Problem:
Score : 200 points

Problem Statement

Find the largest square number not exceeding N. Here, a square number is an integer that can be represented as the square of an integer.

Constraints

1 \leq N \leq 10^9

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the largest square number not exceeding N.

Sample Input 1

10

Sample Output 1

9

10 is not square, but 9 = 3 × 3 is. Thus, we print 9.

Sample Input 2

81

Sample Output 2

81

Sample Input 3

271828182

Sample Output 3

271821169

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

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef unsigned long long ull; 
typedef vector<int> vi;
typedef pair<int,int> pi;

#define F first
#define S second
#define PB push_back
#define MP make_pair
#define FOR(i,a,b) for (int i = a; i < b; i++)
#define SQ(a) (a)*(a)

template <class T, class U> ostream& operator<< (ostream &out, pair<T,U> c) { out << "(" << c.first << " "; out << c.second << ")"; return out; }
template <class T> ostream& operator<< (ostream &out, vector<T> c) { for (auto it = c.begin(); it != c.end(); ++it) { if (it != c.begin()) out << " "; out << *it; } return out; }
template <class T> ostream& operator<< (ostream &out, set<T> c) { for (auto it = c.begin(); it != c.end(); ++it) { if (it != c.begin()) out << " "; out << *it; } return out; }
const double PI  = 3.141592653589793238463;


int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n;
    int beg = 1, end = 1e5;
    ll mid;
    while(end - beg > 1){
        mid = (beg+end)/2;
        if(SQ(mid) == n){
            break;
        }
        else if(SQ(mid) > n){
            end = mid;
        }
        else
        {
            beg = mid;
        }
    }    
    cout << SQ(beg) << endl;
}