问题 L: An Invisible Hand - (2018年第二阶段个人训练赛第三场)
题目描述
Takahashi will begin the travel at town 1, with no apple in his possession. The actions that can be performed during the travel are as follows:
Move: When at town i (i<N), move to town i+1.
Merchandise: Buy or sell an arbitrary number of apples at the current town. Here, it is assumed that one apple can always be bought and sold for Ai yen (the currency of Japan) at town i (1≤i≤N), where Ai are distinct integers. Also, you can assume that he has an infinite supply of money.
For some reason, there is a constraint on merchandising apple during the travel: the sum of the number of apples bought and the number of apples sold during the whole travel, must be at most T. (Note that a single apple can be counted in both.)
During the travel, Takahashi will perform actions so that the profit of the travel is maximized. Here, the profit of the travel is the amount of money that is gained by selling apples, minus the amount of money that is spent on buying apples. Note that we are not interested in apples in his possession at the end of the travel.
Aoki, a business rival of Takahashi, wants to trouble Takahashi by manipulating the market price of apples. Prior to the beginning of Takahashi's travel, Aoki can change Ai into another arbitrary non-negative integer Ai' for any town i, any number of times. The cost of performing this operation is |Ai−Ai'|. After performing this operation, different towns may have equal values of Ai.
Aoki's objective is to decrease Takahashi's expected profit by at least 1 yen. Find the minimum total cost to achieve it. You may assume that Takahashi's expected profit is initially at least 1 yen.
Constraints
1≤N≤105
1≤Ai≤109 (1≤i≤N)
Ai are distinct.
2≤T≤109
In the initial state, Takahashi's expected profit is at least 1 yen.
输入
N T
A1 A2 … AN
输出
样例输入
3 2
100 50 200
样例输出
1
提示
In the initial state, Takahashi can achieve the maximum profit of 150 yen as follows:
1.Move from town 1 to town 2.
2.Buy one apple for 50 yen at town 2.
3.Move from town 2 to town 3.
4.Sell one apple for 200 yen at town 3.
If, for example, Aoki changes the price of an apple at town 2 from 50 yen to 51 yen, Takahashi will not be able to achieve the profit of 150 yen. The cost of performing this operation is 1, thus the answer is 1.
There are other ways to decrease Takahashi's expected profit, such as changing the price of an apple at town 3 from 200 yen to 199 yen.
em题目的意思就是说一个商人可以从一个地方买苹果,然后再下不知道几个地方卖出去,每个地方都有个苹果的价值(且不相等),他想取得最大的利润,(毕竟商人)。而另一个竞争对手想要阻止他,哪怕只令他少赚一块钱,他可以任意修改地方苹果售价,但是要付出相等的代价。求最小的代价。
那我们只需要求出第一个商人最大价值出现了几次(因为地方售价不相等,所以可以不会出现改一个地方售价影响两个最大价值的情况),然后改动他卖出或者出售地方售价就ok,毕竟求最小那么我们就只改动1就好 ,那么最小代价就变成了,最大利润出现的次数。
暴力跑肯定超时的,那么就在输入的时候算出来每个地方的利润,顺便记录最大值即可。
#include<iostream>
#include<math.h>
#include<cstdio> using namespace std; int dp[];
int main()
{
int n,t;
scanf("%d%d",&n,&t);
int minn = 0x3f3f3f3f;
int maxn = ;
for(int i=;i<n;i++)
{
int a;
scanf("%d",&a);
dp[i] = a - minn>=?a - minn:;
minn = min(a,minn);
maxn = max(dp[i],maxn);
}
int ans = ;
for(int i = ;i<n;i++)
{
if(maxn == dp[i])ans++;
}
printf("%d\n",ans);
}
问题 L: An Invisible Hand - (2018年第二阶段个人训练赛第三场)的更多相关文章
- 2018牛客网暑假ACM多校训练赛(第二场)E tree 动态规划
原文链接https://www.cnblogs.com/zhouzhendong/p/NowCoder-2018-Summer-Round2-E.html 题目传送门 - 2018牛客多校赛第二场 E ...
- 2019年第二阶段我要变强个人训练赛第八场 B.序列(seq)
传送门 B.序列(seq) •题目描述 给出一个长度为n的序列a,每次对序列进行一下的某一个操作. •输入 第一行两个整数n,q表示序列长度和操作个数. 接下来一行n个数,表示序列a. 接下来q行表示 ...
- EZ 2018 03 09 NOIP2018 模拟赛(三)
最近挺久没写比赛类的blog了 链接:http://211.140.156.254:2333/contest/59 这次的题目主要考验的是爆搜+打表的能力 其实如果你上来就把所有题目都看过一次就可以知 ...
- UPC 2019年第二阶段我要变强个人训练赛第六场
传送门 A.上学路线 题目描述 小D从家到学校的道路结构是这样的:由n条东西走向和m条南北走向的道路构成了一个n*m的网格,每条道路都是单向通行的(只能从北向南,从西向东走). 已知小D的家在网格的左 ...
- 2018牛客网暑假ACM多校训练赛(第三场)I Expected Size of Random Convex Hull 计算几何,凸包,其他
原文链接https://www.cnblogs.com/zhouzhendong/p/NowCoder-2018-Summer-Round3-I.html 题目传送门 - 2018牛客多校赛第三场 I ...
- 2018牛客网暑假ACM多校训练赛(第三场)G Coloring Tree 计数,bfs
原文链接https://www.cnblogs.com/zhouzhendong/p/NowCoder-2018-Summer-Round3-G.html 题目传送门 - 2018牛客多校赛第三场 G ...
- 2018牛客网暑假ACM多校训练赛(第三场)D Encrypted String Matching 多项式 FFT
原文链接https://www.cnblogs.com/zhouzhendong/p/NowCoder-2018-Summer-Round3-D.html 题目传送门 - 2018牛客多校赛第三场 D ...
- 2018 HDU多校第三场赛后补题
2018 HDU多校第三场赛后补题 从易到难来写吧,其中题意有些直接摘了Claris的,数据范围是就不标了. 如果需要可以去hdu题库里找.题号是6319 - 6331. L. Visual Cube ...
- Lyft Level 5 Challenge 2018 - Final Round (Open Div. 2) (前三题题解)
这场比赛好毒瘤哇,看第四题好像是中国人出的,怕不是dllxl出的. 第四道什么鬼,互动题不说,花了四十五分钟看懂题目,都想砸电脑了.然后发现不会,互动题从来没做过. 不过这次新号上蓝名了(我才不告诉你 ...
随机推荐
- Confluence 6 链接到其他应用
应用链接(Application Links)有时候也被称为 "AppLinks" 是一系列测插件能够允许你在 Atlassian 的应用中互相链接.链接 2 个应用能够允许你在 ...
- 【JDK】JDK模块化(1)-为什么要模块化
Java9发布已经有一年了,跟Java8相比,从目录对比就看得出来差别相当大. 实际上Java9最大的变化就是JDK模块化(Modular). 那么,模块化的目的是什么呢? 官方的说法是: 之前版本的 ...
- Java编程之前的复习和练习
日期:2018.7.14 星期六 博客期:001 今天先是试着写一下博客,最近去青海旅游了,学习时间有点少,但空余时间还是有学习的,不管怎么样吧!先说一下我的这几天的成果——“Bignum”类,虽然很 ...
- day12 函数的嵌套调用 闭包函数,函数对象
函数嵌套: 函数嵌套: 嵌套指的是,一个物体包含另一个物体,函数嵌套就是一个函数包含另一个函数 按照函数的两个阶段 嵌套调用 指的是在函数的执行过程中调用了另一个函数,其好处可以简化外层大函数的代码, ...
- Best Cow Line(POJ3617)
Description FJ is about to take his N (1 ≤ N ≤ 2,000) cows to the annual"Farmer of the Year&quo ...
- laravel 实现一个简单的 RESTful API
创建一个 Article 资源 php artisan make:resource Article 你可以在 app/Http/Resources 目录下看到你刚刚生成的 Article 资源 当然我 ...
- Java 输入一组数字,用穷举的方法列出
import java.util.Scanner; public class TestScanner { public static void main(String[] args) { Scanne ...
- 在centos6.8上源码安装MySQL
1.安装环境:软件包:mysql-5.6.31.tar.gz 需求相关选项: 安装基目录basedir:/mydb/mysql31数据存放目录datadir:/mydb/mysql31/data端口号 ...
- python 内置数据类型之数字
目录: 1.2. 数字 1.2.1. 数字类型 1.2.2. 浮点数 1.2.3. 进制记数 1.2.4. 设置小数精度 1.2.5. 分数 1.2.6. 除法 1.2 数字 1.2.1 数字类型 ...
- python基础知识之zip
names =['zhangning','lsl','lyq','xww']age = [1,2,3,4]for a,b in zip(names,age): print(a,b)S = 'abcde ...