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
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.

题解:按值爆搜交上去就满了。。。

 #include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstring>
#define PAU putchar(' ')
#define ENT putchar('\n')
using namespace std;
const int maxn=+,maxv=,inf=-1u>>;
int vis[maxn],cnt[maxn],stp[maxn],n;
inline int read(){
int x=,sig=;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')sig=-;ch=getchar();}
while(isdigit(ch))x=*x+ch-'',ch=getchar();
return x*=sig;
}
inline void write(int x){
if(x==){putchar('');return;}if(x<)putchar('-'),x=-x;
int len=,buf[];while(x)buf[len++]=x%,x/=;
for(int i=len-;i>=;i--)putchar(buf[i]+'');return;
}
void init(){
n=read();
queue<pair<int,int> >Q;
for(int i=;i<=n;i++){
int num=read();Q.push(make_pair(num,));
while(!Q.empty()){
int x=Q.front().first,y=Q.front().second;Q.pop();
if(x>maxv||vis[x]==i)continue;
vis[x]=i;cnt[x]++;stp[x]+=y;
Q.push(make_pair(x<<,y+));
Q.push(make_pair(x>>,y+));
}
}
int mi=inf;
for(int i=;i<=maxv;i++)if(cnt[i]==n&&stp[i]<mi)mi=stp[i];
write(mi);
return;
}
void work(){
return;
}
void print(){
return;
}
int main(){init();work();print();return ;}

Codeforces Round #312 (Div. 2) C.Amr and Chemistry的更多相关文章

  1. Codeforces Round #312 (Div. 2) C. Amr and Chemistry 暴力

    C. Amr and Chemistry Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/558/ ...

  2. Codeforces Round #312 (Div. 2)B. Amr and The Large Array 暴力

    B. Amr and The Large Array Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contes ...

  3. Codeforces Round #312 (Div. 2) B.Amr and The Large Array

    Amr has got a large array of size n. Amr doesn't like large arrays so he intends to make it smaller. ...

  4. C. Amr and Chemistry(Codeforces Round #312 (Div. 2) 二进制+暴力)

    C. Amr and Chemistry time limit per test 1 second memory limit per test 256 megabytes input standard ...

  5. Codeforces Round #312 (Div. 2) ABC题解

    [比赛链接]click here~~ A. Lala Land and Apple Trees: [题意]: AMR住在拉拉土地. 拉拉土地是一个很漂亮的国家,位于坐标线.拉拉土地是与著名的苹果树越来 ...

  6. Codeforces Round #312 (Div. 2)

    好吧,再一次被水题虐了. A. Lala Land and Apple Trees 敲码小技巧:故意添加两个苹果树(-1000000000, 0)和(1000000000, 0)(前者是位置,后者是价 ...

  7. B. Amr and The Large Array(Codeforces Round #312 (Div. 2)+找出现次数最多且区间最小)

    B. Amr and The Large Array time limit per test 1 second memory limit per test 256 megabytes input st ...

  8. Codeforces Round #312 (Div. 2) A. Lala Land and Apple Trees 暴力

    A. Lala Land and Apple Trees Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/cont ...

  9. Codeforces Round #312 (Div. 2) A.Lala Land and Apple Trees

    Amr lives in Lala Land. Lala Land is a very beautiful country that is located on a coordinate line. ...

随机推荐

  1. Java并发学习之二——获取和设置线程信息

    本文是学习网络上的文章时的总结,感谢大家无私的分享. Thread类的对象中保存了一些属性信息可以帮助我们辨别每个线程.知道它的一些信息 ID:每一个线程的独特标示: Name:线程的名称: Prio ...

  2. PHP 发布两个不用递归的树形数组构造函数(转)

    <?php/** *创建父节点树形数组 * 参数 $ar 数组,邻接列表方式组织的数据 $id 数组中作为主键的下标或关联键名 $pid 数组中作为父键的下标或关联键名 * 返回 多维数组 ** ...

  3. linux 下 apt命令集详解

    apt命令用法 packagename指代为软件包的名称 apt-get update 在修改/etc/apt/sources.list或/etc/apt/preferences之後运行该命令.此外您 ...

  4. [转] C++指针加整数、两个指针相减的问题

    http://blog.csdn.net/onlyou930/article/details/6725051 说来惭愧,写C++有一段时间了.这个问题从来没有认真考虑过,此次标记于此: 考虑如下问题: ...

  5. iOS 如何优雅的处理“回调地狱Callback hell”(一) (下)

    了解完流程之后,就可以开始继续研究源码了.在PromiseKit当中,最常用的当属then,thenInBackground,catch,finally - (PMKPromise *(^)(id)) ...

  6. CDOJ 92 Journey(LCA&RMQ)

    题目连接:http://acm.uestc.edu.cn/#/problem/show/92 题意:给定一棵树,最后给加一条边,给定Q次查询,每次查询加上最后一条边之后是否比不加这条边要近,如果近的话 ...

  7. HDU 5040 Instrusive(BFS+优先队列)

    题意比较啰嗦. 就是搜索加上一些特殊的条件,比如可以在原地不动,也就是在原地呆一秒,如果有监控也可以花3秒的时间走过去. 这种类型的题目还是比较常见的.以下代码b[i][j][x]表示格子i行j列在x ...

  8. 转载:C# HashSet 用法

    原文地址:http://www.cnblogs.com/xiaopin/archive/2011/01/08/1930540.html   感谢博主分享! NET 3.5在System.Collect ...

  9. winapi获取鼠标位置

    using System; using System.Drawing; using System.Runtime.InteropServices; using System.Threading; na ...

  10. SystemConfig.getPropertyValue("test");配置文件已经加了test=abc,但是取得时候空字符串

    1.定位tomcat中System.properties是否配置了,发现配置了 2.定位myeclipse中修改的tomcat是不是自己配置的tomcat.发现是 3.定位如下位置配置是否读取我先在用 ...