer $C$ denoting the number of possible values of $N$. The second line should contain $C$ space-separated integers denoting all possible values of $N$ in increasing order.
It is guaranteed that the sum of $C$ over all test cases does not exceed $10^7$.

-----Constraints-----
- $1 \le T \le 100$
- $2 \le M \le 10^{10}$
- $1 \le A < M$

-----Subtasks-----
Subtask #1 (50 points):
- $M \le 10^6$
Subtask #2 (50 points): original constraints

-----Example Input-----
3
3 35
5 50
4 65

-----Example Output-----
1
10
0

3
13 15 16

I tried it in Python, but could not do it. Can you solve it?
targets: T = int(input())
import math
for i in range(T):
 list = []
 A,M = input().split()
 A = int(A)
 M = int(M)
 L1 = M//A
 L2 = math.ceil(M/(A+1))
 while L2 <= L1 :
  N = L2
  if M - N*A !=0 and N%(M - N*A) == 0 :
   list.append(L2)
  L2 += 1

 print(len(list))
 if len(list) > 0:
  for j in range(len(list)-1):
   print(list[j],end=" ")
  print(list[len(list)-1])
 else:
  print(" ")
inputs: Solve in Python:
The look and say sequence is a sequence in which each number is the result of a "look and say" operation on the previous element.

Considering for example the classical version startin with `"1"`: `["1", "11", "21, "1211", "111221", ...]`. You can see that the second element describes the first as `"1(times number)1"`, the third is `"2(times number)1"` describing the second, the fourth is `"1(times number)2(and)1(times number)1"` and so on.

Your goal is to create a function which takes a starting string (not necessarily the classical `"1"`, much less a single character start) and return the nth element of the series.

## Examples

```python
look_and_say_sequence("1", 1)   == "1"
look_and_say_sequence("1", 3)   == "21"
look_and_say_sequence("1", 5)   == "111221"
look_and_say_sequence("22", 10) == "22"
look_and_say_sequence("14", 2)  == "1114"
```
Trivia: `"22"` is the only element that can keep the series constant.
targets: from itertools import groupby

def look_and_say_sequence(seq, n):
    if seq == "22":
        return "22"
    for _ in range(1, n):
        seq = "".join(f"{len(list(b))}{a}" for a, b in groupby(seq))
    return seq
inputs: Solve in Python:
# Task
 A common way for prisoners to communicate secret messages with each other is to encrypt them. One such encryption algorithm goes as follows.

 You take the message and place it inside an `nx6` matrix (adjust the number of rows depending on the message length) going from top left to bottom right (one row at a time) while replacing spaces with dots (.) and adding dots at the end of the last row (if necessary) to complete the matrix.
 
 Once the message is in the matrix you read again from top left to bottom right but this time going one column at a time and treating each column as one word.

# Example

 The following message `"Attack at noon or we are done for"` is placed in a `6 * 6` matrix :
```
Attack
.at.no
on.or.
we.are
.done.
for...```
Reading it one column at a time we get:

 `A.ow.f tanedo tt..or a.oan. cnrre. ko.e..`

# Input/Output


 - `[input]` string `msg`

  a regular english sentance representing the original message


 - `[output]` a string

  encrypted message
targets: from itertools import zip_longest

def six_column_encryption(msg):
    return ' '.join(map(''.join,
        zip_longest(*(msg[i:i+6].replace(' ', '.') for i in range(0, len(msg), 6)), fillvalue='.')
    ))
inputs: I found an interesting problem on https://codeforces.com/problemset/problem/1203/D2:
The only difference between easy and hard versions is the length of the string.

You are given a string $s$ and a string $t$, both consisting only of lowercase Latin letters. It is guaranteed that $t$ can be obtained from $s$ by removing some (possibly, zero) number of characters (not necessary contiguous) from $s$ without changing order of remaining characters (in other words, it is guaranteed that $t$ is a subsequence of $s$).

For example, the strings "test", "tst", "tt", "et" and "" are subsequences of the string "test". But the strings "tset", "se", "contest" are not subsequences of the string "test".

You want to remove some substring (contiguous subsequence) from $s$ of maximum possible length such that after removing this substring $t$ will remain a subsequence of $s$.

If you want to remove the substring $s[l;r]$ then the string $s$ will be transformed to $s_1 s_2 \dots s_{l-1} s_{r+1} s_{r+2} \dots s_{|s|-1} s_{|s|}$ (where $|s|$ is the length of $s$).

Your task is to find the maximum possible length of the substring you can remove so that $t$ is still a subsequence of $s$.


-----Input-----

The first line of the input contains one string $s$ consisting of at least $1$ and at most $2 \cdot 10^5$ lowercase Latin letters.

The second line of the input contains one string $t$ consisting of at least $1$ and at most $2 \cdot 10^5$ lowercase Latin letters.

It is guaranteed that $t$ is a subsequence of $s$.


-----Output-----

Print one integer — the maximum possible length of the substring you can remove so that $t$ is still a subsequence of $s$.


-----Examples-----
Input
bbaba
bb

Output
3

Input
baaba
ab

Output
2

Input
abcde
abcde

Output
0

Input
asdfasdf
fasd

Output
3

I tried it in Python, but could not do it. Can you solve it?
targets: S = input()
T = input()
N, M = len(S), len(T)

def calc(s, t):
    X = [0] * len(s)
    j = 0
    for i in range(len(s)):
        if j >= len(t):
            X[i] = j
        elif s[i] == t[j]:
            X[i] = j+1
            j += 1
        else:
            X[i] = X[i-1]
    return [0] + X

        
A, B = calc(S, T), calc(S[::-1], T[::-1])[::-1]
# print(A, B)
l, r = 0, N
while r-l>1:
    m = (l+r)//2
    C = [A[i]+B[i+m] for i in range(N-m+1)]
    # print("m, C =", m, C)
    if max(C) >= M:
        l = m
    else:
        r = m
print(l)
inputs: Solve in Python:
One upon a time there were three best friends Abhinav, Harsh, and Akash decided to form a 
team and take part in ICPC from KIIT. Participants are usually offered several problems during 
the programming contest. Long before the start, the friends decided that they will implement a 
problem if at least two of them are sure about the solution. Otherwise, friends won't write the 
problem's solution. 
This contest offers $N$ problems to the participants. For each problem we know, which friend is 
sure about the solution. Help the KIITians find the number of problems for which they will write a 
solution. 
Then n lines contain three integers each, each integer is either 0 or 1. If the first number in the 
line equals 1, then Abhinav is sure about the problem's solution, otherwise, he isn't sure. The 
second number shows Harsh's view on the solution, the third number shows Akash's view. The 
numbers on the lines are 

-----Input:-----
- A single integer will contain $N$, number of problems. 

-----Output:-----
Print a single integer — the number of problems the friends will implement on the contest. 

-----Constraints-----
- $1 \leq N \leq 1000$ 

-----Sample Input:-----
3

1 1 0

1 1 1

1 0 0   

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

-----EXPLANATION:-----
In the first sample, Abhinav and Harsh are sure that they know how to solve the first problem 
and all three of them know how to solve the second problem. That means that they will write 
solutions for these problems. Only Abhinav is sure about the solution for the third problem, but 
that isn't enough, so the group won't take it.
targets: # cook your dish here
n = int(input())
s = 0
for i in range(n):
 lst = list(map(int,input().split()))
 if sum(lst)>=2:
  s += 1
print(s)
inputs: Solve in Python:
A restaurant received n orders for the rental. Each rental order reserve the restaurant for a continuous period of time, the i-th order is characterized by two time values — the start time l_{i} and the finish time r_{i} (l_{i} ≤ r_{i}).

Restaurant management can accept and reject orders. What is the maximal number of orders the restaurant can accept?

No two accepted orders can intersect, i.e. they can't share even a moment of time. If one order ends in the moment other starts, they can't be accepted both.


-----Input-----

The first line contains integer number n (1 ≤ n ≤ 5·10^5) — number of orders. The following n lines contain integer values l_{i} and r_{i} each (1 ≤ l_{i} ≤ r_{i} ≤ 10^9).


-----Output-----

Print the maximal number of orders that can be accepted.


-----Examples-----
Input
2
7 11
4 7

Output
1

Input
5
1 2
2 3
3 4
4 5
5 6

Output
3

Input
6
4 8
1 5
4 7
2 5
1 3
6 8

Output
2
targets: r = lambda: map(int, input().split())

n, = r()
a = sorted(tuple(r()) for _ in range(n))[::-1]

ret, left = 0, 10**9 + 1
for l, r in a:
    if r < left:
        ret += 1
        left = l

print(ret)