Codeforces Round #312 (Div. 2) C.Amr and Chemistry
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?
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 one integer the minimum number of operations required to make all the chemicals volumes equal.
3
4 8 2
2
3
3 5 6
5
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
- 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/ ...
- 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 ...
- 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. ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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
- android shape的使用详解以及常用效果(渐变色、分割线、边框、半透明阴影效果等)
shape使用.渐变色.分割线.边框.半透明.半透明阴影效果. 首先简单了解一下shape中常见的属性.(详细介绍参看 api文档 ) 转载请注明:Rflyee_大飞: http://blog.cs ...
- 查看linux系统状态
就类似你装完xp后,或者你拿到一台新的机器的时候,你通常都是进入系统,看看他的cpu,内存,硬盘使用情况.我也按照这个来看看linux的系统状态.1:top 退出按q,这个就类似windows的任务管 ...
- SPOJ 4053 - Card Sorting 最长不下降子序列
我们的男主现在手中有n*c张牌,其中有c(<=4)种颜色,每种颜色有n(<=100)张,现在他要排序,首先把相同的颜色的牌放在一起,颜色相同的按照序号从小到大排序.现在他想要让牌的移动次数 ...
- Linq101-Set
using System; using System.Collections.Generic; using System.Linq; namespace Linq101 { class Set { / ...
- redisbook笔记——redis内存映射数据结构
虽然内部数据结构非常强大,但是创建一系列完整的数据结构本身也是一件相当耗费内存的工作,当一个对象包含的元素数量并不多,或者元素本身的体积并不大时,使用代价高昂的内部数据结构并不是最好的办法. 为了解决 ...
- 通知(Notification)的使用
新建一个 NotificationTest项目,并修改 activity_main.xml 中的代码,如下所示:<LinearLayout xmlns:android="http:// ...
- (转帖) 如何將值delay n個clock? (SOC) (Verilog)
来源:http://www.cnblogs.com/oomusou/archive/2009/06/15/verilog_dly_n_clk.html /* (C) OOMusou 2009 http ...
- iOS开发UI篇——九宫格坐标计算
一.要求 完成下面的布局 二.分析 寻找左边的规律,每一个uiview的x坐标和y坐标. 三.实现思路 (1)明确每一块用得是什么view (2)明确每个view之间的父子关系,每个视图都只有一个父视 ...
- ASP.NET菜鸟之路之登录系统
背景 我是一个ASP.NET菜鸟,暂时开始学习ASP.NET,在此记录下我个人敲的代码,没有多少参考价值,请看到的盆友们为我点个赞支持我一下,多谢了. 网站介绍 根据书上的例子做了一个比较粗糙的登录例 ...
- emmt html生成
html:5 或 ! html:5 或!:用于HTML5文档类型 html:xt:用于XHTML过渡文档类型 html:4s:用于HTML4严格文档类型 常用过渡文档类型 html:xt 直接c ...