只做了两个就去上课去啦...
A. Company Merging
time limit per test

1 second

memory limit per test

512 megabytes

input

standard input

output

standard output

A conglomerate consists of nn companies. To make managing easier, their owners have decided to merge all companies into one. By law, it is only possible to merge two companies, so the owners plan to select two companies, merge them into one, and continue doing so until there is only one company left.

But anti-monopoly service forbids to merge companies if they suspect unfriendly absorption. The criterion they use is the difference in maximum salaries between two companies. Merging is allowed only if the maximum salaries are equal.

To fulfill the anti-monopoly requirements, the owners can change salaries in their companies before merging. But the labor union insists on two conditions: it is only allowed to increase salaries, moreover all the employees in one company must get the same increase.

Sure enough, the owners want to minimize the total increase of all salaries in all companies. Help them find the minimal possible increase that will allow them to merge companies into one.

Input

The first line contains a single integer nn — the number of companies in the conglomerate (1≤n≤2⋅1051≤n≤2⋅105). Each of the next nn lines describes a company.

A company description start with an integer mimi — the number of its employees (1≤mi≤2⋅1051≤mi≤2⋅105). Then mimi integers follow: the salaries of the employees. All salaries are positive and do not exceed 109109.

The total number of employees in all companies does not exceed 2⋅1052⋅105.

Output

Output a single integer — the minimal total increase of all employees that allows to merge all companies.

Example
input

Copy
3
2 4 3
2 2 1
3 1 1 1
output

Copy
13
Note

One of the optimal merging strategies is the following. First increase all salaries in the second company by 22, and merge the first and the second companies. Now the conglomerate consists of two companies with salaries [4,3,4,3][4,3,4,3] and [1,1,1][1,1,1]. To merge them, increase the salaries in the second of those by 33. The total increase is 2+2+3+3+3=132+2+3+3+3=13.

思路:一个大集团下面有n个公司,每个公司m个员工,每个员工的工资mi,现在要合并公司.两两合并,最后只剩一个才行,合并要求是,两个公司之间的员工最大工资相等才行.

只要记录每个公司的最大工资,然后排序,再开始从小到大挨着两两合并.

#include <iostream>
#include <algorithm>
#include <cstdlib> using namespace std; struct everycom{
int maxx;
int num;
}; struct everycom ec[]; int cmp(struct everycom a,struct everycom b){
return a.maxx<b.maxx;
} int main(){
int n,m;
int now=;
int maxx=;
//while(scanf("%d",&n)){
scanf("%d",&n);
long long sum=;
long long nownum=;
for(int i=;i<n;i++){
cin>>m;
ec[i].num=m;
maxx=;
for(int j=;j<m;j++){
cin>>now;
maxx=now>maxx?now:maxx;
}
ec[i].maxx=maxx;
}
sort(ec,ec+n,cmp);
for(int i=;i<n-;i++){
long long tmp=ec[i+].maxx-ec[i].maxx;
nownum+=ec[i].num;
sum+=tmp*nownum;
}
cout<<sum<<endl;
}

M. The Pleasant Walk

time limit per test

1 second

memory limit per test

512 megabytes

input

standard input

output

standard output

There are nn houses along the road where Anya lives, each one is painted in one of kk possible colors.

Anya likes walking along this road, but she doesn't like when two adjacent houses at the road have the same color. She wants to select a long segment of the road such that no two adjacent houses have the same color.

Help Anya find the longest segment with this property.

Input

The first line contains two integers nn and kk — the number of houses and the number of colors (1≤n≤1000001≤n≤100000, 1≤k≤1000001≤k≤100000).

The next line contains nn integers a1,a2,…,ana1,a2,…,an — the colors of the houses along the road (1≤ai≤k1≤ai≤k).

Output

Output a single integer — the maximum number of houses on the road segment having no two adjacent houses of the same color.

Example
input

Copy
8 3
1 2 3 3 2 1 2 2
output

Copy
4
Note

In the example, the longest segment without neighboring houses of the same color is from the house 4 to the house 7. The colors of the houses are [3,2,1,2][3,2,1,2] and its length is 4 houses.

思路:让找最长的序列,序列中任何两个相邻的元素不能相等.

#include <iostream>

using namespace std;

int main(){
int n,k;
//int a[100005];
int ai;
while(cin>>n>>k){
int la=;
int now=;
int maxx=;
cin>>ai;
la=ai;
for(int i=;i<n;i++){
//cin>>a[i];
cin>>ai;
if(ai!=la){
now++;
}else{
now=;
}
la=ai;
maxx= now>maxx?now:maxx;
}
cout<<maxx<<endl;
}
}

CF_2018-2019 Russia Open High School Programming Contest (Unrated, Online Mirror, ICPC Rules, Teams Preferred)的更多相关文章

  1. 2018-2019 Russia Open High School Programming Contest (Unrated, Online Mirror, ICPC Rules, Teams Preferred)

    前言 有一场下午的cf,很滋磁啊,然后又和dalao(见右面链接)组队打了,dalao直接带飞我啊. 这是一篇题解,也是一篇总结,当然,让我把所有的题目都写个题解是不可能的了. 按照开题顺序讲吧. 在 ...

  2. 2019-2020 ICPC, Asia Jakarta Regional Contest (Online Mirror, ICPC Rules, Teams Preferred)

    2019-2020 ICPC, Asia Jakarta Regional Contest (Online Mirror, ICPC Rules, Teams Preferred) easy: ACE ...

  3. 2019-2020 ICPC, NERC, Southern and Volga Russian Regional Contest (Online Mirror, ICPC Rules, Teams Preferred)【A题 类型好题】

    A. Berstagram Polycarp recently signed up to a new social network Berstagram. He immediately publish ...

  4. D. Dog Show 2017-2018 ACM-ICPC, NEERC, Southern Subregional Contest, qualification stage (Online Mirror, ACM-ICPC Rules, Teams Preferred)

    http://codeforces.com/contest/847/problem/D 巧妙的贪心 仔细琢磨... 像凸包里的处理 #include <cstdio> #include & ...

  5. 2016-2017 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred)

    A 思路: 贪心,每次要么选两个最大的,要么选三个,因为一个数(除了1)都可以拆成2和3相加,直到所有的数都相同就停止,这时就可以得到答案了; C: 二分+bfs,二分答案,然后bfs找出距离小于等于 ...

  6. 2014-2015 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred)

    I. Sale in GameStore(贪心) time limit per test 2 seconds memory limit per test 512 megabytes input sta ...

  7. 2016-2017 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred) 几道简单题的题解

    A. Toda 2 题意:给你n个人,每个人的分数是a[i],每次可以从两个人到五个人的使得分数减一,使得最终的分数相等: 思路:假设答案为m:每个人的分数与答案m的差值为d[i],sum为d[i]的 ...

  8. 2016-2017 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred) J dp 背包

    J. Bottles time limit per test 2 seconds memory limit per test 512 megabytes input standard input ou ...

  9. 2016-2017 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred) G 优先队列

    G. Car Repair Shop time limit per test 2 seconds memory limit per test 512 megabytes input standard ...

随机推荐

  1. LFYZ-OJ ID: 1009 阶乘和

    思路 循环n次,每次计算i的阶乘i!,并加入sum中. n的范围从1~100,这里一定要使用高精度运算,涉及到"高精度乘低精度","高精度加高精度". 避免每次 ...

  2. [物理学与PDEs]第4章习题1 反应力学方程组形式的化约 - 动量方程与未燃流体质量平衡方程

    试证明: 利用连续性方程, 可将动量方程 (2. 14) 及未燃流体质量平衡方程 (2. 16) 分别化为 (2. 19) 与 (2. 20) 的形式. 证明: 注意到 $$\beex \bea \c ...

  3. [译]Ocelot - Delegating Handlers

    原文 可以为HttpClient添加delegating handlers. Usage 为了添加delegating handler需要做两件事. 首先如下一样创建一个类. public class ...

  4. .net 委托多线程 实时更新界面

    Thread thread = new Thread(() => { button2.Invoke(new EventHandler(delegate { button2.Enabled = t ...

  5. 什么是UDP

  6. HTTP报错401和403详解及解决办法

    一.401: 1. HTTP 401 错误 - 未授权: (Unauthorized) 您的Web服务器认为,客户端发送的 HTTP 数据流是正确的,但进入网址 (URL) 资源 , 需要用户身份验证 ...

  7. Spring注解@Configuration和Java Config

    1.从Spring 3起,JavaConfig功能已经包含在Spring核心模块,它允许开发者将bean定义和在Spring配置XML文件到Java类中.但是,仍然允许使用经典的XML方式来定义bea ...

  8. 深度学习在graph上的使用

    原文地址:https://zhuanlan.zhihu.com/p/27216346 本文要介绍的这一篇paper是ICML2016上一篇关于 CNN 在图(graph)上的应用.ICML 是机器学习 ...

  9. jQuery选择器 :eq() 不能识别变量参数的问题解决方案

    问题: js语法中,引号内变量会直接解释为字符串,因此使用:eq()时参数将被识别为字符串而不是变量指代的内容 如下错误写法: $(".circle span:eq(count-1)&quo ...

  10. Java集合图谱

    比较 是否有序 是否允许元素重复 Collection 否 是 List 是 是 Set AbstractSet 否 否 HashSet TreeSet 是(用二叉排序树) Map AbstractM ...