Stripies
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 21506   Accepted: 9478

Description

Our chemical biologists have invented a new very useful form of life called stripies (in fact, they were first called in Russian - polosatiki, but the scientists had to invent an English name to apply for an international patent). The stripies are transparent amorphous amebiform creatures that live in flat colonies in a jelly-like nutrient medium. Most of the time the stripies are moving. When two of them collide a new stripie appears instead of them. Long observations made by our scientists enabled them to establish that the weight of the new stripie isn't equal to the sum of weights of two disappeared stripies that collided; nevertheless, they soon learned that when two stripies of weights m1 and m2 collide the weight of resulting stripie equals to 2*sqrt(m1*m2). Our chemical biologists are very anxious to know to what limits can decrease the total weight of a given colony of stripies. 
You are to write a program that will help them to answer this question. You may assume that 3 or more stipies never collide together. 

Input

The first line of the input contains one integer N (1 <= N <= 100) - the number of stripies in a colony. Each of next N lines contains one integer ranging from 1 to 10000 - the weight of the corresponding stripie.

Output

The output must contain one line with the minimal possible total weight of colony with the accuracy of three decimal digits after the point.

Sample Input

3
72
30
50

Sample Output

120.000

贪心和优先队列,先选择最大的两个数合并,然后加入队列,可以用priority_queue(),默认为最大堆。
C++代码(只能用C++编辑器过,G++却会WA)
#include<iostream>
#include<queue>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;
int main(){
priority_queue<double> pq;
int N;
scanf("%d",&N);
double m;
int t = N;
while(t--){
scanf("%lf",&m);
pq.push(m);
}
int k = N - ;
double a,b;
while(k--){
a = pq.top();
pq.pop();
b = pq.top();
pq.pop();
pq.push( * sqrt(a*b));
}
printf("%.3lf\n",pq.top());
pq.pop();
return ;
}
#include<iostream>
#include<queue>
#include<algorithm>
#include<cstdio>
#include<cmath>
using namespace std;
int main(){
priority_queue<double> pq;
int N;
scanf("%d",&N);
double m;
int t = N;
while(t--){
scanf("%lf",&m);
pq.push(m);
}
double a,b,tmp;
while(!pq.empty()){
a = pq.top();
pq.pop();
if(pq.empty()){
printf("%.3lf\n",a);
break; //要加break
}
b = pq.top();
pq.pop();
tmp = * sqrt(a*b);
pq.push(tmp);
}
return ;
}

(贪心和优先队列) POJ1862 Stripies的更多相关文章

  1. Luogu 1090 合并果子(贪心,优先队列,STL运用)

    Luogu 1090 合并果子(贪心,优先队列,STL运用) Description 在一个果园里,多多已经将所有的果子打了下来,而且按果子的不同种类分成了不同的堆.多多决定把所有的果子合成一堆. 每 ...

  2. bzoj1528 sam-Toy Cars(贪心,优先队列)

    「BZOJ1528」[POI2005] sam – Toy Cars Description Jasio 是一个三岁的小男孩,他最喜欢玩玩具了,他有n 个不同的玩具,它们都被放在了很高的架子上所以Ja ...

  3. POJ1862 Stripies 贪心 B

    POJ 1862 Stripies https://vjudge.net/problem/POJ-1862 题目:     Our chemical biologists have invented ...

  4. CF #374 (Div. 2) D. 贪心,优先队列或set

    1.CF #374 (Div. 2)   D. Maxim and Array 2.总结:按绝对值最小贪心下去即可 3.题意:对n个数进行+x或-x的k次操作,要使操作之后的n个数乘积最小. (1)优 ...

  5. sgu548 Dragons and Princesses   贪心+优先队列

    题目链接:http://acm.sgu.ru/problem.php?contest=0&problem=548 题目意思: 有一个骑士,要经过n个房间,开始在第一个房间,每个房间里面有龙或者 ...

  6. poj3190区间类贪心+优先队列

    题意:每个奶牛产奶的时间为A到B,每个奶牛产奶时要占用一间房子,问n头奶牛产奶共需要多少房子,并输出每头奶牛用哪间房子 分析:这题就是一个裸的贪心,将奶牛按开始时间进行排序即可,但考虑一下数据范围,我 ...

  7. 【贪心算法】POJ-1862 简单哈夫曼

    一.题目 Description Our chemical biologists have invented a new very useful form of life called stripie ...

  8. 【BZOJ 3661】 Hungry Rabbit (贪心、优先队列)

    3661: Hungry Rabbit Time Limit: 100 Sec  Memory Limit: 512 MBSec  Special JudgeSubmit: 67  Solved: 4 ...

  9. 2015多校第6场 HDU 5360 Hiking 贪心,优先队列

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5360 题意:给定n个人,现在要邀请这些人去远足,但每个人同意邀请的条件是当前已经同意去远足的人数c必须 ...

随机推荐

  1. Supervisord管理进程实践

    今天凑空研究了下Supervisord,这是一款linux进程管理工具,使用python开发,主要用于在后台维护进程(类似master守护进程),可以实现监控进程的状态.自动重启进程等操作,便于一些服 ...

  2. 前端bug记录

    记录一下前端入门时期遇到的一些bug. 1.同步.异步问题 背景:Javascript语言的执行环境是单线程.即一次只能完成一个任务.若有多个任务则需排队逐个执行——前一个任务完成,再执行后一个任务. ...

  3. HTTP协议那些事儿(Web开发补充知识点)

    HTTP协议 超文本传输协议(英文:HyperText Transfer Protocol,缩写:HTTP)是一种用于分布式.协作式和超媒体信息系统的应用层协议.HTTP是万维网的数据通信的基础. H ...

  4. 了解AutoCAD对象层次结构 —— 3 ——数据库

    数据库的结构是什么样的?对象是如何存储在数据库中的?这些问题我们需要搞明白.在此我们可以借助工具ArxDbg或MgdDbg来查看数据库结构及其内容.下图就是利用MgdDbg工具查看到的内容,我们可以看 ...

  5. java Builder模式创建不可变类

    package com.geostar.gfstack.operationcenter.logger.manager.common; /** * Created by Nihaorz on 2017/ ...

  6. Codeforces Round #419 Div. 1

    A:暴力枚举第一列加多少次,显然这样能确定一种方案. #include<iostream> #include<cstdio> #include<cmath> #in ...

  7. 一个服务器创建两个MySQL

    转载:http://www.2cto.com/database/201412/357863.html 将已安装的数据库文件夹复制到另一个目录下 打开复制目录下的my.ini文件修改 [client] ...

  8. Codeforces1023E Down or Right 【贪心】

    题目分析: 从起点开始询问终点连通性,优先右走.从终点开始询问起点连通性,优先上走. 代码: #include<bits/stdc++.h> using namespace std; in ...

  9. Matplotlib学习---用matplotlib画柱形图,堆积柱形图,横向柱形图(bar chart)

    这里利用Nathan Yau所著的<鲜活的数据:数据可视化指南>一书中的数据,学习画图. 数据地址:http://datasets.flowingdata.com/hot-dog-cont ...

  10. tp5 日志管理

    日志驱动 日志可以通过驱动支持不同的方式写入,默认日志会记录到文件中,系统已经内置的写入驱动包括 File.Socket,如果要临时关闭日志写入,可以设置日志类型为Test即可,例如: 'log' = ...