POJ 2370 Democracy in danger(简单贪心)
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 3388 | Accepted: 2508 |
Description
The essence of the reform is as follows. From the moment of its coming into effect all the citizens were divided into K (may be not equal) groups. Votes on every question were to be held then in each group, moreover, the group was said to vote "for" if more than half of the group had voted "for", otherwise it was said to vote "against". After the voting in each group a number of group that had voted "for" and "against" was calculated. The answer to the question was positive if the number of groups that had voted "for" was greater than the half of the general number of groups.
At first the inhabitants of the island accepted this system with pleasure. But when the first delights dispersed, some negative properties became obvious. It appeared that supporters of the party, that had introduced this system, could influence upon formation of groups of voters. Due to this they had an opportunity to put into effect some decisions without a majority of voters "for" it.
Let's consider three groups of voters, containing 5, 5 and 7 persons, respectively. Then it is enough for the party to have only three supporters in each of the first two groups. So it would be able to put into effect a decision with the help of only six votes "for" instead of nine, that would .be necessary in the case of general votes.
You are to write a program, which would determine according to the given partition of the electors the minimal number of supporters of the party, sufficient for putting into effect of any decision, with some distribution of those supporters among the groups.
Input
Output
Sample Input
3
5 7 5
Sample Output
6
Source
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
using namespace std;
int main()
{
int a[];
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=;i<n;i++)
scanf("%d",&a[i]);
sort(a,a+n);
int sum=;
for(int i=;i<n/+;i++)
sum+=a[i]/+;
printf("%d\n",sum);
}
return ;
}
POJ 2370 Democracy in danger(简单贪心)的更多相关文章
- CF 628C --- Bear and String Distance --- 简单贪心
CF 628C 题目大意:给定一个长度为n(n < 10^5)的只含小写字母的字符串,以及一个数d,定义字符的dis--dis(ch1, ch2)为两个字符之差, 两个串的dis为各个位置上字符 ...
- poj 3069 Saruman's Army (贪心)
简单贪心. 从左边开始,找 r 以内最大距离的点,再在该点的右侧找到该点能覆盖的点.如图. 自己的逻辑有些混乱,最后还是参考书上代码.(<挑战程序设计> P46) /*********** ...
- Uva 11729 Commando War (简单贪心)
Uva 11729 Commando War (简单贪心) There is a war and it doesn't look very promising for your country. N ...
- CDOJ 1502 string(简单贪心)
题目大意:原题链接 相邻两个字母如果不同,则可以结合为前一个字母,如ac可结合为a.现给定一个字符串,问结合后最短可以剩下多少个字符串 解体思路:简单贪心 一开始读题时,就联想到之前做过的一道题,从后 ...
- ACM_发工资(简单贪心)
发工资咯: Time Limit: 2000/1000ms (Java/Others) Problem Description: 作为广财大的老师,最盼望的日子就是每月的8号了,因为这一天是发工资的日 ...
- ACM_Ruin of Titanic(简单贪心)
Ruin of Titanic Time Limit: 2000/1000ms (Java/Others) Problem Description: 看完Titanic后,小G做了一个梦.梦见当泰坦尼 ...
- POJ 2209 The King(简单贪心)
The King Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 7499 Accepted: 4060 Descript ...
- POJ 2376 (区间问题,贪心)
题目链接:http://poj.org/problem?id=2376 题目大意:选择一些区间使得能够覆盖1-T中的每一个点,并且区间数最少 题目分析:这道题目很明显可以用贪心法来解决.但题目没有看起 ...
- hdu 2037简单贪心--活动安排问题
活动安排问题就是要在所给的活动集合中选出最大的相容活动子集合,是可以用贪心算法有效求解的很好例子.该问题要求高效地安排一系列争用某一公共资源的活动.贪心算法提供了一个简单.漂亮的方法使得尽可能多的活动 ...
随机推荐
- ES6之Promise
Promise是一个对象,用来传递异步操作的消息,他有两个特点:第一对象的状态不受外界的影响,第二一旦状态改变就不会在变,任何时候都可以得到这个结果,他有两个参数分别是resolve(他的作用是将Pr ...
- 用C#写入Excel表并保存
想用C#操作Excel表,首先要做一些准备工作. 如果要操作 microsoft office Excel 2003表,就需要引入Microsoft office 11.0 object librar ...
- tensorflow ckpt文件转caffemodel时遇到的坑
p.p1 { margin: 0.0px 0.0px 0.0px 0.0px; font: 12.0px ".PingFang SC"; color: #454545 } p.p2 ...
- 程序员的自我救赎---1.4.1:核心框架讲解(DAL)
<前言> (一) Winner2.0 框架基础分析 (二)PLSQL报表系统 (三)SSO单点登录 (四) 短信中心与消息中心 (五)钱包系统 (六)GPU支付中心 (七)权限系统 (八) ...
- bzoj 3551: [ONTAK2010]Peaks加强版
Description [题目描述]同3545 Input 第一行三个数N,M,Q. 第二行N个数,第i个数为h_i 接下来M行,每行3个数a b c,表示从a到b有一条困难值为c的双向路径. 接下来 ...
- MySQL数据库入门(建库和建表)--陈远波
建库.建表 1.建库 (1)SQL语句命令建库: Create database数据库名称 (该方法创建的数据库没有设置编码乱码) 1 2 3 4 5 -- 创建数据库时,设置数据库的编码方式 -- ...
- cleanMyMac
想看外国网站可以找我,facebook.youtube.XX大片等,只要8元钱,无限制用到服务器关闭.看大片流畅不成问题 需要cleanMyMac的请加微信只要10或直接拍 http://a.p6ff ...
- pycharm2017.3专业版激活注册码
pycharm作为一个不错的python编程的ide很有用处 这里拔出一段专业版的注册码,社区版用起来确实着实让人着急. 2017-12-1921:40:38 EB101IWSWD-eyJsaWNlb ...
- 【自问自答】关于 Swift 的几个疑问
感觉自己给自己释疑,也是一个极为有趣的过程.这次,我还新增了"猜想"一栏,来尝试回答一些暂时没有足够资料支撑的问题. Swift 版本是:4.0.3.不同版本的 Swift,可能无 ...
- 前端学习_01_css网页布局
引子 之前也自己陆陆续续地学了一些web方面的知识,包括前段和后端都有涉及到,自己也比较感兴趣,感谢peter老师,愿意无偿提供从零开始的教学,之前也看过peter老师的一些视频,节奏非常适合我,决心 ...