B. Vanya and Food Processor
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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:

  1. 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.
  2. 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.

Input

The first line of the input contains integers nh 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.

Output

Print a single integer — the number of seconds required to smash all the potatoes following the process described in the problem statement.

Examples
input
5 6 3
5 4 3 2 1
output
5
input
5 6 3
5 5 5 5 5
output
10
input
5 6 3
1 2 1 1 1
output
2
Note

Consider the first sample.

  1. 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.
  2. Now Vanya puts the piece of potato of height 4. At the end of the second there is amount of height 3 remaining.
  3. Vanya puts the piece of height 3 inside and again there are only 3 centimeters remaining at the end of this second.
  4. 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.
  5. 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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. [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 ...

  6. [ 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< ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. django 运行python manage.py sqlall books 时报错 app has migration

    出现这个问题的原因是版本之前的不兼容,我用的django版本是1.8.6 而 这条python manage.py sqlall books 是基于django1.0版本的. 在django1.8.6 ...

  2. Can&#39;t open named pipe to host: . pipe: MySQL

    今天遇到mysql连接odbc时报例如以下错误:Can't open named pipe to host: . pipe: MySQL 错误截图例如以下: 依照网上说的方法包含mysql的官方说法都 ...

  3. [单元測试]_[VC2010使用gtest单元測试入门]

    场景: 1. gtest作为C++的单元測试工具非常优秀了,它集成了非常多标准assert所没有的功能,比方让流程继续运行的EXPECT,仅仅測试特定測试用例的--gtest_filter, 输出xm ...

  4. 03 xml封装通信接口

    <?php class Response_xml{ /** *按xml方式输出通信 *@param integet $code 状态码 *@param string $message 提示信息 ...

  5. new和delete的基本用法

    前言 new和delete是C++中用来动态管理内存分配的运算符,其用法较为灵活.如果你对它们的几种不同用法感到困惑,混淆,那么接着看下去吧. 功能一:动态管理单变量/对象空间 下面例子使用new为单 ...

  6. android启动另一应用

    http://www.2cto.com/kf/201203/122910.html Android SDK中有这样一个API: public abstract Intent getLaunchInte ...

  7. Android 调用QQ登录

    调用QQ登录        在如今的项目开发.调用第三方登录.差点儿是必须的,而调用QQ登录也是不可缺少的,这里把相关代码分享出来,希望能拿去就能够用,降低项目开发的时间.希望对大家实用. 1,去QQ ...

  8. 怎样解决 no jzmq in java.library.path

    1. 确保zmq的各种library有安装正确.检查方法:查看/usr/local/lib, 看里面有没有libjzmq.a, libjzmq.dylib, libzmq.a, libjzmq.dyl ...

  9. c++中class与struct的区别

    0. 结构是一种用关键字struct声明的自定义数据类型.与类相似,也可以包含构造函数,常数,字段,方法,属性,索引器,运算符和嵌套类型. 1.结构在堆栈中创建,是值类型,而类是引用类型.每当需要一种 ...

  10. UVA10518 How Many Calls? —— 矩阵快速幂

    题目链接:https://vjudge.net/problem/UVA-10518 题解: 问:求斐波那契数f[n]的时候调用了多少次f[n] = f[n-1] + f[n-2],没有记忆化,一直递归 ...