Codeforces Round #200 (Div. 1) C. Read Time 二分
C. Read Time
Time Limit: 1 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/343/problem/C
Description
Mad scientist Mike does not use slow hard disks. His modification of a hard drive has not one, but n different heads that can read data in parallel.
When viewed from the side, Mike's hard drive is an endless array of tracks. The tracks of the array are numbered from left to right with integers, starting with 1. In the initial state the i-th reading head is above the track number hi. For each of the reading heads, the hard drive's firmware can move the head exactly one track to the right or to the left, or leave it on the current track. During the operation each head's movement does not affect the movement of the other heads: the heads can change their relative order; there can be multiple reading heads above any of the tracks. A track is considered read if at least one head has visited this track. In particular, all of the tracks numbered h1, h2, ..., hn have been read at the beginning of the operation.
Mike needs to read the data on m distinct tracks with numbers p1, p2, ..., pm. Determine the minimum time the hard drive firmware needs to move the heads and read all the given tracks. Note that an arbitrary number of other tracks can also be read.
i(Ai≤106) 分,之后每过一分钟这道题目的分值会减少 BiB_iBi 分,并且保证到比赛结束时分值不会减少为负值。比如,一个人在第 xxx 分钟结束时做出了第 iii 道题目,那么他/她可以得到 Ai−Bi∗xA_i - B_i * xAi−Bi∗x 分。
若一名选手在第 xxx 分钟结束时做完了一道题目,则他/她可以在第 x+1x+1x+1 分钟开始时立即开始做另一道题目。
参加省队选拔的选手 dxy 具有绝佳的实力,他可以准确预测自己做每道题目所要花费的时间,做第 iii 道需要花费 Ci(Ci≤t)C_i(C_i \leq t)Ci(Ci≤t) 分钟。由于 dxy 非常神,他会做所有的题目。但是由于比赛时间有限,他可能无法做完所有的题目。他希望安排一个做题的顺序,在比赛结束之前得到尽量多的分数
Input
The first line of the input contains two space-separated integers n, m (1 ≤ n, m ≤ 105) — the number of disk heads and the number of tracks to read, accordingly. The second line contains n distinct integers hi in ascending order (1 ≤ hi ≤ 1010, hi < hi + 1) — the initial positions of the heads. The third line contains m distinct integers pi in ascending order (1 ≤ pi ≤ 1010, pi < pi + 1) - the numbers of tracks to read.
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is recommended to use the cin, cout streams or the %I64d specifier.,Bi,Ci,即此题的初始分值、每分钟减少的分值、dxy做这道题需要花费的时间。
Output
Print a single number — the minimum time required, in seconds, to read all the needed tracks.
Sample Input
3 4
2 5 6
1 3 6 8
Sample Output
2
HINT
题意
有n个刷子,每个刷子都可以左移动和右移动,每移动一格是1秒
然后问你最少多少秒,可以让这n个刷子把这m个接口都刷完
题解:
二分时间之后check就好
check注意可以先往左再往右,也可以先往右再往左的
注意一下就好啦
代码:
#include<stdio.h>
#include<iostream>
#include<math.h>
using namespace std; long long a[];
long long b[];
int n,m;
long long abs(long long a)
{
if(a<)return -a;
return a;
}
int judge(long long t)
{
int tot = ;
int l=,r=;
for(int i=;i<=n;i++)
{
long long ans = abs(b[r]-b[l]) + min(abs(b[r]-a[i]),abs(b[l]-a[i]));
while(r<=m&&ans<=t)
{
r++;
ans = abs(b[r]-b[l]) + min(abs(b[r]-a[i]),abs(b[l]-a[i]));
}
l=r;
if(r==m+)return ;
}
return ;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=;i<=n;i++)
scanf("%lld",&a[i]);
for(int i=;i<=m;i++)
scanf("%lld",&b[i]);
long long l = ,r = 2e10;
while(l<=r)
{
long long mid = (l+r)/2LL;
if(judge(mid))r = mid - ;
else l = mid + ;
}
cout<<l<<endl;
}
Codeforces Round #200 (Div. 1) C. Read Time 二分的更多相关文章
- Codeforces Round #200 (Div. 2) E. Read Time(二分)
题目链接 这题,关键不是二分,而是如果在t的时间内,将n个头,刷完这m个磁盘. 看了一下题解,完全不知怎么弄.用一个指针从pre,枚举m,讨论一下.只需考虑,每一个磁盘是从右边的头,刷过来的(左边来的 ...
- Codeforces Round #200 (Div. 1)A. Rational Resistance 数学
A. Rational Resistance Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/343 ...
- Codeforces Round #200 (Div. 1)D. Water Tree dfs序
D. Water Tree Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/343/problem/ ...
- Codeforces Round #200 (Div. 1) B. Alternating Current 栈
B. Alternating Current Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/343 ...
- Codeforces Round #200 (Div. 2) C. Rational Resistance
C. Rational Resistance time limit per test 1 second memory limit per test 256 megabytes input standa ...
- Codeforces Round #200 (Div. 1) BCD
为了锻炼个人能力奋力div1 为了不做原题从200开始 B 两个电线缠在一起了 能不能抓住两头一扯就给扯分开 很明显当len为odd的时候无解 当len为偶数的时候 可以任选一段长度为even的相同字 ...
- Codeforces Round #200 (Div. 2)D. Alternating Current (堆栈)
D. Alternating Current time limit per test 1 second memory limit per test 256 megabytes input standa ...
- Codeforces Round #200 (Div. 1) D. Water Tree(dfs序加线段树)
思路: dfs序其实是很水的东西. 和树链剖分一样, 都是对树链的hash. 该题做法是:每次对子树全部赋值为1,对一个点赋值为0,查询子树最小值. 该题需要注意的是:当我们对一棵子树全都赋值为1的 ...
- Codeforces Round #200 (Div. 1) D Water Tree 树链剖分 or dfs序
Water Tree 给出一棵树,有三种操作: 1 x:把以x为子树的节点全部置为1 2 x:把x以及他的所有祖先全部置为0 3 x:询问节点x的值 分析: 昨晚看完题,马上想到直接树链剖分,在记录时 ...
随机推荐
- spring+springMVC+JPA配置详解(使用缓存框架ehcache)
SpringMVC是越来越火,自己也弄一个Spring+SpringMVC+JPA的简单框架. 1.搭建环境. 1)下载Spring3.1.2的发布包:Hibernate4.1.7的发布包(没有使用h ...
- 利用IE/FF的不同识别CSS来使用浏览器兼容问题
区别IE6与FF: background:orange;*background:blue; 区别IE6与IE7: background:green !important;background:blue ...
- MySQL SQL优化之字符串索引隐式转换
之前有用户很不解:SQL语句非常简单,就是select * from test_1 where user_id=1 这种类型,而且user_id上已经建立索引了,怎么还是查询很慢? test_1的表结 ...
- .Net课程体系
.Net课程体系
- C# 中 string.Empty、""、null的区别
原文C# 中 string.Empty."".null的区别 一.string.Empty 和 "" 1.Empty是string类中的一个静态的只读字段,它是 ...
- put a ContextMenu into the header of a TabPage z
publicclassMyTabControl:TabControl { protected override void OnMouseUp(MouseEventArgs e){ if(e.Butto ...
- A Pretty Good Splash Screen in C#
http://www.codeproject.com/Articles/5454/A-Pretty-Good-Splash-Screen-in-C
- [Tommas] 测试用例覆盖率(一)
一.测试用例的切面设计 所谓测试切面设计,其实就是测试用例大项的划分.测试用例划分的经典方法是瀑布模型,也就是从上到下,逐渐细分,大模块包括小模块,小模块包括更小的模块.但仅仅如此是不够的,我们还要从 ...
- FOR 循环 索引从n 开始
RF 中FOR 循环默认是从0开始,如果想从任意n开始如下所示: 方法一: 结果,如你所愿输出1-6: 方法二,利用FOR遍历list来实现: 结果: 这里注意是输出1-9而不是1-10
- MVC框架模式技术实例(用到隐藏帧、json、仿Ajax、Dom4j、jstl、el等)
前言: 刚刚学完了MVC,根据自己的感悟和理解写了一个小项目. 完全按照MVC模式,后面有一个MVC的理解示意图. 用MVC模式重新完成了联系人的管理系统: 用户需求: 多用户系统,提供用户注册.登录 ...