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. 笔记09 WS,WCF

    http://blog.csdn.net/avi9111/article/details/5655563 http://www.cnblogs.com/tearer/archive/2013/04/2 ...

  2. canvas 五角星之回顾【初中三角函数】

    当程序中遇到三角函数的时候我是懵逼的,于是百度了“初中三角函数”, 忘了这几个公式的,自己打脸. 目的是通过Canvas画一个五角星, 突破口:只要能通过给定的两个外圈点的半径,和内圈点的半径,借助上 ...

  3. RESTful API 设计原则

    http://www.ruanyifeng.com/blog/2014/05/restful_api.html http://www.ruanyifeng.com/blog/2011/09/restf ...

  4. php跳转

    header("Location: http://bbs. lampbrother.net"); header("refresh:0;url=./login.php&qu ...

  5. 富文本编辑器 - RichEditor

    基本功能 RichEditor 是一个继承自 WebView 的自己定义 view,枚举类型 Type 定了它所支持的排版格式: public enum Type { BOLD, ITALIC, SU ...

  6. Vue使用axios

    main.js-------------------   import axios from "axios"; import qs from "qs"; imp ...

  7. java中两字符串比较--compareTo方法

    java.lang.String.compareTo() 方法比较两个字符串的字典,比较是基于字符串中的每个字符的Unicode值 String n1 = "1"; String ...

  8. System.IO.File类和System.IO.FileInfo类

    1.System.IO.File类 ※文件create, copy,move,SetAttributes,open,exists ※由于File.Create方法默认向所有用户授予对新文件的完全读写. ...

  9. govendor

    cd  到工程目录. govendor init : 初始化 govendor fetch : 拉取包 go 1.6以后编译go代码会优先从vendor目录先寻找依赖包: controllers\ar ...

  10. leetcode 792. Number of Matching Subsequences

    Given string S and a dictionary of words words, find the number of words[i] that is a subsequence of ...