So easy!So easy?《So easy》

言归正传，只要找二进制的1了

#include <bits/stdc++.h>
using namespace std;
int min_pow(int a,int b)
{
    int j = 0;
    while (b / a > 0)
    {
        b /= a;
        j++;
    }
    return pow(a,j);

}
int main()
{
    long long n,t[10005],l = 0,p = 0;
    cin >> n;
    for (int i = 1;i <= n;i++)
    {
        cin >> t[i];
        l += t[i];
    }
    while (l > 0)
    {
        p++;
        l -= min_pow(2,l);
    }
    cout << p;
}