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

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)


#include <iostream>
#include <vector>
#include <array>
#include <set>
#include <cmath>
#include <string>
#include <sstream>
#include <queue> 
#include <list>
#include <stack>
#include <unordered_map>
#include <algorithm> 
#include <numeric>  
#include <functional>
#include <cassert>
#include <bitset>
#include <cctype>  
#include <iomanip> 
#include <limits>  
#include <regex>
// std::numeric_limits<int>::max(), min()
// std::setprecision(X)
// std::islower(), std::isupper(), std::tolower, std::toupper
// std::accumulate(m.begin(), m.end(), 0LL);
// std::sort(begin(),end()) // 0, 1, 2, 3 ,4 (default, less);
// std::priority_queue<int, vector<int>, less<int>> que; // 4, 3, 2, 1 (defualt, less) : 
// if ( std::regex_match(S, std::regex("(dream|dreamer|erase|eraser)*")) ) std::cout << "YES" << std::endl;

#define rep(i,n)  for(int i = 0; i < (int)(n);    i++)
#define sz(x)     ((int)(x).size())
#define all(x)    (x).begin(),(x).end()

using ll = long long;

ll gcd ( ll a, ll b ) { if ( b == 0 ) return a; else return gcd ( b, a%b ); }
ll lcm ( ll a, ll b ) { return a * b / gcd(a, b); }

ll check(ll x) {
  ll l = -1000, r = 1000;
  while( r - l > 1 ) {
    ll mid = (l+r)/2;
    ll c = std::pow(mid,5);
    if ( c < x ) l = mid;
    else r = mid;
  }
  return r;
}

int main (void) {
  std::cin.tie(0);  std::ios::sync_with_stdio(false);
  ll X; std::cin >> X;

  for(ll b=-1000; b<=1000; b++) {
    ll a5 = b*b*b*b*b + X;
    ll a = check(a5);
    if ( a5 == a*a*a*a*a ) {
      std::cout << a << " " << b << std::endl;
      return 0;
    }
  }
  assert(0);
  return 0;
}