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
- ZOJ3329之经典概率DP
One Person Game Time Limit: 1 Second Memory Limit: 32768 KB Special Judge There is a very ...
- Spring 3整合Quartz 2实现定时任务--转
常规整合 http://www.meiriyouke.net/?p=82 最近工作中需要用到定时任务的功能,虽然Spring3也自带了一个轻量级的定时任务实现,但感觉不够灵活,功能也不够强大.在考虑之 ...
- Android 开源项目 eoe 社区 Android 客户端(转)
本文内容 环境 开源项目 eoe 社区 Android 客户端 本文介绍 eoe 社区 Android 客户端.它是一个开源项目,功能相对简单,采用侧边菜单栏.可以学习一下.点击此处查看 GitHub ...
- Linux开发工具之gdb(上)
三.gdb调试(上) 01.gdb:gdb是GNU debugger的缩写,是编程调试工作. 功能: 启动程序,可以按照用户自定义的要求随心所欲的运行程序: 可让被调试的程序在用户所指定的调试 ...
- 模板-->Guass消元法(求解多元一次方程组)
如果有相应的OJ题目,欢迎同学们提供相应的链接 相关链接 所有模板的快速链接 简单的测试 None 代码模板 /* * TIME COMPLEXITY:O(n^3) * PARAMS: * a The ...
- python-property、docstring--笔记
已经有人总结的非常详细,而且写得不错,就直接贴过来用了 property作为装饰器函数,负责把一个方法变成属性调用的 廖雪峰关于property的讲解 http://www.liaoxuefeng.c ...
- HTML5 FileAPI读取实例---(一)
在HTML5中,提供了一个关于文件操作的API,通过这个API,对于从web页面上访问本地文件系统的相关处理变得十分简单.到目前为止只有部分浏览器对它提供支持. 1.FileList对象和File对象 ...
- Ext checkbox
Ext.require([ 'Ext.grid.*', 'Ext.data.*', 'Ext.util.*', 'Ext.grid.PagingScroller', ...
- app发布流程详解
https://developer.apple.com 1. 点击 Member Center 2. 创建应用ID 3. 创建项目 4. 在AppStore创建对应的应用 5. 创建授权文件 6. 配 ...
- Linux文件编程实例
//捕获fopen调用中的错误 #include <stdio.h> #include <errno.h> #include <string.h> #define ...