B. Vanya and Food Processor【转】
B. Vanya and Food Processortime limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Vanya smashes potato in a vertical food processor. At each moment of time the height of the potato in the processor doesn't exceed hand the processor smashes k centimeters of potato each second. If there are less than k centimeters remaining, than during this second processor smashes all the remaining potato.
Vanya has n pieces of potato, the height of the i-th piece is equal to ai. He puts them in the food processor one by one starting from the piece number 1 and finishing with piece number n. Formally, each second the following happens:
- If there is at least one piece of potato remaining, Vanya puts them in the processor one by one, until there is not enough space for the next piece.
- Processor smashes k centimeters of potato (or just everything that is inside).
Provided the information about the parameter of the food processor and the size of each potato in a row, compute how long will it take for all the potato to become smashed.
InputThe first line of the input contains integers n, h and k (1 ≤ n ≤ 100 000, 1 ≤ k ≤ h ≤ 109) — the number of pieces of potato, the height of the food processor and the amount of potato being smashed each second, respectively.
The second line contains n integers ai (1 ≤ ai ≤ h) — the heights of the pieces.
OutputPrint a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.
Examplesinput5 6 3
5 4 3 2 1output5input5 6 3
5 5 5 5 5output10input5 6 3
1 2 1 1 1output2NoteConsider the first sample.
- First Vanya puts the piece of potato of height 5 into processor. At the end of the second there is only amount of height 2 remaining inside.
- Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
- Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
- Vanya finally puts the pieces of height 2 and 1 inside. At the end of the second the height of potato in the processor is equal to 3.
- During this second processor finally smashes all the remaining potato and the process finishes.
In the second sample, Vanya puts the piece of height 5 inside and waits for 2 seconds while it is completely smashed. Then he repeats the same process for 4 other pieces. The total time is equal to 2·5 = 10 seconds.
In the third sample, Vanya simply puts all the potato inside the processor and waits 2 seconds.
题解代码:
#include <iostream>
#include <stdio.h>
using namespace std;
typedef long long ll;
ll i,n,h,ans,x,cur_h,k;
int main()
{
cin >> n >> h >> k;
ans = ;
cur_h = ;
for (i = ; i < n; i++)
{
scanf("%I64d", &x);
if (cur_h + x <= h)
cur_h += x;
else
ans++, cur_h = x;
ans += cur_h/k;
cur_h %= k;
}
ans += cur_h/k;
cur_h %= k;
ans += (cur_h>);
cout << ans << endl;
return ;
}
B. Vanya and Food Processor【转】的更多相关文章
- codeforces 677B B. Vanya and Food Processor(模拟)
题目链接: B. Vanya and Food Processor time limit per test 1 second memory limit per test 256 megabytes i ...
- Codeforces Round #355 (Div. 2) B. Vanya and Food Processor 水题
B. Vanya and Food Processor 题目连接: http://www.codeforces.com/contest/677/problem/B Description Vanya ...
- Codeforces Round #355 (Div. 2)-B. Vanya and Food Processor,纯考思路~~
B. Vanya and Food Processor time limit per test 1 second memory limit per test 256 megabytes input s ...
- 暑假练习赛 006 A Vanya and Food Processor(模拟)
Description Vanya smashes potato in a vertical food processor. At each moment of time the height of ...
- [Codeforces677B]Vanya and Food Processor(模拟,数学)
题目链接:http://codeforces.com/contest/677/problem/B 题意:n个土豆,每个土豆高ai.现在有个加工机,最高能放h,每次能加工k.问需要多少次才能把土豆全加工 ...
- [ An Ac a Day ^_^ ] CodeForces 677B Vanya and Food Processor 模拟
题意: 你有一个榨汁机 还有n个土豆 榨汁机可以容纳h高的土豆 每秒可以榨k高的东西 问按顺序榨完土豆要多久 思路: 直接模拟 一开始以为是最短时间排了个序 后来发现多余了…… #include< ...
- Codeforces Round #355 (Div. 2) B. Vanya and Food Processor
菜菜菜!!!这么撒比的模拟题,听厂长在一边比比比了半天,自己想一想,然后纯模拟一下,中间过程检测一下,妥妥的就可以过. 题意:有N个东西要去搞碎,每个东西有一个高度,然后有一台机器支持里面可以达到的最 ...
- Codeforces Round #355 (Div. 2)-B
B. Vanya and Food Processor 题目链接:http://codeforces.com/contest/677/problem/B Vanya smashes potato in ...
- CF Round #355 Div.2
http://codeforces.com/contest/677 B. Vanya and Food Processor 题意:有一个食物加工器,每次能加工不超过h高度的土豆,且每秒加工至多k高度的 ...
随机推荐
- sass的使用(载)
一.什么是SASSSASS是一种CSS的开发工具,提供了许多便利的写法,大大节省了设计者的时间,使得CSS的开发,变得简单和可维护.本文总结了SASS的主要用法.我的目标是,有了这篇文章,日常的一般使 ...
- 服务管理-Apache
WEB服务器介绍 web server 有两个意思: 一台负责提供网页的服务器,通过HTTP协议传给客户端(一般是指网页浏览器). 一个提供网页的服务器程序. 常见的WEB服务器 Apache是世界使 ...
- 谈谈 T 型人才
谈谈 T 型人才 昨天的图片发模糊了,正好我把这个话题展开聊一聊吧.这个话题是关于复合型人才的,我把它称作 T 型人才. 「全栈」工程师 前一段时间,「全栈」工程师的概念很火,不过大多数时候,「全 ...
- Canvas学习笔记——缓动
当你驾车在高速公路上行驶时,速度是很快的,而快到收费站时,则开始减速直到停下.将这个例子转换成物理模型就是当物体向终点运动时,开始速度会很快,而在快要到达终点时,速度会逐渐放缓直至0,整个运动过程就是 ...
- (转)Java web 项目中文件路径
文件路径分为绝对路径和相对路径,在项目中页面跳转.配置文件读写.文件上传下载等等许多地方都涉及到文件路径问题. 一篇好文转载于此:http://blog.csdn.net/shendl/archive ...
- EasyDarwin开发出类似于美拍、秒拍的短视频拍摄SDK:EasyVideoRecorder
EasyVideoRecorder Github:https://github.com/EasyDarwin/EasyVideoRecorder EasyVideoRecorder作为一款短视频拍摄的 ...
- 九度OJ 1132:与7无关的数 (数字特性)
时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:1619 解决:1037 题目描述: 一个正整数,如果它能被7整除,或者它的十进制表示法中某个位数上的数字为7, 则称其为与7相关的数.现求所 ...
- Use Apache HBase™ when you need random, realtime read/write access to your Big Data.
Apache HBase™ is the Hadoop database, a distributed, scalable, big data store. Use Apache HBase™ whe ...
- 5 Ways to Send Email From Linux Command Line
https://tecadmin.net/ways-to-send-email-from-linux-command-line/ We all know the importance of email ...
- 什么是 CAS 机制?
1.Synchronized关键字 Synchronized关键字会让没有得到锁资源的线程进入BLOCKED状态,而后在争夺到锁资源后恢复为RUNNABLE状态,这个过程中涉及到操作系统用户模式和内核 ...