Democracy in danger
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 3388   Accepted: 2508

Description

In one of the countries of Caribbean basin all decisions were accepted by the simple majority of votes at the general meeting of citizens (fortunately, there were no lots of them). One of the local parties, aspiring to come to power as lawfully as possible, got its way in putting into effect some reform of the election system. The main argument was that the population of the island recently had increased and it was to longer easy to hold general meetings.
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

The input of this problem contains two lines. In the first line an only natural number K <= 101 — a quantity of groups — is written. In the second line there are written K natural numbers, separated with a space. Those numbers define a number of voters in each group. In order to simplify the notion of "the majority of votes" we'll say that the number of groups also as the number of voters in each group is odd. You may also consider, that the population of the island does not exceeds 10001 persons.

Output

You should write an only natural number — a minimal quantity of supporters of the party, that can put into effect any decision.

Sample Input

3
5 7 5

Sample Output

6

Source

题目链接:http://poj.org/problem?id=2370
题解:以前挂的一些贪心的题没有做,有位大佬叫我写下题解,有些看不懂题意,我恭敬不如从命了,写点吧,算是复习下贪心吧!
题目大意是关于投票,已知k个组,这k个组中只要有一半以上通过了,就算通过了所以取k/2+1;要想去最少的通过人数,就想办法使得这k/2+1这些组的人数都是最少的,这时可以进行排序,然后取前k/2+1个组;每个组中只要有一半以上的人通过了,就算通过了,所以只要这些k/2+1组的每组超过一半的人通过了,就通过了;及a/2+1,a为每组的人数!
下面给出AC代码:
 #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

  1. 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 ...

  2. 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) /*********** ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. POJ 2209 The King(简单贪心)

    The King Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7499   Accepted: 4060 Descript ...

  8. 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 ...

  9. 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

  1. NoSQL数据库

    NoSQL数据库 1.NoSQL简介 最初表示"反SQL"运动,用新型的非关系型数据库取代关系数据库:现在表示"Not only SQL"关系和非关系型数据库各 ...

  2. 云计算之路-阿里云上-2017年最错误的选择: 生产环境使用 docker swarm

    2017年12月29日 10:18 ~ 11:00 左右,由于整个 docker swarm 集群宕机,造成我们迁移至 .net core 跑在 docker swram 上的所有站点无法正常访问,由 ...

  3. 解决vue.js修改数据无法触发视图

    data:{checkValue:{}}that.checkValue[key] = [] 赋值无法实时改变变量:(数据其实最终被修改,但是并没有触发检测从而更新视图)原因:Vue 不能检测到对象属性 ...

  4. openstack ocata版本简化安装

    Network Time Protocol (NTP) Controller Node apt install chrony Edit the /etc/chrony/chrony.conf 添加如下 ...

  5. LODOP打印控件示例

    一.lodop打印预览效果图 LODOP.PRINT_SETUP();打印维护效果图 LODOP.PREVIEW();打印预览图 二.写在前面 最近项目用到了LODOP的套打,主要用到两个地方,一是物 ...

  6. MYSQL忘记root密码后如何修改

    方法1: 用SET PASSWORD命令 首先登录MySQL. 格式:mysql> set password for 用户名@localhost = password('新密码'); 例子:my ...

  7. fdisk 命令详解

    fdisk  作用: 查看磁盘实体使用情况,也可对硬盘分区. 选项:  -b 分区大小 -l  列出指定的外围设备的分区表状况 -s 分区编号, 将指定的分区大小输出到标准输出上, 单位为区块 -u ...

  8. pinyin utils

    package cn.itcast.bos.utils;   import java.util.Arrays;   import net.sourceforge.pinyin4j.PinyinHelp ...

  9. SQL SERVER 日期转换大全

    博客转自:http://blog.csdn.net/baiduandxunlei/article/details/9180075 CONVERT(data_type,expression[,style ...

  10. jQuery基础 (四)——使用jquery-cookie 实现点赞功能

    jquery-cookie 下载地址:https://github.com/carhartl/jquery-cookie 直接上代码 html <span class="jieda-z ...