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: Runtime Error)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <float.h>
 
#pragma region 定義
#define i_cin(X) scanf("%d", &X)
#define i_cin2(X,Y) scanf("%d %d",&X,&Y)
#define i_cin4(X,Y,Z,W) scanf("%d %d %d %d",&X,&Y,&Z,&W)
#define i_cin3(X,Y,Z) scanf("%d %d %d",&X,&Y,&Z)
#define l_cin(X) scanf("%ld",&X)
#define f_cin(X) scanf("%f",&X)
#define l_cin2(X,Y) scanf("%ld %ld",&X,&Y)
#define s_cin(X) scanf("%s",&X)
#define ll_cin(X) scanf("%lld", &X)
 
#define  i_cout(X) printf("%d\n",X)
#define  f_cout(X) printf("%g\n",X)
#define  i_cout2(X,Y) printf("%d %d\n",X,Y)
#define  l_cout(X) printf("%ld\n",X)
#define  s_cout(X) printf("%s\n",X)
#define  ll_cout(X) printf("%lld\n",X)
 
typedef long long  ll;
#define rept(x, s, n) for (int x = s; x < n; x++)
 
#define S_RTN(S)  s_cout(S);return 0
 
 
#define _itoa(A,N) sprintf(A, "%d", N);
 
int i_cins(int n, int* A);
int l_cins2(int n, long* A, long* B);
int s_dsort(const void* a, const void* b);
int s_asort(const void* a, const void* b);
 
int  _gcd(int a, int b);
int  _swp(int* a, int* b);
int _cknum(char* a, int n);
int _atoi(char* s, int len);
 
 
int s_asorts(const void* a, const void* b);
 
//昇順
int s_asorts(const void* a, const void* b) {
    return(strcmp((char*)a, (char*)b));
}
int s_dsort(const void* a, const void* b) {
    return(*(int*)b - *(int*)a);
}
int s_asort(const void* a, const void* b) {
    return(*(int*)a - *(int*)b);
}
int l_cins2(int n, long* a, long* b) {
    int i;
    rept(i, 0, n) {
        l_cin2(*(a + i), *(b + i));
    }
    return 0;
}
int i_cins(int n, int* a) {
    int i;
    for (i = 0; i < n; i++) {
        i_cin(*(a + i));
    }
 
    return 0;
}
int _gcd(int a, int b) {
    int r, tmp;
    if (a < b) { tmp = a; a = b; b = tmp; }
    r = a % b;
    while (r) {
        if (!r) break;
        a = b; b = r;
        r = a % b;
    }
    return(b);
}
 
long _max(long a, long b) {
    return a > b ? a : b;
}
long _min(long a, long b) {
    return a < b ? a : b;
}
int _swp(int* a, int* b)
{
    int tmp;
    tmp = *b; *b = *a; *a = tmp;
    return 0;
}


int main(){
    int x;
    int p=0,q;
    int ans1=0,ans2;
    scanf("%d",&x);
    for(int i=2;i<x;i++){
        if(x%i==0){
            p=i;
        }
    }
    
    q=x/p;
    ans2=p+ans1;
    for(ans1=1;ans1<x;ans1++){
        if((pow(ans1,4)+pow(ans1,3)*pow(ans2,1)+pow(ans1,2)*pow(ans2,2)+pow(ans2,3)*pow(ans1,1)+pow(ans2,4))==q)
            printf("%d%d",ans1,ans2);
    }
    
}