题干：
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 ailiters. 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.
Examples
Input
3 4 8 2
Output
2
Input
3 3 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.
题目大意：
给你n个整数,每一个整数可以进行两种操作，除2(取整)或者乘2.每个整数可以进行任意次这样的操作。
使这n个整数都变为相同的整数最少需要多少次操作。
解题报告：
bfs考虑每一个数可以到达的地方以及他可以该地方的所有的步数。注意一个细节就是每次bfs不要都清空vis，可能会超时，因为肯定memset了很多本来就是0的值，可以选择vis[i]=i来判断是否走过。
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
const int MAX = 2e5 + 5;
const int INF = 0x3f3f3f3f;
int cnt[MAX], vis[MAX], steps[MAX];
int main() {int n, res = INF, x, y;scanf("%d", &n);for(int i=1; i<=n; i++) {scanf("%d", &x);queue<pair<int, int> > q;q.push(pm(x, 0));while(!q.empty()) {x = q.front().first;y = q.front().second;q.pop();if(x > 100003) continue;if(vis[x] == i) continue;vis[x] = i;steps[x]+=y;cnt[x]++;q.push(pm(x * 2, y + 1));q.push(pm(x / 2, y + 1));}}for(int i=0; i<=100000; i++)if(cnt[i] == n)if(res > steps[i])res = steps[i];printf("%d", res);return 0;
}
)
