Problem:


Score : 400 points

Problem Statement

Give a pair of integers (A, B) such that A^5-B^5 = X.
It is guaranteed that there exists such a pair for the given integer X.

Constraints

1 \leq X \leq 10^9

X is an integer.

There exists a pair of integers (A, B) satisfying the condition in Problem Statement.

Input

Input is given from Standard Input in the following format:

X

Output

Print A and B, with space in between.
If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.

A B

Sample Input 1

33

Sample Output 1

2 -1

For A=2 and B=-1, A^5-B^5 = 33.

Sample Input 2

1

Sample Output 2

0 -1



Sample input:
33


Sample output:
2 -1

Submission status: Accepted

Solution code:
#!/usr/bin/env python
# -*- coding: utf-8 -*-

import sys
sys.setrecursionlimit(10**7)
from pprint import pprint as pp
from pprint import pformat as pf

import math
#from sortedcontainers import SortedList, SortedDict, SortedSet # no in atcoder
import bisect


def calc(a, b):
    return a ** 5 - b ** 5

def solve(x):
    for a in range(-120, 120):
        for b in range(-120, 120):
            if x == calc(a, b):
                print(a, b)
                return


if __name__ == '__main__':
    x = int(input())
    solve(x)

    #print('\33[32m' + 'end' + '\033[0m') # debug

Submission status: Accepted

Solution code:
i=0
exec('j=-i;exec("i**5-%s-j**5or exit(print(i,j));j+=1;"*i*2);i+=1;'%input()*200)

Submission status: Accepted

Solution code:
from itertools import product

X = int(input())
for a, b in product(range(-200, 201), repeat=2):
    if a**5 - b**5 == X: print(a, b); break

Submission status: Accepted

Solution code:
X = int(input())

import sys

for a in range(-120,120+1):
    for b in range(-120,120+1):
        if a**5 - b**5 == X:
            print(a,b)
            sys.exit()

Submission status: Accepted

Solution code:
#!/usr/bin/env python3
n = int(input())
list_5 = []
a = 0
b = 0
for _ in range(-121,121):
    list_5.append(_**5)
for i in range(-121,121):
    if (n + list_5[i]) in list_5:
        b = i  
        a = list_5.index(n + list_5[i])       
print(a-121, b-121)

Submission status: Accepted

Solution code:
exec('x=int(input());i=0;'+'j=-i;exec("i**5-x-j**5or exit(print(i,j));j+=1;"*i*2);i+=1;'*200)

Submission status: Accepted

Solution code:
X = int(input())


def is_ok(A, B, X):
    return A ** 5 - B ** 5 == X


def spiral():
    i = 1
    p = [0, 0]
    yield p
    while True:
        for j in range(i):
            p[0] += 1
            yield p
        for j in range(i):
            p[1] -= 1
            yield p
        i += 1
        for j in range(i):
            p[0] -= 1
            yield p
        for j in range(i):
            p[1] += 1
            yield p
        i += 1


for a, b in spiral():
    if is_ok(a, b, X):
        print(a, b)
        break

Submission status: Accepted

Solution code:
x=int(input())
for a in range(-118,120):
  for b in range(-119,119):
    if a**5-b**5==x:
      print(a,b)
      exit()

Submission status: Accepted

Solution code:
X = int(input())

A = -119
while A<= 120:
    B = -119
    while B<= 120:
        if A**5 - B**5 == X:
            print('%d %d'%(A, B))
            break
        B += 1
    else:
        A+= 1
        continue
    break

Submission status: Accepted

Solution code:
x = int(input())
for i in range(-300, 300):
    for j in range(-300, 300):
        if pow(i, 5) - pow(j, 5) == x:
            a = i
            b = j
            break

print(a, b)

Submission status: Accepted

Solution code:

def solve():
    x=int(input())
    for i in range(-1000,1001):
        for j in range(-1000,1001):
            if(i**5-j**5==x):
                print(i,j)
                return
solve()

Submission status: Accepted

Solution code:
x = int(input())

for i in range(-1000,1000):
    for j in range(-1000,1000):
        if i ** 5 - j ** 5 == x:
            print(i,j)
            quit()

Submission status: Accepted

Solution code:
x = int(input())
for i in range(-120, 120):
    for j in range(-120, 120):
        if i ** 5 - j ** 5 == x:
            print(i, j)
            exit()

Submission status: Accepted

Solution code:
import math

X = int(input())
A, B = 0, 0

for i in range(1000):
    for j in range(-1000, 1000):
        x = int(math.pow(i, 5) - math.pow(j, 5))
        if X == x:
            A, B = i, j
            break
print(A,B)

Submission status: Accepted

Solution code:
# -*- coding: utf-8 -*-
"""
Created on Sun May  3 20:56:48 2020
"""

import sys
#import numpy as np

sys.setrecursionlimit(10 ** 9)
#def input():
#    return sys.stdin.readline()[:-1]
mod = 10**9+7

X = int(input())

for i in range(-5000,5001):
    for j in range(-5000,5001):
        if pow(i,5) - pow(j,5) == X:
            print(i,j)
            sys.exit()

Submission status: Accepted

Solution code:
# -*- coding: utf-8 -*-
"""
Created on Sun May  3 20:56:48 2020
"""

import sys
#import numpy as np

sys.setrecursionlimit(10 ** 9)
#def input():
#    return sys.stdin.readline()[:-1]
mod = 10**9+7

X = int(input())

for i in range(-1000,1001):
    for j in range(-1000,1001):
        if pow(i,5) - pow(j,5) == X:
            print(i,j)
            sys.exit()

Submission status: Accepted

Solution code:
import sys
X=int(input())

for i in range(0,120):
    for j in range(-120,120):
        if i**5 - j**5 ==X:
            A = i
            B = j
            print(A,B)
            
            sys.exit()

Submission status: Accepted

Solution code:
from scipy import optimize

X = int(input())

def make_divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n//i)
    divisors.sort()
    return divisors

divisors = make_divisors(X)
divisors = divisors + [-x for x in divisors]
for p in divisors:
    def f(x):
        return (x+p)**5 - x**5 - X 
    sol = optimize.fsolve(f, X**(1./5.))
    B = int(round(sol[0]))
    A = B + p
    if A**5 - B**5 == X:
        print(A, B)
        break

Submission status: Accepted

Solution code:
from scipy import optimize

X = int(input())

def make_divisors(n):
    divisors = []
    for i in range(1, int(n**0.5)+1):
        if n % i == 0:
            divisors.append(i)
            if i != n // i:
                divisors.append(n//i)
    divisors.sort()
    return divisors

divisors = make_divisors(X)
divisors = divisors + [-x for x in divisors]
for p in divisors:
    def f(x):
        return (x+p)**5 - x**5 - X 
    sol = optimize.fsolve(f, X**(1./5.))
    #print(sol[0])
    B = int(round(sol[0]))
    A = B + p
    #print(p, A, B)
    if A**5 - B**5 == X:
        print(A, B)
        break

Submission status: Accepted

Solution code:
x = int(input())
for a in range(-118,119):
    for b in range(-119,118):
        if a**5 - b**5 == x:
            print(a,b)
            exit()

Submission status: Accepted

Solution code:
x = int(input())
max_cand = 120
for a in range(max_cand+1):
    for b in range(max_cand+1):
        if (a**5 - b**5) == x:
            print(a, b)
            break
        elif (a**5 + b**5) == x:
            print(a, -b)
            break
    else:
        continue
    break

Submission status: Accepted

Solution code:
X = int(input())
for i in range(-200, 200):
    for j in range(-200, 200):
        if i**5 - j**5 == X:
            print(i,j)
            exit()

Submission status: Accepted

Solution code:
x,r=int(input()),range(-99,120);[i**5-x-j**5or exit(print(i,j))for i in r for j in r]

Submission status: Accepted

Solution code:
x = int(input())

for a in range(-1000,1000):
    for b in range(-1000,1000):
         if (a**5)-(b**5) == x:
             print(a,b)
             exit()

Submission status: Accepted

Solution code:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-


def main():
    x = int(input())

    flag = False

    for i in range(-118, 120):
        for j in range(-119, 119):
            if i**5- j**5 == x:
                print(i, j)
                flag = True
                break
        if flag:
            break




if __name__ == '__main__':
    main()

Submission status: Accepted

Solution code:
def solve():
    x = int(input())

    for a in range(-118, 120):
        for b in range(-119, 119):
            if a**5 - b**5 == x:
                return a, b





print(*solve())

Submission status: Accepted

Solution code:
x = int(input())
a = 0
b = 0
for i in range(-120, 120):
    for j in range(-120, 120):
        if i ** 5 - j ** 5 == x:
            a = i
            b = j
print(a, b)

Submission status: Accepted

Solution code:
x = int(input())
a = 0
b = 0
for i in range(-200, 200):
    for j in range(-200, 200):
        if i ** 5 - j ** 5 == x:
            a = i
            b = j
print(a, b)

Submission status: Accepted

Solution code:
x=int(input())
for a in range(-1000,1000):
    for b in range(-1000,1000):
        if a**5-b**5==x:
            print(a,b)
            break
    else:
        continue
    break

Submission status: Accepted

Solution code:
x = int(input())

for a in range(1, 120):
    for b in range(-a, a):
        if pow(a, 5) - pow(b, 5) == x:
            break
    if pow(a, 5) - pow(b, 5) == x:
         break
            
print(a, b)

Submission status: Accepted

Solution code:
X=int(input())
Ni=0
while (Ni+1) ** 5 - Ni** 5 <= X:
    Ni+=1
    
for i in range(1,Ni+1):
    for ii in range(-i,i):
        if i**5 - ii**5==X:
            break
    else:
        continue
    break

print(i,ii)

Submission status: Accepted

Solution code:
x = int(input())

for a in range(-118,119):
  for b in range(-119,118):
    if a**5 - b**5 == x:
      print(a,b)
      exit()

Submission status: Accepted

Solution code:
X = int(input())

for A in range(-1000, 1001):
    for B in range(-1000, 1001):
        if A**5 - B**5 == X:
            print(A, B)
            quit()

Submission status: Accepted

Solution code:
import sys
stdin = sys.stdin
sys.setrecursionlimit(10 ** 7)

def LI(): return list(map(int, stdin.readline().split()))
def LS(): return list(stdin.readline())

import math
x = int(input())
for i in range(-200, 200):
    for j in range(-200, 200):
        if i**5-j**5==x:
            print(i, j)
            exit()

Submission status: Accepted

Solution code:
X=int(input())

for a in range(-118,120):
  for b in range(-119,119):
    if X==a**5-b**5:
      print(a,b)
      exit()

Submission status: Accepted

Solution code:
x = int(input())

divisors = []
for i in range(1, int(x**0.5)+1):
  if x % i == 0:
    divisors.append(i)
    if i != x // i:
      divisors.append(x//i)

boption = []
a = 1
while boption == []:
  for i in divisors:
    b = a - i
    if a**5-b**5 == x:
      boption.append(b)
  a += 1

a -= 1
print("{} {}".format(int(a),int(boption[0])))

Submission status: Accepted

Solution code:
## D
X = int(input())
lp = 1000
for a in range(-lp, lp):
    for b in range(-lp, lp):
        if a**5 - b**5 == X:
            print(a, b)
            break
    else:
        continue
    break

Submission status: Accepted

Solution code:
# D - I hate Factorization
def main():
    X = int(input())
    for i in range(-120, 120):
        for j in range(-120, 120):
            if i ** 5 - j ** 5 == X:
                print(i, j)
                return


if __name__ == "__main__":
    main()

Submission status: Accepted

Solution code:
x = int(input())
for i in range(-200,200,1):
    for j in range(-200,200,1):
        if i**5-j**5==x:
          print(i,j)
          exit()

Submission status: Accepted

Solution code:
import math
import sys

x = int(input())
r = math.ceil(x ** 0.2)
# print(r)
def reverse_num(x):
    return r - x
def calc_b(aa):
    bb = aa - x
    bbb = abs(bb) ** 0.2
    if bb < 0:
        b = -1
    else:
        b = 1
    # print(bb,b * int(bbb.real + bbb.imag))
    return b * int(bbb.real + bbb.imag)
def check(a):
    aa = a ** 5
    b = calc_b(aa)
    # print(a,b)
    if aa - b ** 5 == x:
        print(a,b)
        return True
    else:
        return False
for a in range(r // 2 + 1):
    if check(a) or check(reverse_num(a)):
        sys.exit()
a = r + 1
while True:
    if check(a):
        sys.exit()
    a += 1

Submission status: Accepted

Solution code:
x=int(input())
r=range(-99,200)
[i**5-x-j**5or exit(print(i,j))for i in r for j in r]

Submission status: Accepted

Solution code:
x = int(input())
for a in range(-118, 120):
    for b in range(-119, 119):
        if a**5 - b**5 == x:
            print(a, b)
            exit()

Submission status: Accepted

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 = mod_factorial_list(n)
    INF = 10 ** 18
    MOD = 10 ** 9 + 7
    AtoZ = list('abcdefghijklmnopqrsruvwxyz')
    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]

    # Graph: https://en.wikipedia.org/wiki/Directed_graph
    # 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]

    """ 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
    #%%
    X = r_int()
    for i in range(-1000, 1000):
        for j in range(-1000, 1000):
            if i**5 - j**5 == X:
                print(i, j)
                return
    

    #%%
    # 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: Accepted

Solution code:
x = int(input())

flg = 'n'
for a in range(-120, 121, 1):
    for b in range(-120, 121, 1):
        if a ** 5 - b ** 5 == x:
            flg = 'y'
            break
    if flg == 'y':
        break
print(a, b)

Submission status: Accepted

Solution code:
import itertools

s = 119
X = int(input())

l = [i ** 5 for i in range(-1 * s, s)]

for i, j in itertools.permutations(l, 2):
    if(i + j == X):
        print (l.index(i) - s, l.index(j) - s)
        exit(0)

    if(i - j == X):
        print (l.index(i) - s, l.index(j) - s)
        exit(0)

Submission status: Accepted

Solution code:
x = int(input())

for a in range(-118, 120):
    for b in range(-118, 120):
        if a**5 - b**5 == x:
            print(a, b)
            exit()

Submission status: Accepted

Solution code:
X = int(input())

list = []
for i in range(-119,120):
    list.append(i)
for j in range(len(list)):
    for k in range(len(list)):
        if X == (list[j]) ** 5 - list[k] ** 5:
            A = list[j]
            B = list[k]
            break

print(str(A)+" "+str(B))

Submission status: Accepted

Solution code:
x=int(input())
[i**5-x-j**5or exit(print(i,j))for i in range(200)for j in range(-i,i)]

Submission status: Accepted

Solution code:
# D - I hate Factorization

x = int(input())

# a ** 5 - b ** 5 == x となる組 (a,b) を1つ答える.
# 1 <= x <= 10 ** 9
# a >= 0と仮定してよい. b < aと仮定してよい. bが減ると左辺は増加.

a = 0
while True:
    b = a - 1
    while a ** 5 - b ** 5 < x:
        b -= 1
    if a ** 5 - b ** 5 == x:
        print(a, b)
        exit()
    a += 1

Submission status: Accepted

Solution code:
x=int(input())
[i**5-x-j**5or print(i,j)+exit()for i in range(200)for j in range(-i,i)]

Submission status: Accepted

Solution code:
import sys
x = int(input())

for a in range(-118, 120):
    for b in range(-119, 119):
        if a**5 - b**5 == x:
            print(a, b)
            sys.exit()

Submission status: Accepted

Solution code:
def eval_pair(A, B, x):
  return A**5-B**5 == x

def get_pair(x):
  # exit early for x a power of 5.
  # apart from this, A,B don't need to be zero.
  A = int(round(x**(1/5)))
  if A**5 == x: return (A, 0)
  # try A, B:
  # +,+ and +,-
  # don't need -,- (same as +,-)
  # can't do -,+ because x is constrained positive
  # for +,+ A>=120 would give x>10**9
  # so 1<=A<120 and 1<=B<119
  valid = [(A,B) for A in range(1,120) for B in range(1,119) if eval_pair(A,B,x)]
  if valid: return valid.pop()
  # for +,- both are positive so A>=64 would give x=10**9
  # so 1<=A<64 and -64<B<-1
  valid = [(A,B) for A in range(1,64) for B in range(-1,-64, -1) if eval_pair(A,B,x)]
  if valid: return valid.pop()

x = int(input())
valid_pair = get_pair(x)
print(*valid_pair)

Submission status: Accepted

Solution code:
target = int(input())

for A in range(-118, 120):
	for B in range(-119, 119):
		if target == A ** 5 - B ** 5:
			print(f'{A} {B}')
			exit(0)

Submission status: Accepted

Solution code:
x = int(input())

for a in range(-122, 122):
    for b in range(-122, 122):
        if a**5 - b**5 == x:
            print(a, b)
            exit(0)

Submission status: Accepted

Solution code:
X = int(input())
pin = 0
for i in range(-120, 120):
    if pin == 1:
        break
    for j in range(-120, 120):
        if pow(i, 5) - pow(j, 5) == X :
            pin =1
            print(i, j)
            break