Amr loves Chemistry, and specially doing experiments. He is preparing for a new interesting experiment.
Amr has n different types of chemicals. Each chemical i has an initial volume of ai liters. For this experiment, Amr has to mix all the chemicals together, but all the chemicals volumes must be equal first. So his task is to make all the chemicals volumes equal.
To do this, Amr can do two different kind of operations.
- Choose some chemical i and double its current volume so the new volume will be 2ai
- Choose some chemical i and 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
The first line contains one number n (1 ≤ n ≤ 105), the number of chemicals.
The second line contains 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
34 8 2
output
2
input
33 5 6
output
5
Note
In the first sample test, the optimal solution is to divide the second chemical volume by two, and multiply the third chemical volume by two to make all the volumes equal 4.
In the second sample test, the optimal solution is to divide the first chemical volume by two, and divide the second and the third chemical volumes by two twice to make all the volumes equal 1.
/*
题意：把n个数变成相同所需要走的最小的步数
易得到结论，两个奇数不同，一直×2不可能有重叠
枚举每个数可能到得所有值，以及统计达到该值的时候已经走的步数
最终答案就是1到up中num[i]最小的数
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAX = 1e5 + 10;
int num[MAX];
int cnt[MAX];
int a[MAX];
const int up = 1e5;
const int inf = 0x3f3f3f3f;
int main()
{
int n;
while(~scanf("%d", &n)){
memset(num, 0, sizeof(num));
memset(cnt, 0, sizeof(cnt));
for(int i = 1; i <= n ; i++)
scanf("%d", &a[i]);
for(int i = 1; i <= n ; i++){
int x = a[i];
int pre = 0;
while(x){
int s = 0;
while(x % 2 == 0){
x /= 2;
s++;//从当前偶数到最后的奇数移动的步数
}
int y = x;
int x1 = 0;
while(y <= up){
cnt[y]++;//可以得到的值
num[y] += pre + abs(s - x1);
x1++;
y *= 2;
}
pre += s + 1;//达到该值已经走过的步数,在接着处理一步+1
x /= 2;
}
}
int ans = inf;
for(int i = 1; i <= up ;i++){
if(cnt[i] == n){
ans = min(ans, num[i]);
}
}
printf("%d\n", ans);
}
return 0;
}