Problem:
Score : 200 points

Problem Statement

You are given a positive integer X.
Find the largest perfect power that is at most X.
Here, a perfect power is an integer that can be represented as b^p, where b is an integer not less than 1 and p is an integer not less than 2.

Constraints

1 ≤ X ≤ 1000

X is an integer.

Input

Input is given from Standard Input in the following format:

X

Output

Print the largest perfect power that is at most X.

Sample Input 1

10

Sample Output 1

9

There are four perfect powers that are at most 10: 1, 4, 8 and 9.
We should print the largest among them, 9.

Sample Input 2

1

Sample Output 2

1

Sample Input 3

999

Sample Output 3

961

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

/**
 *      ABC 97 B
 *      author  : kyomukyomupurin
 *      created : 2018/08/23
**/

#include <bits/stdc++.h>
using namespace std;

int main(){
    int X; cin >> X;
    vector<int> v;
    v.push_back(1);
    for (int i = 2; i < 10; ++i) {
        v.push_back(pow(2, i));
    }
    for (int i = 2; i < 7; ++i) {
        v.push_back(pow(3, i));
    }
    for (int i = 2; i < 5; ++i) {
        v.push_back(pow(5, i));
    }
    for (int i = 6; i < 9; ++i) {
        for (int j = 2; j < 4; ++j) {
            v.push_back(pow(i, j));
        }
    }
    for (int i = 2; i < 4; ++i) {
        v.push_back(pow(10, i));
    }
    for (int i = 11; i < 32; ++i) {
        v.push_back(pow(i, 2));    
    }
    sort(v.begin(), v.end());
    for (int i = 0; i < v.size(); ++i){
        if (X > v.at(i) && X <= v.at(i + 1)){
            cout << v.at(i + 1); break;
        }
    }
    
    return 0;
}