题解 | #游游的除2操作#

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<ll, ll>PII;
const int N = 2e5 + 10;
const int MOD = 998244353;
const int INF = 0X3F3F3F3F;
const int dx[] = {-1, 1, 0, 0, -1, -1, +1, +1};
const int dy[] = {0, 0, -1, 1, -1, +1, -1, +1};
const int M = 1e6 + 10;

ll a[N];
int main() {
    int n;
    cin >> n;
    priority_queue<ll>q;
    for (int i = 1; i <= n; i ++) {
        cin >> a[i];
        q.push(a[i]);
    }
    sort(a + 1, a + 1 + n);
    int cnt = 0;
    ll mins = a[1], maxs = a[n];//保证最大的跟最小的相同即可
    if (mins == maxs) cout << 0 << endl;
    else {
        while (1) {
            ll k = q.top();
            q.pop();
            if (k == mins) break;
            else {
                cnt ++;
                k /= 2;
                if (k < mins) mins = k;
                q.push(k);
            }
        }
        cout << cnt << endl;
    }
    return 0;
}