CF 558C(Amr and Chemistry-构造法)

C. Amr and Chemistry

time limit per test

memory limit per test

input

output

Amr loves Chemistry, and specially doing experiments. He is preparing for a new interesting experiment.

n different types of chemicals. Each chemical i has an initial volume of ai

To do this, Amr can do two different kind of operations.

• iand double its current volume so the new volume will be 2ai
• iand divide its volume by two (integer division) so the new volume will be

Suppose that each chemical is contained in a vessel of infinite volume. Now Amr wonders what is the minimum number of operations required to make all the chemicals volumes equal?

Input

n (1 ≤ n ≤ 105), the number of chemicals.

n space separated integers ai (1 ≤ ai ≤ 105), representing the initial volume of the i-th chemical in liters.

Output

Output one integer the minimum number of operations required to make all the chemicals volumes equal.

Sample test(s)

input

3 4 8 2

output

2

input

3 3 5 6

output

5

Note

4.

1.

将每个数写为2进制，计算公共前缀

答案必为 [最长公共前缀(LCS)]+00...00

证：

若LCS后有1

则每个数都要在相应位置有1，

故LCS可增长，矛盾

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (100000+10)
typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,a[MAXN],a2[MAXN],h[MAXN]={0},q[MAXN]={0};
int main()
{
  freopen("C.in","r",stdin);
//  freopen(".out","w",stdout);
  
  cin>>n;
  For(i,n) scanf("%d",&a[i]);
  
  sort(a+1,a+1+n);
  memcpy(a2,a,sizeof(a));
  
  int p=a[1];
  For(i,n)
  {
    while (p^a[i])
    {
      if (p<a[i]) a[i]>>=1;
      else p>>=1;
    }
  }
  
  memcpy(a,a2,sizeof(a));
  
  For(i,n)
  {
    while (a[i]^p) 
    {
      h[i]--;
      if (a[i]&1) q[i]=h[i];
      a[i]>>=1;
      
    }
  }
  
  ll ans=INF;
  
  
  Rep(t,100)
  {
    ll p=0;
    For(i,n)
    {
      if (q[i]==0||h[i]+t<=q[i]) p+=abs(h[i]+t);
      else p+=abs(q[i])+abs(q[i]-(h[i]+t));
    }
    ans=min(ans,p);
    
  }
  cout<<ans<<endl;
  
//  For(i,n) cout<<h[i]<<' ';
//  For(i,n) cout<<q[i]<<' ';
  
  
  
  return 0;
}