Problem:


Score : 500 points

Problem Statement

We have N logs of lengths A_1,A_2,\cdots A_N.

We can cut these logs at most K times in total. When a log of length L is cut at a point whose distance from an end of the log is t (0<t<L), it becomes two logs of lengths t and L-t.

Find the shortest possible length of the longest log after at most K cuts, and print it after rounding up to an integer.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq K \leq 10^9

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \cdots A_N

Output

Print an integer representing the answer.

Sample Input 1

2 3
7 9

Sample Output 1

4

First, we will cut the log of length 7 at a point whose distance from an end of the log is 3.5, resulting in two logs of length 3.5 each.

Next, we will cut the log of length 9 at a point whose distance from an end of the log is 3, resulting in two logs of length 3 and 6.

Lastly, we will cut the log of length 6 at a point whose distance from an end of the log is 3.3, resulting in two logs of length 3.3 and 2.7.

In this case, the longest length of a log will be 3.5, which is the shortest possible result. After rounding up to an integer, the output should be 4.

Sample Input 2

3 0
3 4 5

Sample Output 2

5

Sample Input 3

10 10
158260522 877914575 602436426 24979445 861648772 623690081 433933447 476190629 262703497 211047202

Sample Output 3

292638192



Sample input:
2 3
7 9


Sample output:
4

Submission status: Time Limit Exceeded

Solution code:
from math import ceil, floor
n,k = map(int,input().split())
a = list(map(int,input().split()))
l = 0
r = 10**9
m = (l+r)//2+1
for i in range(100):
    # print(l,m,r)
    cnt = 0
    for j in a:
        cnt += (ceil(j/m)-1)
    # print(cnt)
    # print(l,m,r,cnt)
    if cnt <= k:
        r = m
        m = ceil((l+r)/2)
    else:
        l = m
        m = ceil((l+r)/2)
print(m)

Submission status: Time Limit Exceeded

Solution code:
import heapq
import math

N,K=input().split()
N,K=int(N),int(K)
_A=input().split()
A=[]

for a in _A:
    A.append(int(a))

t=[]
for idx,i in enumerate(A):
    t.append((-i,idx))
heapq.heapify(t)

toubun=[1]*N

for i in range(K):
    a=heapq.heappop(t)
#    print(a)
    idx=a[1]
    l=a[0]
    toubun[idx]+=1
    heapq.heappush(t,(-A[idx]/toubun[idx],idx))

print(math.ceil(-heapq.heappop(t)[0]))

Submission status: Time Limit Exceeded

Solution code:
import sys
import heapq
from math import ceil


def _ia(): return map(int, sys.stdin.readline().strip().split())


def main():
    _, k = _ia()
    a = list(map(lambda x: [-x, 0], _ia()))
    heapq.heapify(a)
    for _ in range(k):
        l, nn = heapq.heappop(a)
        if -l < 1:
            break
        nn += 1
        heapq.heappush(a, [l * nn / (nn + 1), nn])

    ret, _ = heapq.heappop(a)
    return ceil(-ret)


if __name__ == "__main__":
    print(main())

Submission status: Time Limit Exceeded

Solution code:
def resolve():
    import heapq
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    q = []
    for index, i in enumerate(a):
        heapq.heappush(q, (-i, index, 1))
    for _ in range(k):
        length, index, count = heapq.heappop(q)
        heapq.heappush(q, (-((a[index] + count) //
                             (count + 1)), index, count + 1))
    length, index, count = heapq.heappop(q)
    print((a[index] + count - 1) // count)


if __name__ == '__main__':
    resolve()

Submission status: Time Limit Exceeded

Solution code:
import heapq 
import math

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

lst = []
for i in range(n):
  lst.append([-a[i], 1])
  
heapq.heapify(lst)

for i in range(k):
  pop = heapq.heappop(lst)
  pop = [pop[0]*(pop[1])/(pop[1]+1), pop[1]+1]
  heapq.heappush(lst, pop)
  
pop = heapq.heappop(lst)
print(math.ceil(-pop[0]))

Submission status: Time Limit Exceeded

Solution code:
from math import ceil
from heapq import heapify, heappop, heappush

N, K, *A = map(int, open(0).read().split())

a = [(-i, i, 1) for i in A]
heapify(a)
for _ in range(K):
    i, j, k = heappop(a)
    k += 1
    i = -(j / k)
    heappush(a, (i, j, k))
print(ceil(heappop(a)[0] * -1))

Submission status: Time Limit Exceeded

Solution code:
from math import ceil

N, K = map(int, input().split())
A = list(map(int, input().split()))
# N, K = 2*10**5, 1
# A = [10**9]*N

def solve(x):
    m = 0
    for a in A:
        m += a//x - (a%x == 0)
    # print(x, m)
    return m <= K

ok, ng = max(A), 0
while abs(ok-ng > 0.1**6):
    mid = (ok+ng)/2
    # print(mid)
    if solve(mid):
        ok = mid
    else:
        ng = mid

print(ceil(mid))

Submission status: Time Limit Exceeded

Solution code:
from decimal import Decimal

n, k = map(int, input().split())
a = list(map(Decimal, input().split()))

for i in range(n):
  a[i] = [(-1) * a[i], (-1) * a[i], 1]

import heapq
heapq.heapify(a)

for _ in range(k):
  i = heapq.heappop(a)
  i[2] += 1
  i[0] = i[1] / i[2]
  heapq.heappush(a, i)

from math import ceil
print(ceil((-1) * heapq.heappop(a)[0]))

Submission status: Time Limit Exceeded

Solution code:
def main(sample_file = ''):

    """ convenient functions
    # for i, a in enumerate(iterable)
    # q, mod = divmod(a, b)
    # divmod(x, y) returns the tuple (x//y, x%y)
    # Higher-order function: reduce(operator.mul, xyz_count, 1)
    # manage median(s) using two heapq https://atcoder.jp/contests/abc127/tasks/abc127_f
    """

    """convenient decorator
    # @functools.lru_cache():
    # to facilitate use of recursive function
        # ex:
        # from functools import lru_cache
        # import sys
        # sys.setrecursionlimit(10**9)
        # @lru_cache(maxsize=None)
        # def fib(n):
        #     if n < 2:
        #         return n
        #     return fib(n-1) + fib(n-2)
        # print(fib(1000))
    """
    
    # import numpy as np
    import sys
    sys.setrecursionlimit(10**7)
    from itertools import accumulate, combinations, permutations, product # https://docs.python.org/ja/3/library/itertools.html
    # accumulate() returns iterator! to get list: list(accumulate())
    from math import factorial, ceil, floor, sqrt
    def factorize(n):
        """return the factors of the Arg and count of each factor
        
        Args:
            n (long): number to be resolved into factors
        
        Returns:
            list of tuples: factorize(220) returns [(2, 2), (5, 1), (11, 1)]
        """
        fct = []  # prime factor
        b, e = 2, 0  # base, exponent
        while b * b <= n:
            while n % b == 0:
                n = n // b
                e = e + 1
            if e > 0:
                fct.append((b, e))
            b, e = b + 1, 0
        if n > 1:
            fct.append((n, 1))
        return fct
    def combinations_count(n, r):   
        """Return the number of selecting r pieces of items from n kinds of items.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        # TODO: How should I do when n - r is negative?
        if n < 0 or r < 0:
            raise Exception('combinations_count(n, r) not defined when n or r is negative')
        if n - r < r: r = n - r
        if r < 0: return 0
        if r == 0: return 1
        if r == 1: return n
        numerator = [n - r + k + 1 for k in range(r)]
        denominator = [k + 1 for k in range(r)]
        for p in range(2,r+1):
            pivot = denominator[p - 1]
            if pivot > 1:
                offset = (n - r) % p
                for k in range(p-1,r,p):
                    numerator[k - offset] /= pivot
                    denominator[k] /= pivot
        result = 1
        for k in range(r):
            if numerator[k] > 1:
                result *= int(numerator[k])
        return result
    def combinations_with_replacement_count(n, r):
        """Return the number of selecting r pieces of items from n kinds of items allowing individual elements to be repeated more than once.
        
        Args:
            n (long): number
            r (long): number
        
        Raises:
            Exception: not defined when n or r is negative
        
        Returns:
            long: number
        """
        if n < 0 or r < 0:
            raise Exception('combinations_with_replacement_count(n, r) not defined when n or r is negative')
        elif n == 0:
            return 1
        else:
            return combinations_count(n + r - 1, r)
    from bisect import bisect_left, bisect_right
    from collections import deque, Counter, defaultdict # https://docs.python.org/ja/3/library/collections.html#collections.deque
    from heapq import heapify, heappop, heappush, heappushpop, heapreplace,nlargest,nsmallest # https://docs.python.org/ja/3/library/heapq.html
    from copy import deepcopy, copy # https://docs.python.org/ja/3/library/copy.html
    import operator
    from operator import itemgetter #sort 
    # ex1: List.sort(key=itemgetter(1))
    # ex2: sorted(tuples, key=itemgetter(1,2))
    from functools import reduce, lru_cache
    def chmin(x, y):
        """change minimum
        if x > y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current minimum value
            y (long): potential minimum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x > y:
            x = y
            return (x, True)
        else:
            return (x, False)
    def chmax(x, y):
        """change maximum
        if x < y, x = y and return (x, True).
        convenient when solving problems of dp[i]
        
        Args:
            x (long): current maximum value
            y (long): potential maximum value
        
        Returns:
            (x, bool): (x, True) when updated, else (x, False)
        """
        if x < y:
            x = y
            return (x, True)
        else:
            return (x, False)

    from fractions import gcd # Deprecated since version 3.5: Use math.gcd() instead.
    def gcds(numbers):
        return reduce(gcd, numbers)
    def lcm(x, y):
        return (x * y) // gcd(x, y)
    def lcms(numbers):
        return reduce(lcm, numbers, 1)
    def make_divisors(n, reversed=False):
        """create list of divisors
        
        Args:
            number (int): number from which list of divisors is created
            reversed (bool, optional): ascending order if False. Defaults to False.
        
        Returns:
            list: list of divisors
        """
        divisors = set()
        for i in range(1, int(n**0.5)+1):
            if n % i == 0:
                divisors.add(i)
                divisors.add(n//i)
        return sorted(list(divisors),reverse=reversed)
    
    # first create factorial_list
    # fac_list = modfactorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    AtoZ = list('abcdefghijklmnopqrsruvwxyz') # 26
    modpow = lambda a, n, p = MOD: pow(a, n, p) # Recursive function in python is slow!
    def modinv(a, p = MOD):
        # evaluate reciprocal using Fermat's little theorem:
        # a**(p-1) is identical to 1 (mod p) when a and p is coprime
        return modpow(a, p-2, p)
    def modinv_list(n, p = MOD):
        if n <= 1:
            return [0,1][:n+1]
        else:
            inv_t = [0,1]
            for i in range(2, n+1):
                inv_t += [inv_t[p % i] * (p - int(p / i)) % p]
            return inv_t
    def modfactorial_list(n, p = MOD):
        if n == 0:
            return [1]
        else:
            l = [0] * (n+1)
            tmp = 1
            for i in range(1, n+1):
                tmp = tmp * i % p
                l[i] = tmp
            return l
    def modcomb(n, k, fac_list = [], p = MOD):
        # fac_list = modfactorial_list(100)
        # print(modcomb(100, 5, modfactorial_list(100)))
        from math import factorial
        if n < 0 or k < 0 or n < k: return 0
        if n == 0 or k == 0: return 1
        if len(fac_list) <= n:
            a = factorial(n) % p
            b = factorial(k) % p
            c = factorial(n-k) % p
        else:
            a = fac_list[n]
            b = fac_list[k]
            c = fac_list[n-k]
        return (a * modpow(b, p-2, p) * modpow(c, p-2, p)) % p
    def modadd(a, b, p = MOD):
        return (a + b) % MOD
    def modsub(a, b, p = MOD):
        return (a - b) % p
    def modmul(a, b, p = MOD):
        return ((a % p) * (b % p)) % p
    def moddiv(a, b, p = MOD):
        return modmul(a, modpow(b, p-2, p))

    class UnionFindTree:
        """union find tree class
        TODO: fix this description...
        how to use (example):
        >>  uf = UnionFindTree(N) 
        >>  if uf.find_root(a) == uf.find_root(b):
        >>      do something
        >>  else:
        >>      do something
        >>      uf.unite(a, b)
        """
        def __init__(self, N):
            self.root = [-1] * (N+1)
            self.rank   = [0]  * (N+1)
            self.connected_num = [1] * (N+1)
    
        def find_root(self,x):
            root = self.root
            while root[x] != -1:
                x = root[x]
            return x
    
        def unite(self,x,y):
            root = self.root
            rank = self.rank
            connected_num = self.connected_num
            find_root = self.find_root
    
            rx = find_root(x)
            ry = find_root(y)
            if rx != ry:
                if rank[rx] < rank[ry]:
                    root[rx] = ry
                    rx,ry = ry,rx
                else:
                    if rank[rx] == rank[ry]:
                        rank[rx] += 1
                    root[ry] = rx       
                connected_num[rx] += connected_num[ry]

    class SegTree:
        """
        ref: https://qiita.com/takayg1/items/c811bd07c21923d7ec69
        init(init_val, ide_ele): 配列init_valで初期化 O(N)
        update(k, x): k番目の値をxに更新 O(logN)
        query(l, r): 区間[l, r)をsegfuncしたものを返す O(logN)
        """
        def __init__(self, init_val, segfunc, ide_ele):
            """
            init_val: 配列の初期値
            segfunc: 区間にしたい操作
            ide_ele: 単位元
            n: 要素数
            num: n以上の最小の2のべき乗
            tree: セグメント木(1-index)
            """
            n = len(init_val)
            self.segfunc = segfunc
            self.ide_ele = ide_ele
            self.num = 1 << (n - 1).bit_length()
            self.tree = [ide_ele] * 2 * self.num
            # 配列の値を葉にセット
            for i in range(n):
                self.tree[self.num + i] = init_val[i]
            # 構築していく
            for i in range(self.num - 1, 0, -1):
                self.tree[i] = self.segfunc(self.tree[2 * i], self.tree[2 * i + 1])

        def update(self, k, x):
            """
            k番目の値をxに更新
            k: index(0-index)
            x: update value
            """
            k += self.num
            self.tree[k] = x
            while k > 1:
                self.tree[k >> 1] = self.segfunc(self.tree[k], self.tree[k ^ 1])
                k >>= 1

        def query(self, l, r):
            """
            [l, r)のsegfuncしたものを得る
            l: index(0-index)
            r: index(0-index)
            """
            res = self.ide_ele
            l += self.num
            r += self.num
            while l < r:
                if l & 1:
                    res = self.segfunc(res, self.tree[l])
                    l += 1
                if r & 1:
                    res = self.segfunc(res, self.tree[r - 1])
                l >>= 1
                r >>= 1
            return res

    # ************ Graph ************
    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # 最短経路問題: https://ja.wikipedia.org/wiki/%E6%9C%80%E7%9F%AD%E7%B5%8C%E8%B7%AF%E5%95%8F%E9%A1%8C
    
    # Bellman-Ford: O(|V||E|). Use this if there exists an edge with negative length in the graph
    # After N steps, the shortest path has converded if there doesn't exist an cycle of edges with negative
    # Watch out: d[N] == d[2*N] doesn't necessarily mean the graph doesn't have negative cycle
    # ref: https://www.youtube.com/watch?v=1Z6ofKN03_Y

    def BellmanFord(N, M, ABC, vertex_start, vertex_end, value_if_inf = -1, find_shortest = False):
        """to calculate furthest or shortest length between vertex_start and vertex_end using BellmanFord algorithm
        
        Args:
            N (int): number of vertices
            M (int): number of edges
            ABC (list): [(ai, bi, ci) for _ in range(N)] where i-th edge is directed from vertex ai to vertex bi and the length is ci 
            vertex_start (int): start vertex. usually use 0.
            vertex_end (int): end vertex. usually use N-1.
            value_if_inf (int or string as you like, optional): value you want when the furthest (or shortest) distance is infinite (or -infinite). Defaults to -1.
            find_shortest (bool, optional): choose False to find furthest path. Defaults to False.
        
        Returns:
            int or string: normally int (but can be str if you set value_if_inf to str)
            
        Example:
                N, M, P = R()
                ABC = [R() for _ in range(M)]
                ABC = [(a-1, b-1, c-P) for a, b, c in ABC]
                print(BellmanFord(N, M, ABC, 0, N-1, value_if_inf = 'inf'))

        """

        def make_reachable_list(N, M, ABC, vertex_start, vertex_end):
            reachable_to_direct = defaultdict(list)
            reachable_from_direct = defaultdict(list)
            reachable_from_start = [False] * N
            reachable_to_end = [False] * N
            reachable_from_start[vertex_start] = True
            reachable_to_end[vertex_end] = True
            reachable_from_both_sides = [False] * N
            dfs_from_start = []
            dfs_to_end = []
            for a, b, c in ABC:
                reachable_to_direct[a].append(b)
                reachable_from_direct[b].append(a)
                if a == vertex_start:
                    dfs_from_start.append(b)
                    reachable_from_start[b] = True
                if b == vertex_end:
                    dfs_to_end.append(a)
                    reachable_to_end[a] = True
            while dfs_from_start:
                v = dfs_from_start.pop()
                for i in reachable_to_direct[v]:
                    if not reachable_from_start[i]:
                        reachable_from_start[i] = True
                        dfs_from_start.append(i)
            while dfs_to_end:
                v = dfs_to_end.pop()
                for i in reachable_from_direct[v]:
                    if not reachable_to_end[i]:
                        reachable_to_end[i] = True
                        dfs_to_end.append(i)
            for i in range(N):
                reachable_from_both_sides[i] = reachable_from_start[i] and reachable_to_end[i]
            return reachable_from_both_sides
        
        reachable_from_both_sides = make_reachable_list(N, M, ABC, vertex_start, vertex_end)
        
        if find_shortest:
            dist = [INF for i in range(N)]
        else:
            dist = [-INF for i in range(N)]
        
        dist[vertex_start] = 0
        for i in range(N):
            updated = False
            for a, b, c in ABC:
                if not reachable_from_both_sides[a]:
                    continue
                elif find_shortest:
                    update_condition = dist[a] + c < dist[b]
                else:
                    update_condition = dist[a] + c > dist[b]
                if dist[a] != INF and update_condition:
                    dist[b] = dist[a] + c
                    updated = True
                    if i == N-1:
                        return value_if_inf
            if not updated:
                break
        return dist[vertex_end]

    # Warshall Floyd O(V**3)
    def warshall_floyd(number_vertex, XYD, shift_one_for_vertex=False):
        # ref: https://juppy.hatenablog.com/entry/2018/11/01/%E8%9F%BB%E6%9C%AC_python_%E5%85%A8%E7%82%B9%E5%AF%BE%E6%9C%80%E7%9F%AD%E7%B5%8C%E8%B7%AF%E6%B3%95%EF%BC%88%E3%83%AF%E3%83%BC%E3%82%B7%E3%83%A3%E3%83%AB%E3%83%95%E3%83%AD%E3%82%A4%E3%83%89%E6%B3%95
        # number_vertex: len(vertex)
        # XYD: list of distance from x to y
        # shift_one_for_vertex: XYDの値が1~Nの時(0~N-1でない時)に調整するフラグ。
        # d[i][j]: iからjへの最短距離
        d = [[float("inf")]*number_vertex for i in range(number_vertex)]
        for x, y, dist in XYD:
            if shift_one_for_vertex:
                x = x - 1
                y = y - 1
            d[x][y] = dist
            d[y][x] = dist
        for i in range(number_vertex):
            d[i][i] = 0 # distance of same vertex is 0
        for k in range(number_vertex):
            for i in range(number_vertex):
                for j in range(number_vertex):
                    d[i][j] = min(d[i][j],d[i][k] + d[k][j])
        return d


    """ initialize variables and set inputs
    # initialize variables
        # to initialize list, use [0] * n
        # to initialize two dimentional array:
            # ex) [[0] * N for _ in range(N)]
            # ex2) dp = [[0] * (N+1) for _ in range(W*2)]
    # set inputs
        # put inputs between specific values (almost as quickly)
        # ex) S = [-INF] + [int(r()) for _ in range(A)] + [INF]
        # open(0).read() is sometimes useful:
        # ex) n, m, *x = map(int, open(0).read().split())
        #     min(x[::2]) - max(x[1::2])
        # ex2) *x, = map(int, open(0).read().split())
        #     don't forget to add comma after *x if only one variable is used
    # preprocessing
        # transpose = [x for x in zip(*data)]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [(1, 4, 7), (2, 5, 8), (3, 6, 9)]
        # flat = [flatten for inner in data for flatten in inner]
        # ex) [[1, 2, 3], [4, 5, 6], [7, 8, 9]] => [1, 2, 3, 4, 5, 6, 7, 8, 9]
    # calculate and output
        # output pattern
        # ex1) print(*l) => when l = [2, 5, 6], printed 2 5 6
    """

    # functions to read input
    r = lambda: sys.stdin.readline().strip()
    r_int = lambda: int(r())
    r_float = lambda: float(r())
    R = lambda: list(map(int, r().split()))
    R_str = lambda: r().split()
    R_map = lambda: map(int, r().split())
    R_float = lambda: list(map(float, r().split()))
    R_tuple = lambda: tuple(map(int, r().split()))

    """ how to treat input
    # single int: int(r())
    # single string: r()
    # single float: float(r())
    # line int: R()
    # line string: r().split()
    # line (str, int, int): [j if i == 0 else int(j) for i, j in enumerate(r().split())]
    # lines int: [R() for _ in range(n)]
    """

    # for test
    if sample_file:
        sys.stdin = open(sample_file)
        
    # ----------------------------------
    # main
    #%%

    # H, W, K = R()
    # C = [R_str() for _ in range(H)]
    # D = []
    # ans = 0
    
    # for i in range(2**H):
    #     for j in range(2**W):
    #         C_ = C.copy()
    #         bin_h = format(i, '08b')
    #         bin_w = format(j, '08b')
    #         for k in range(H):
    #             if bin_h[-k-1] == "1":
    #                 for l in range(W):
    #                     C_[k][l] = "."
    #         for k in range(W):
    #             if bin_w[-k-1] == "1":
    #                 for l in range(H):
    #                     C_[l][k] = "."
    #         cnt = 0
    #         for i in range(H):
    #             for j in range(W):
    #                 if C_[i][j] == "#":
    #                     cnt += 1
    #         if cnt == K:
    #             ans += 1
    # print(ans)
    
    N, K = R()
    A = R()
    h = []
    l = [1] * N
    for i,a in enumerate(A):
        heappush(h, (-a, i))
    
    for i in range(K):
        a, j = heappop(h)
        l[j] += 1
        heappush(h, (-A[j]/l[j], j))
    
    a, j = heappop(h)
    print(ceil(-a))
    
    
    # D = r_int()
    # C = R()
    # S = [R() for _ in range(D)]
    # # T = [r_int() for _ in range(D)]
    
    # def calc_score(D, C, S, T):
    #     sat = 0
    #     satl = [0] * D
    #     DI = [0] * 26 # daily irritation
    #     T_ = [t-1 for t in T]
    #     for i, t in enumerate(T_):
    #         for j in range(26):
    #             if j != t:
    #                 DI[j] += 1
    #                 sat -= C[j] * DI[j]
    #             else: # j == t
    #                 DI[j] = 0
    #                 sat += S[i][j]
    #         satl[i] = sat
    #     # return satl
    #     return sat
    
    # def donyoku(D, C, S):
    #     # return T
    #     DI = [0] * 26
    #     T = [0] * D
    #     for i in range(D):
    #         candidate = -1
    #         candidate_value = -INF
    #         for j in range(26):
    #             tmp_j = (DI[j] + 1) * C[j] + S[i][j] # merit
    #             if candidate_value < tmp_j:
    #                 candidate_value = tmp_j
    #                 candidate = j
    #         T[i] = candidate
    #     return [t+1 for t in T]

    # def emphasize_c(D, C, S, emp = 0.9):
    #     # return T
    #     DI = [0] * 26
    #     T = [0] * D
    #     for i in range(D):
    #         candidate = -1
    #         candidate_value = -INF
    #         for j in range(26):
    #             tmp_j = (DI[j] + 1) * C[j] + S[i][j] * emp # merit
    #             if candidate_value < tmp_j:
    #                 candidate_value = tmp_j
    #                 candidate = j
    #         T[i] = candidate
    #     return [t+1 for t in T]

    # def update_one(T, score):
    #     target_t = random.randint(0, D-1)
    #     after_abc = random.randint(1, 24)
    #     before_abc = T[target_t]
    #     T[target_t] = after_abc
    #     tmp = calc_score(D, C, S, T)
    #     if score < tmp:
    #         score = tmp
    #     else:
    #         T[target_t] = before_abc
    #     return T, score
    
    # def change_two(T, score):
    #     target_1, target_2 = random.sample(range(0, D), 2)
    #     before_1 = T[target_1]
    #     before_2 = T[target_2]
    #     T[target_1] = before_2
    #     T[target_2] = before_1
    #     tmp = calc_score(D, C, S, T)
    #     if score < tmp:
    #         score = tmp
    #     else:
    #         T[target_1] = before_1
    #         T[target_2] = before_2
    #     return T, score

    # import time
    # start = time.perf_counter()
    
    # T = [1] * D
    # score = -INF
    # Cand = [donyoku(D, C, S), emphasize_c(D, C, S), emphasize_c(D, C, S, 0.5), emphasize_c(D, C, S, 0)]
    # for c in Cand:
    #     tmp = calc_score(D, C, S, c)
    #     if tmp > score:
    #         T = c
    #         score = tmp
    
    # import random
    # while time.perf_counter() - start < 1.8:
    #     if random.random() < 0.5:
    #         T, score = update_one(T, score)
    #     else:
    #         T, score = change_two(T, score)

    # print(*T, sep='\n')


    #%%
    # end of main
    # ----------------------------------

    """memo: how to use defaultdict of list
    # initialize
    Dic = defaultdict(list)
    # append / extend
    Dic[x].append(y)
    # three methods for loop: keys(), values(), items()
    for k, v in Dic.items():
    """
    
    """memo: how to solve binary problems
    # to make binary digits text
    >>> a = 5
    >>> bin_str_a = format(a, '#06b')
    >>> print(bin_str_a)
    0b0101 # first 2 strings (='0b') indicates it is binary
    """
    
    """memo: how to solve the problem
    creating simple test/answer
    greed
    simple dp
    graph
    """

if __name__ == '__main__':
    main()

Submission status: Time Limit Exceeded

Solution code:
import copy
import math

n,k = map(int,input().split())
a = list(map(int,input().split()))
a2 = copy.copy(a)
x = [0] * n
for i in range(k):
    tmax = max(a2)
    tindex = a2.index(tmax)
    x[tindex] += 1
    a2[tindex] = a[tindex]/(x[tindex]+1)
max0 = 0
for i in range(n):
    max0 = max(max0,math.ceil(a[i]/(x[i]+1)))
print(max0)

Submission status: Time Limit Exceeded

Solution code:
import copy
import math
n,k = map(int, input().split())
a = list(map(int, input().split()))

num_cut = [1]*n
cut_list = copy.copy(a)

for i in range(k):
    max_index = cut_list.index(max(cut_list))
    num_cut[max_index] += 1
    cut_list[max_index] = a[max_index] / num_cut[max_index]

print(math.ceil(max(cut_list)))

Submission status: Time Limit Exceeded

Solution code:
rule=input().split()
N=int(rule[0])
K=int(rule[1])
data=input().split()
data_int=[int(s) for s in data]
data_count=[0]*len(data)
for i in range(K):
    x=data_int.index(max(data_int))
    data_int[x]=data_int[x]*(data_count[x]+1)/(data_count[x]+2)
    data_count[x]+=1
import math
print(math.ceil(max(data_int)))

Submission status: Time Limit Exceeded

Solution code:
import sys
import heapq
from math import ceil


def _ia(): return map(int, sys.stdin.readline().strip().split())


def main():
    _, k = _ia()
    a = list(map(lambda x: [-x, 0], _ia()))
    heapq.heapify(a)
    for _ in range(k):
        l, nn = heapq.heappop(a)
        nn += 1
        heapq.heappush(a, [l * nn / (nn + 1), nn])

    ret, _ = heapq.heappop(a)
    return ceil(-ret)


if __name__ == "__main__":
    print(main())

Submission status: Time Limit Exceeded

Solution code:
import math
N,K=map(int,input().split())
import heapq
A=list(map(int,input().split()))
h=[]
for i in range(len(A)):
  heapq.heappush(h,(-A[i],1))
for i in range(K):
  a=heapq.heappop(h)
  b=a[0]
  c=a[1]
  heapq.heappush(h,(c*b/(c+1),c+1))


ans=heapq.heappop(h)
print(math.ceil(-ans[0]))