Problem:
Score : 200 points

Problem Statement

Takahashi loves numbers divisible by 2.

You are given a positive integer N. Among the integers between 1 and N (inclusive), find the one that can be divisible by 2 for the most number of times. The solution is always unique.

Here, the number of times an integer can be divisible by 2, is how many times the integer can be divided by 2 without remainder.

For example,

6 can be divided by 2 once: 6 -> 3.

8 can be divided by 2 three times: 8 -> 4 -> 2 -> 1.

3 can be divided by 2 zero times.

Constraints

1 ≤ N ≤ 100

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer.

Sample Input 1

7

Sample Output 1

4

4 can be divided by 2 twice, which is the most number of times among 1, 2, ..., 7.

Sample Input 2

32

Sample Output 2

32

Sample Input 3

1

Sample Output 3

1

Sample Input 4

100

Sample Output 4

64

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

#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long int ll;

#define rep(i, n) for(int i = 1; i < (int)(n); i++)
#define YES cout << "YES" << endl
#define Yes cout << "Yes" << endl
#define yes cout << "yes" << endl
#define NO cout << "NO" << endl
#define No cout << "No" << endl
#define no cout << "no" << endl

const int INF = 1001001001;
const int mod = 1000000007;
#define PI 3.14159265359;

void P(int x) {cout << x << endl;}
void P(long x) {cout << x << endl;}
void P(double x) {cout << x << endl;}
void P(ll x) {cout << x << endl;}
void P(string x) {cout << x << endl;}
void P(char x) {cout << x << endl;}

ll gcd(ll a, ll b) { return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}

int main() {
  int n;
  cin >> n;
  int maxCount = 0;
  int ans = 0;
  rep (i, n) {
    int tmp = i;
    int count = 0;
    while (tmp != 0 && tmp % 2 == 0) {
      tmp /= 2;
      count++;
    }
    if (maxCount < count) {
      maxCount = count;
      ans = i;
    }
  }



  return 0;
}