Codeforces 263C. Appleman and Toastman
2 seconds
256 megabytes
standard input
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?
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.
Print a single integer — the largest possible score.
3
3 1 5
26
1
10
10
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
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- Codeforces 263B. Appleman and Card Game
B. Appleman and Card Game time limit per test 1 second memory limit per test 256 megabytes input ...
- Codeforces 263A. Appleman and Easy Task
A. Appleman and Easy Task time limit per test 1 second memory limit per test 256 megabytes input ...
- 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 ...
- 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
- .NET 的WebSocket开发包详细比较(2)
AlchemyWebSocket http://alchemywebsockets.net/ 当我想到websocket库时,这个让人不可思议.没错这是真的.它可以排在Fleck后面,它非常容易使用, ...
- Black Rock Shooter
在人气动漫 Black Rock shooter 中,当加贺里对麻陶 说出了"滚回去"以后,与此同时,在另一个心灵世界里, BRS 也遭到了敌人的攻击.此时,一共有 n 个攻击排成 ...
- eclipse中Kotlin的基础应用
最近逛网站时无意中发现有一门新语言谈论很广-- kotlin ,能够完全兼容Java.这就引起了楼主的好奇心,据所周知,Java就是因为多平台的支持 才流行起来.OK,闲话不多说,下面看图讲代码. 1 ...
- [Apple开发者帐户帮助]六、配置应用服务(5.1)推送通知(APN):使用身份验证令牌与APN通信
您可以使用一个APN签名密钥为多个应用程序验证令牌.签名密钥适用于开发和生产环境.签名密钥不会过期,但可以撤消. 首先在Xcode项目中启用推送通知.接下来创建并下载启用了APN 的私钥. 然后获取密 ...
- [Swift通天遁地]九、拔剑吧-(16)搭建卡片页面:Card Peek/Pop动态切换界面
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★➤微信公众号:山青咏芝(shanqingyongzhi)➤博客园地址:山青咏芝(https://www.cnblogs. ...
- Java高质量20问
问题一:在多线程环境中使用HashMap会有什么问题?在什么情况下使用get()方法会产生无限循环? HashMap本身没有什么问题,有没有问题取决于你是如何使用它的.比如,你在一个线程里初始化了一个 ...
- JavaScript判断对象数组中是否存在某个对象【转】
1. 如果要判断数组中是否存在某个元素的话很好判断,直接用数组的indexOf方法就好,存在返回当前索引不存在返回-1 var arr=[1,2,3,4] arr.indexOf(3) arr.ind ...
- HTML--文本域,支持多行文本输入
当用户需要在表单中输入大段文字时,需要用到文本输入域. 语法: <textarea rows="行数" cols="列数">文本</texta ...
- j建立一个小的servlet小程序
我们建立一个最简单的servlet程序,这个servelt程序只是单纯的输出helloworld. 步骤如下:如图:在Eclipse中选择新建一个项目,其中选择tomcat project然后点击下一 ...
- 332 Reconstruct Itinerary 重建行程单
Given a list of airline tickets represented by pairs of departure and arrival airports [from, to], r ...