A. Laptops
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

One day Dima and Alex had an argument about the price and quality of laptops. Dima thinks that the more expensive a laptop is, the better it is. Alex disagrees. Alex thinks that there are two laptops, such that the price of the first laptop is less (strictly smaller) than the price of the second laptop but the quality of the first laptop is higher (strictly greater) than the quality of the second laptop.

Please, check the guess of Alex. You are given descriptions of n laptops. Determine whether two described above laptops exist.

Input

The first line contains an integer n (1 ≤ n ≤ 105) — the number of laptops.

Next n lines contain two integers each, ai and bi (1 ≤ ai, bi ≤ n), where ai is the price of the i-th laptop, and bi is the number that represents the quality of the i-th laptop (the larger the number is, the higher is the quality).

All ai are distinct. All bi are distinct.

Output

If Alex is correct, print "Happy Alex", otherwise print "Poor Alex" (without the quotes).

Examples
input
2
1 2
2 1
output
Happy Alex

题意:n个物品 a为价格 b为物品的质量 若满足价格越高质量越好 输出Poor Alex 反之输出Happy Alex

题解:水  注意结构体排序的一个细节....orz

 /******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^ ^ ^
O O
******************************/
#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<algorithm>
#include<queue>
#define ll __int64
using namespace std;
int n;
struct node
{
int a,b;
}N[];
bool cmp(struct node aa,struct node bb)
{
return aa.a<bb.a;
}
int main()
{
scanf("%d",&n);
int flag=;
for(int i=;i<n;i++){
scanf("%d %d",&N[i].a,&N[i].b);
if(N[i].a!=N[i].b)
flag=;
}
if(flag)
cout<<"Happy Alex"<<endl;
else
cout<<"Poor Alex"<<endl;
return ;
}
B. Fedya and Maths
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Fedya studies in a gymnasium. Fedya's maths hometask is to calculate the following expression:

(1n + 2n + 3n + 4nmod 5

for given value of n. Fedya managed to complete the task. Can you? Note that given number n can be extremely large (e.g. it can exceed any integer type of your programming language).

Input

The single line contains a single integer n (0 ≤ n ≤ 10105). The number doesn't contain any leading zeroes.

Output

Print the value of the expression without leading zeros.

Examples
input
4
output
4
input
124356983594583453458888889
output
0
Note

Operation x mod y means taking remainder after division x by y.

Note to the first sample:

题意:计算(1n + 2n + 3n + 4nmod 5  n为次幂 n为大数

枚举n的值可以发现规律 只要n%4==0则输出4 否则输出0 剩下的问题就是大数对小数取模了

题解:大数对小数取模 从高位到低位 具体看代码

 /******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^ ^ ^
O O
******************************/
#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<algorithm>
#include<queue>
#define ll __int64
using namespace std;
int n;
char a[];
int main()
{
cin>>a;
n=strlen(a);
int exm=;
for(int i=;i<n;i++)
{
exm=exm*+a[i]-'';
exm%=;
}
if(exm==)
cout<<""<<endl;
else
cout<<""<<endl;
return ;
}
C. Boredom
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Alex doesn't like boredom. That's why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.

Given a sequence a consisting of n integers. The player can make several steps. In a single step he can choose an element of the sequence (let's denote it ak) and delete it, at that all elements equal to ak + 1 and ak - 1 also must be deleted from the sequence. That step brings ak points to the player.

Alex is a perfectionist, so he decided to get as many points as possible. Help him.

Input

The first line contains integer n (1 ≤ n ≤ 105) that shows how many numbers are in Alex's sequence.

The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105).

Output

Print a single integer — the maximum number of points that Alex can earn.

Examples
input
2
1 2
output
2
input
3
1 2 3
output
4
input
9
1 2 1 3 2 2 2 2 3
output
10
Note

Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this[2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.

题意:给你n个值 每次取出一个值ak  则删除所有的ak +1 ,ak -1 一直到最后一个数

输出 取出的值的和的最大值

题解:这n个值的范围为1 ≤ ai ≤ 10^5 先标记 记录每个数i出现的次数dis[i]

转移方程 dp[i]=max(dp[i-2]+i*dis[i],dp[i-1])   也就是相当于判断当前这个数是被删除?还是被取出?

i*dis[i]表示数i对结果的贡献 dp[i]代表以i为结尾的所要求的最大值

 /******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^ ^ ^
O O
******************************/
#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#include<algorithm>
#include<queue>
#define ll __int64
using namespace std;
int n;
ll a[];
ll dis[];
ll ans[];
int main()
{
scanf("%d",&n);
memset(dis,,sizeof(dis));
memset(ans,,sizeof(ans));
for(int i=;i<=n;i++)
{
scanf("%I64d",&a[i]);
dis[a[i]]++;
}
ans[]=dis[];
for(int i=;i<=1e5;i++)
ans[i]=max(ans[i-]+i*dis[i],ans[i-]);
printf("%I64d\n",ans[]);
return ;
}

Codeforces Round #260 (Div. 2) A B C 水 找规律(大数对小数取模) dp的更多相关文章

  1. Codeforces Round #327 (Div. 2) C Median Smoothing(找规律)

    分析: 三个01组合只有八种情况: 000 s001 s010 0011 s100 s101 1110 s111 s 可以看出只有010,101是不稳定的.其他都是稳定的,且连续地出现了1或0,标记为 ...

  2. Codeforces Round #272 (Div. 2) D.Dreamoon and Sets 找规律

    D. Dreamoon and Sets   Dreamoon likes to play with sets, integers and .  is defined as the largest p ...

  3. DP Codeforces Round #260 (Div. 1) A. Boredom

    题目传送门 /* 题意:选择a[k]然后a[k]-1和a[k]+1的全部删除,得到点数a[k],问最大点数 DP:状态转移方程:dp[i] = max (dp[i-1], dp[i-2] + (ll) ...

  4. 递推DP Codeforces Round #260 (Div. 1) A. Boredom

    题目传送门 /* DP:从1到最大值,dp[i][1/0] 选或不选,递推更新最大值 */ #include <cstdio> #include <algorithm> #in ...

  5. Codeforces Round #367 (Div. 2) A. Beru-taxi (水题)

    Beru-taxi 题目链接: http://codeforces.com/contest/706/problem/A Description Vasiliy lives at point (a, b ...

  6. Codeforces Round #603 (Div. 2) A. Sweet Problem(水.......没做出来)+C题

    Codeforces Round #603 (Div. 2) A. Sweet Problem A. Sweet Problem time limit per test 1 second memory ...

  7. Codeforces Round #334 (Div. 2) A. Uncowed Forces 水题

    A. Uncowed Forces Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/604/pro ...

  8. Codeforces Round #285 (Div. 2) A, B , C 水, map ,拓扑

    A. Contest time limit per test 1 second memory limit per test 256 megabytes input standard input out ...

  9. Codeforces Round #260 (Div. 2)AB

    http://codeforces.com/contest/456/problem/A A. Laptops time limit per test 1 second memory limit per ...

随机推荐

  1. 一模 (1) day2

    第一题:(水题) 题目大意:就是给出扫雷的图,然后统计每个九宫格的雷的个数. 解题过程: 1.好久没做这样的水题了.直接模拟水过.. 第二题: 题目大意:给出一个长度小于1000的数k,要求一个尽可能 ...

  2. json数组,随便测试

    Pid := '1001411225514227,926792194654225'; json := SA([]); json.AsArray.Add(SO(pid)); ShowMessage( j ...

  3. HighAvailability和LoadBalancer

    HighAvailability                         LoadBalancer 红帽RHCS                                lvs(三种工作 ...

  4. <转>2015-7-14面试题

    由于一些原因,最近打算换一份工作,主要目标是大型的互联网公司.在经历了上周三天小公司试水后,昨天终于开始正式的面试之旅了(其实接到面试通知的就几家公司

  5. 番茄工作法和Bullet Journal笔记法

    番茄工作法是一套时间管理方法,Bullet journal是一套笔记记录的方法,最近通过学习这2个方法来管理自己的时间,让自己战胜拖延症. 番茄工作法 番茄工作法的整体思路就是将时间分割成一个个小时间 ...

  6. [Unity3D]调用Android接口

    简介 有一些手机功能,Unity没有提供相应的接口,例如震动,例如不锁屏,例如GPS,例如... 有太多的特殊功能Unity都没有提供接口,这时候,我们就需要通过使用Android原生的ADT编辑器去 ...

  7. C++ primer的第三章的主要内容

    第三章主要介绍了C++中标准库类型.主要讲到string和vector类型.在string类型中,能够很方便的操作字符串,应该要注意的地方就是它的字符串中元素的位置的类型是:size_type类型的数 ...

  8. 《day16_多线程细节_Eclipse使用》

    多线程的一些细节: 1,面试题:sleep方法和wait方法异同点是什么? 相同点:可以让线程处于冻结状态. 不同点: 1, sleep必须指定时间. wait可以指定时间,也可以不指定时间. 2, ...

  9. (spring-第16回【AOP基础篇】)基本概念

    AOP(Aspect Oriented Programing),面向切面方程.介绍具体定义前,先看一个例子: package com.baobaotao.concept; public class F ...

  10. iOS interface guidelines (界面设计指南)<一>

    一.      为iOS而设计 1.iOS体现的主题: (1)Deference(顺从):UI的存在就是为了让顾客更加容易理解和进行交互,而不是要和顾客玩智力游戏 (2)Clarity(清晰):在每个 ...