A. Anastasia and pebbles
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

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.

Input

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.

Output

The only line of output contains one integer — the minimum number of days Anastasia needs to collect all the pebbles.

Examples
Input
3 2
2 3 4
Output
3
Input
5 4
3 1 8 9 7
Output
5
Note

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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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, ...

  5. 【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 ...

  6. 【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 ...

  7. Codeforces Round #407 (Div. 2)

    来自FallDream的博客,未经允许,请勿转载,谢谢. ------------------------------------------------------ A.Anastasia and ...

  8. 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 ...

  9. 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

  1. MVC模式浅谈

    MVC模式浅谈 一.MVC模式概述 模型-视图-控制器(MVC模式)是一种非常经典的软件架构模式,在UI框架和UI设计思路中扮演着非常重要的角色.从设计模式的角度来看,MVC模式是 一种复合模式,它将 ...

  2. tabel 选中行变色和取当前选中行值等问题

    先把代码贴出来 $("#tableId tbody tr").mousedown(function () {        $('#tableId  tr').each(funct ...

  3. 关于git经常忘记的:远程仓库关联。

    我们有时习惯建立好工程后再传到git上,这是时候就忘记咋弄啦, 其实,只要配置远程仓库就行: git remote add +url...具体看网上哦,这里提醒下 Git clone远程分支 Git ...

  4. SAP 自定义进度条

    *&---------------------------------------------------------------------* *& Report ZCHENH028 ...

  5. 04_web基础(六)之请求转发与重定向

    1.交互方式 Web组件之间跳转: 从AServlet 跳转到 BServlet. 三种类型: 1:请求转发(forward) 2:URL重定向(redirect) 3:请求包含(include) 3 ...

  6. JIT物料在途未清PO作为供给

    以下是本应在TEMP中处理,目前是写到FP_CHECK_SAP_DATA中 ----JIT物料在途未清PO作为供给 add by landor on 20180809 INSERT INTO SAP_ ...

  7. Android Studio 插件-Android Styler 的使用 (转)

    作用:把 xml文件 转为 style 截图保留 使用方法 使用方法:选中xml代码 按下 Ctrl+Shift+D 转自:http://blog.csdn.net/zxwd2015/article/ ...

  8. vmware 完全关闭时间同步

    参考 http://blog.51cto.com/hezhang/1535577 修改.vmx文件 tools.syncTime = "FALSE" time.synchroniz ...

  9. centos搭建OPENldap

    LDAP是轻量目录访问协议,英文全称是Lightweight Directory Access Protocol,一般都简称为LDAP.它是基于X.500标准的,但是简单多了并且可以根据需要定制.与X ...

  10. PAT1131(dfs)

    In the big cities, the subway systems always look so complex to the visitors. To give you some sense ...