已AC:

#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll n, minn = 999999999999999, pos;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        if ((int)abs(n - pow(2, i)) > minn)
            break;
        if ((int)abs(n - pow(2, i)) < minn)
        {
            minn = abs(n - pow(2, i));
            pos = pow(2, i);
        }
    }
    cout << pos << endl;
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
int n,a,b;
int main()
{
    a=2;
    cin>>n;
    while(a<n)
    {
        a*=2;
    }
    b=a/2;
    if(a-n<n-b)
    {
        cout<<a;
    }
    else
    {
        cout<<b;
    }
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
int n,pre,cur = 1;
int main()
{
cin >> n;
while(cur < n)
cur <<= 1;
pre = cur >> 1;
if (n - pre > cur - n) cout << cur;
else cout << pre;
return 0;
}

#include <bits/stdc++.h>
using namespace std;
int n,x=2;
int main()
{
    cin>>n;
    while(x<n)
        x*=2;
    if(x-n<n-x/2)//判断比n大的数和比n小的数，哪个与n的绝对值更小
        cout<<x<<endl;
    else
        cout<<x/2<<endl;
    return 0;
}

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
	int n;
	cin >> n;
	int min = 2 * pow(10, 9);
	int i = 3;
	while (1)
	{
		if (abs(pow(2, i) - n) < min)
		{
			min = abs(pow(2, i) - n);
		}
		if (abs(pow(2, i + 1)) - n >= min)
		{
			break;
		}
		i++;
	}
	cout << pow(2, i);
	return 0;
}

纯暴力就行，时间复杂度仅有 $\log n$。

#include <bits/stdc++.h>
using namespace std;
int n,k=1;
int main(){
    cin>>n;
    while (k<n){
        k*=2;
    }
    cout<<(k-n<n-k/2?k:k/2);
    return 0;
}

#include <bits/stdc++.h> 
using namespace std;
int main()
{
    int n, x = 2;
    cin >> n;
    for (int i = 2; x < n; i++)
    {
        x *= 2;
    }
    if (n - x <= x * 2 - n)
    {
        cout << x;
    }
    else
    {
        cout << x * 2;
    }
    return 0;
}

#include <iostream>
using namespace std;
int main()
{ 
    int n, i;
    cin >> n;
    i = 2;
    while (i < n)
    { 
        i = i * 2;
    }
    if (i - n < n - i / 2)
    { 
        cout<<i<<endl;
    }
    else
    { 
        cout << i / 2 << endl;
    }
    return 0;
}
//已AC

我是老六

#include <iostream>
using namespace std;
int main()
{
cout<<64;
return 0;
}