789A Anastasia and pebbles
1 second
256 megabytes
standard input
standard output
Anastasia loves going for a walk in Central Uzhlyandian Park. But she became uninterested in simple walking, so she began to collect Uzhlyandian pebbles. At first, she decided to collect all the pebbles she could find in the park.
She has only two pockets. She can put at most k pebbles in each pocket at the same time. There are n different pebble types in the park, and there are wi pebbles of the i-th type. Anastasia is very responsible, so she never mixes pebbles of different types in same pocket. However, she can put different kinds of pebbles in different pockets at the same time. Unfortunately, she can't spend all her time collecting pebbles, so she can collect pebbles from the park only once a day.
Help her to find the minimum number of days needed to collect all the pebbles of Uzhlyandian Central Park, taking into consideration that Anastasia can't place pebbles of different types in same pocket.
The first line contains two integers n and k (1 ≤ n ≤ 105, 1 ≤ k ≤ 109) — the number of different pebble types and number of pebbles Anastasia can place in one pocket.
The second line contains n integers w1, w2, ..., wn (1 ≤ wi ≤ 104) — number of pebbles of each type.
The only line of output contains one integer — the minimum number of days Anastasia needs to collect all the pebbles.
3 2
2 3 4
3
5 4
3 1 8 9 7
5
In the first sample case, Anastasia can collect all pebbles of the first type on the first day, of second type — on the second day, and of third type — on the third day.
Optimal sequence of actions in the second sample case:
- In the first day Anastasia collects 8 pebbles of the third type.
- In the second day she collects 8 pebbles of the fourth type.
- In the third day she collects 3 pebbles of the first type and 1 pebble of the fourth type.
- In the fourth day she collects 7 pebbles of the fifth type.
- In the fifth day she collects 1 pebble of the second type.
题意:有两个口袋,每个口袋只能放一种石头,共有 K 种石头, 每种分别w[i] 种
解题思路:计算共需要多少口袋装完, (ans + 1) / 2 即为所求
#include <iostream>
#define ll long long
using namespace std;
int main()
{
ll n, k;
cin >> n >> k;
int ans = ;
for(int i = ;i <= n; i++)
{
int w;
cin >> w;
ans += w / k;
if(w % k) ans++;
}
cout << (ans + ) / << endl;
}
789A Anastasia and pebbles的更多相关文章
- Codeforces 789A Anastasia and pebbles(数学,思维题)
A. Anastasia and pebbles time limit per test:1 second memory limit per test:256 megabytes input:stan ...
- Codeforces 789A Anastasia and pebbles( 水 )
链接:传送门 题意:这个人每次都去公园捡石子,她有两个口袋,每个口袋最多装 k 个石子,公园有 n 种石子,每种石子 w[i] 个,询问最少几天能将石子全部捡完 思路:排个序,尽量每天都多装,如果 k ...
- codeforces 789 A. Anastasia and pebbles
链接 A. Anastasia and pebbles 题意 这个人有两个口袋,有n种类型的鹅卵石,每种鹅卵石有wi个,每次可以放同一种最多k个,每次不能把不同类型的鹅卵石放进同一个口袋,但是她可以同 ...
- CF789A. Anastasia and pebbles
/* CF789A. Anastasia and pebbles http://codeforces.com/contest/789/problem/A 水题 题意:有两个背包,每次分别可取k个物品, ...
- 【codeforces 789A】Anastasia and pebbles
[题目链接]:http://codeforces.com/contest/789/problem/A [题意] 有n种物品,每种物品有wi个; 你有两个口袋,每个口袋最多装k个物品; 且口袋里面只能装 ...
- 【贪心】Codeforces Round #407 (Div. 2) A. Anastasia and pebbles
贪心地一个一个尽可能往口袋里放,容易发现和顺序无关. #include<cstdio> #include<iostream> using namespace std; type ...
- Codeforces Round #407 (Div. 2)
来自FallDream的博客,未经允许,请勿转载,谢谢. ------------------------------------------------------ A.Anastasia and ...
- Codeforces Round #407 div2 题解【ABCDE】
Anastasia and pebbles 题意:你有两种框,每个框可以最多装k重量的物品,但是你每个框不能装不一样的物品.现在地面上有n个物品,问你最少多少次,可以把这n个物品全部装回去. 题解:其 ...
- cf-789A (思维)
A. Anastasia and pebbles time limit per test 1 second memory limit per test 256 megabytes input stan ...
随机推荐
- 解决:python 连接Oracle 11g 报错:ORA-12514: TNS: 监听程序当前无法识别连接描述符中请求的服务
其次,将查询到的service_name替换sid即可:conn=cx_Oracle.connect('hr/admin@localhost:1521/EE.oracle.docker')
- Pandas数据存取
pd.read_excel('foo.xlsx', 'Sheet1', index_col=None, na_values=['NA']) Pandas数据存取 Pandas可以存取多种介质类型数据, ...
- javascript中scrollTop和offsetTop的区别
scrollTop是指某个可滚动区块向下滚动的距离,offsetTop则是元素的上边框与父元素的上边框的绝对距离. 1.offsetTop : 当前对象到其上级层顶部的距离. 不能对其进行赋值.设 ...
- Centos7安装出现的问题:找不到安装源或者检查软件配置出错
安装启动时到以下界面 此时,按一下Tab键,将会出现在屏幕下方出现这一串文字 vmlinuz initrd=initrd.img inst.stage2=hd:LABEL=CentOS\x207\x2 ...
- 判断元素16种方法expected_conditions
前言 标签(空格分隔): 判断元素 经常有小伙伴问,如何判断一个元素是否存在,如何判断alert弹窗出来了,如何判断动态的元素等等一系列的判断,在selenium的expected_condition ...
- ssh X协议转发
X协议的作用是远程登录Linux运行GUI界面 主机2开启ssh服务service ssh start 主机1 ssh连接主机2:ssh -X root@192.168.1.110 -p 53 然后可 ...
- 解题4(NumberToEnglish )
题目描述 Jessi初学英语,为了快速读出一串数字,编写程序将数字转换成英文: 如22:twenty two,123:one hundred and twenty three. 说明: 数字为正整数, ...
- 使用 Actuator 监控
参考文章:https://www.jianshu.com/p/ba85f56a2013 Actuator 提供对自身应用的监控.配置查看等. 步骤一:导入actuator 依赖 <depende ...
- day17 正则表达式 re模块和hashlib模块
今日内容 1. re&正则表达式(*****) 注:不要将自定义文件命名为re import re re.findall(正则表达式,被匹配的字符串) 拿着正则表达式去字符串中找,返回一个列表 ...
- Java获得数据库查询结果的列数和行数,打印查询结果
Java连接数据库及简单操作见我以前的一篇随笔:http://www.cnblogs.com/meitian/p/5036332.html 一.获取查询结果的行数和列数 查询结果为ResultSe ...