C. Appleman and Toastman
time limit per test 

2 seconds

memory limit per test 

256 megabytes

input 

standard input

output 

standard output

Appleman and Toastman play a game. Initially Appleman gives one group of nnumbers to the Toastman, then they start to complete the following tasks:

  • Each time Toastman gets a group of numbers, he sums up all the numbers and adds this sum to the score. Then he gives the group to the Appleman.
  • Each time Appleman gets a group consisting of a single number, he throws this group out. Each time Appleman gets a group consisting of more than one number, he splits the group into two non-empty groups (he can do it in any way) and gives each of them to Toastman.

After guys complete all the tasks they look at the score value. What is the maximum possible value of score they can get?

Input

The first line contains a single integer n (1 ≤ n ≤ 3·105). The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 106) — the initial group that is given to Toastman.

Output

Print a single integer — the largest possible score.

Sample test(s)
input
3
3 1 5
output
26
input
1
10
output
10
Note

Consider the following situation in the first example. Initially Toastman gets group [3, 1, 5] and adds 9 to the score, then he give the group to Appleman. Appleman splits group [3, 1, 5] into two groups: [3, 5] and [1]. Both of them should be given to Toastman. When Toastman receives group [1], he adds 1 to score and gives the group to Appleman (he will throw it out). When Toastman receives group [3, 5], he adds 8 to the score and gives the group to Appleman. Appleman splits [3, 5] in the only possible way: [5] and [3]. Then he gives both groups to Toastman. When Toastman receives [5], he adds 5 to the score and gives the group to Appleman (he will throws it out). When Toastman receives [3], he adds 3 to the score and gives the group to Appleman (he will throws it out). Finally Toastman have added 9 + 1 + 8 + 5 + 3 = 26 to the score. This is the optimal sequence of actions.

解题:贪心,当时乱搞了下,居然对了。

 #include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <climits>
#include <vector>
#include <queue>
#include <cstdlib>
#include <string>
#include <set>
#include <stack>
#define LL long long
#define pii pair<int,int>
#define INF 0x3f3f3f3f
using namespace std;
const int maxn = ;
LL d[maxn],sum,ans;
int n;
int main() {
while(~scanf("%d",&n)){
for(int i = sum = ; i < n; i++){
cin>>d[i];
sum += d[i];
}
sort(d,d+n);
ans = sum;
for(int i = ; i+ < n; i++){
ans += sum;
sum -= d[i]; }
cout<<ans<<endl;
}
return ;
}

Codeforces 263C. Appleman and Toastman的更多相关文章

  1. codeforces 462C Appleman and Toastman 解题报告

    题目链接:http://codeforces.com/problemset/problem/461/A 题目意思:给出一群由 n 个数组成的集合你,依次循环执行两种操作: (1)每次Toastman得 ...

  2. 贪心 Codeforces Round #263 (Div. 2) C. Appleman and Toastman

    题目传送门 /* 贪心:每次把一个丢掉,选择最小的.累加求和,重复n-1次 */ /************************************************ Author :R ...

  3. Codeforces 461B Appleman and Tree(木dp)

    题目链接:Codeforces 461B Appleman and Tree 题目大意:一棵树,以0节点为根节点,给定每一个节点的父亲节点,以及每一个点的颜色(0表示白色,1表示黑色),切断这棵树的k ...

  4. CodeForces 462B Appleman and Card Game(贪心)

    题目链接:http://codeforces.com/problemset/problem/462/B Appleman has n cards. Each card has an uppercase ...

  5. Codeforces461A Appleman and Toastman 贪心

    题目大意是Appleman每次将Toastman给他的Ni个数拆分成两部分后再还给Toastman,若Ni == 1则直接丢弃不拆分.而Toastman将每次获得的Mi个数累加起来作为分数,初始时To ...

  6. Codeforces 263B. Appleman and Card Game

    B. Appleman and Card Game time limit per test  1 second memory limit per test  256 megabytes input  ...

  7. Codeforces 263A. Appleman and Easy Task

    A. Appleman and Easy Task time limit per test  1 second memory limit per test  256 megabytes input  ...

  8. Codeforces 461B. Appleman and Tree[树形DP 方案数]

    B. Appleman and Tree time limit per test 2 seconds memory limit per test 256 megabytes input standar ...

  9. Codeforces 461B Appleman and Tree

    http://codeforces.com/problemset/problem/461/B 思路:dp,dp[i][0]代表这个联通块没有黑点的方案数,dp[i][1]代表有一个黑点的方案数 转移: ...

随机推荐

  1. 异常java.lang.UnsupportedOperationException: The application must supply JDBC connections

    转自:https://blog.csdn.net/q952420873/article/details/81355586 先上图  根据这个错误溯源 于是 我来到了数据库连接部分的代码 ,发现多了一个 ...

  2. 43. ExtJs控件属性配置详细

    转自:https://www.cnblogs.com/mannixiang/p/6558225.html 序言:    1.本文摘自网络,看控件命名像是4.0以前的版本,但控件属性配置仍然可以借鉴(不 ...

  3. 3.4 目录和spooling

    文件管理部分主要讲文件目录.文件目录它是用于检索文件的.文件目录它是一种文件系统实现按0存取的一种重要手段.一个文件目录它由若干个目录项组成的.每一个目录项它记录了一个文件的相关信息.这个文件信息指明 ...

  4. [App Store Connect帮助]三、管理 App 和版本(5)添加平台以创建通用购买

    您可以为 App 添加一个平台以创建通用购买.例如,为现有的 iOS App 添加相关的 Apple TVOS App,从而将该 Apple TVOS App 和 iOS App 一同出售. 与创建新 ...

  5. Akka源码分析-Actor发消息(续)

    上一篇博客我们分析道mailbox同时也是一个forkjointask,run方法中,调用了processMailbox处理一定数量的消息,然后最终调用dispatcher的registerForEx ...

  6. 音频处理中的尺度--Bark尺度与Mel尺度

    由于人耳对声音的感知(如:频率.音调)是非线性的,为了对声音的感知进行度量,产生了一系列的尺度(如:十二平均律),这里重点说下Bark尺度与Mel尺度.刚开始的时候,我自己也没弄明白这两个尺度的区别. ...

  7. NetCore Netty 框架 BT.Netty.RPC 系列随讲 二 WHO AM I 之 NETTY/NETTY 与 网络通讯 IO 模型之关系?

    一:NETTY 是什么? Netty 是什么?  这个问题其实百度上一搜一堆. 这是官方话的描述:Netty 是一个基于NIO的客户.服务器端编程框架,使用Netty 可以确保你快速和简单的开发出一个 ...

  8. linux tmux基本操作

    1. 安装工具 Centos : yum install tmux 2. 基本操作 新建会话:tmux new -s session-name 查看会话:tmux ls 进入会话:tmux a -t ...

  9. mysql中类型转换

    MySQL 的CAST()和CONVERT()函数可用来获取一个类型的值,并产生另一个类型的值 CAST(xxx AS 类型), CONVERT(xxx,类型) 二进制,同带binary前缀的效果 : ...

  10. MVC系列学习(十二)-服务端的验证

    在前一讲,提到过,客户端的东西永远可以造假,所以我们还要在服务端进行验证 注意:先加载表单,后添加js文件,才能有效:而先加载js,后添加表单,是没有效果的 1.视图与Model中的代码如下 2.一张 ...