Codeforces Round #345 (Div. 2)【A.模拟,B,暴力,C,STL,容斥原理】
A. Joysticks
Friends are going to play console. They have two joysticks and only one charger for them. Initially first joystick is charged at a1 percent and second one is charged at a2 percent. You can connect charger to a joystick only at the beginning of each minute. In one minute joystick either discharges by 2 percent (if not connected to a charger) or charges by 1 percent (if connected to a charger).
Game continues while both joysticks have a positive charge. Hence, if at the beginning of minute some joystick is charged by 1 percent, it has to be connected to a charger, otherwise the game stops. If some joystick completely discharges (its charge turns to 0), the game also stops.
Determine the maximum number of minutes that game can last. It is prohibited to pause the game, i. e. at each moment both joysticks should be enabled. It is allowed for joystick to be charged by more than 100 percent.
The first line of the input contains two positive integers a1 and a2 (1 ≤ a1, a2 ≤ 100), the initial charge level of first and second joystick respectively.
Output the only integer, the maximum number of minutes that the game can last. Game continues until some joystick is discharged.
3 5
6
4 4
5
In the first sample game lasts for 6 minute by using the following algorithm:
- at the beginning of the first minute connect first joystick to the charger, by the end of this minute first joystick is at 4%, second is at 3%;
- continue the game without changing charger, by the end of the second minute the first joystick is at 5%, second is at 1%;
- at the beginning of the third minute connect second joystick to the charger, after this minute the first joystick is at 3%, the second one is at 2%;
- continue the game without changing charger, by the end of the fourth minute first joystick is at 1%, second one is at 3%;
- at the beginning of the fifth minute connect first joystick to the charger, after this minute the first joystick is at 2%, the second one is at 1%;
- at the beginning of the sixth minute connect second joystick to the charger, after this minute the first joystick is at 0%, the second one is at 2%.
After that the first joystick is completely discharged and the game is stopped.
题目链接:http://codeforces.com/contest/651/problem/A
题意:现有两部手机,可是只有一根充电线。两个手机不可能同时充电。如果手机不充电的话,每分钟会降2%的点,如果充电的话可以每分钟可以涨1%的电。一旦其中有一部手机没电了就直接game over了。给出这两部手机的初始电量(初始电量不超过100,但你给他们充电的话电量可以超过100)。问两部手机最久能撑多久?
分析:直接模拟一下过程就好了,找出两部手机电量最小的不小于2即可!
下面给出AC代码:
#include <bits/stdc++.h>
using namespace std;
int main()
{
int a1,a2;
cin>>a1>>a2;
int ans=;
while()
{
if(a1<=a2)
swap(a1,a2);
if(a1<)
break;
a1-=;
a2+=;
ans++;
if(a1<=||a2<=)
break;
}
cout<<ans<<endl;
}
B. Beautiful Paintings
There are n pictures delivered for the new exhibition. The i-th painting has beauty ai. We know that a visitor becomes happy every time he passes from a painting to a more beautiful one.
We are allowed to arranged pictures in any order. What is the maximum possible number of times the visitor may become happy while passing all pictures from first to last? In other words, we are allowed to rearrange elements of a in any order. What is the maximum possible number of indices i (1 ≤ i ≤ n - 1), such that ai + 1 > ai.
The first line of the input contains integer n (1 ≤ n ≤ 1000) — the number of painting.
The second line contains the sequence a1, a2, ..., an (1 ≤ ai ≤ 1000), where ai means the beauty of the i-th painting.
Print one integer — the maximum possible number of neighbouring pairs, such that ai + 1 > ai, after the optimal rearrangement.
5
20 30 10 50 40
4
4
200 100 100 200
2
In the first sample, the optimal order is: 10, 20, 30, 40, 50.
In the second sample, the optimal order is: 100, 200, 100, 200.
题目链接:http://codeforces.com/contest/651/problem/B
题意:把a这个数组重新排列,使得序列中 满足条件ai+1>ai 的i尽可能的多,输出符合条件的i的数目。
分析:暴力可做,找出对应美丽值相同的最大个数t,n-t即为所求值!
下面给出AC代码:
#include <bits/stdc++.h>
using namespace std;
int a[];
int main()
{
int n,x;
int ans=,t=-;
cin>>n;
for(int i=;i<=n;i++)
{
cin>>x;
a[x]++;
}
for(int i=;i<=;i++)
t=max(t,a[i]);
cout<<n-t<<endl;
return ;
}
C. Watchmen
Watchmen are in a danger and Doctor Manhattan together with his friend Daniel Dreiberg should warn them as soon as possible. There are n watchmen on a plane, the i-th watchman is located at point (xi, yi).
They need to arrange a plan, but there are some difficulties on their way. As you know, Doctor Manhattan considers the distance between watchmen i and j to be |xi - xj| + |yi - yj|. Daniel, as an ordinary person, calculates the distance using the formula .
The success of the operation relies on the number of pairs (i, j) (1 ≤ i < j ≤ n), such that the distance between watchman i and watchmen j calculated by Doctor Manhattan is equal to the distance between them calculated by Daniel. You were asked to compute the number of such pairs.
The first line of the input contains the single integer n (1 ≤ n ≤ 200 000) — the number of watchmen.
Each of the following n lines contains two integers xi and yi (|xi|, |yi| ≤ 109).
Some positions may coincide.
Print the number of pairs of watchmen such that the distance between them calculated by Doctor Manhattan is equal to the distance calculated by Daniel.
3
1 1
7 5
1 5
2
6
0 0
0 1
0 2
-1 1
0 1
1 1
11
In the first sample, the distance between watchman 1 and watchman 2 is equal to |1 - 7| + |1 - 5| = 10 for Doctor Manhattan and for Daniel. For pairs (1, 1), (1, 5) and (7, 5), (1, 5) Doctor Manhattan and Daniel will calculate the same distances.
题意:
钟表匠们的好基友马医生和蛋蛋现在要执行拯救表匠们的任务。在平面内一共有n个表匠,第i个表匠的位置为(xi, yi).
他们需要安排一个任务计划,但是确发现了一些问题很难解决。马医生从第i个表匠到第j个表匠所需要的时间为|xi - xj| + |yi - yj|。然而蛋蛋所需要的时间为
要想成功完成任务,必须保证两人从第i个表匠出发,同时到达第j个表匠。现在请你计算最多有多少组表匠的位置满足条件
分析:map去搞一搞就好了!
下面给出AC代码:【这个代码编译器运行不了,我也不知道为啥,队友写的,参考参考】
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll n,x,y,ans;
map<ll,ll> a,b;
map<pair<ll,ll>,ll> c;
int main()
{
for(cin>>n;cin>>x>>y;)
ans+=(a[x]++)+(b[y]++)-(c[make_pair(x,y)]++);
cout<<ans<<endl;
return ;
}
它解:容斥原理
#include <bits/stdc++.h>
using namespace std;
struct Node
{
int a,b;
}num[];
bool cmp1(Node a,Node b) //第一次排序
{
if(a.a==b.a)
{
return a.b<b.b;
}
return a.a<b.a;
}
bool cmp2(Node a,Node b) //第二次排序
{
if(a.b==b.b)
{
return a.a<b.a;
}
return a.b<b.b;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=;i<n;i++)
{
scanf("%d%d",&num[i].a,&num[i].b);
}
sort(num,num+n,cmp1);
__int64 temp=; //几个是相同的
__int64 tt=; //重复的个数
__int64 res=;
for(int i=;i<n;i++)
{
if(num[i].a==num[i-].a) //xi == xi-1
{
temp++;
if(num[i].b==num[i-].b)
{
tt++;
}
else
{
res-=tt*(tt-)/; //去重
tt=;
}
}
else
{
res+=temp*(temp-)/; //排列组合,从temp个两两组合的个数
res-=tt*(tt-)/;
tt=;
temp=;
}
}
if(tt!=) //判断结尾是不是有些没有去重
{
res-=tt*(tt-)/;
}
tt=;
if(temp!=) //判断结尾有些是不是没有计算
{
res+=temp*(temp-)/;
}
temp=;
sort(num,num+n,cmp2); //第二次排序
for(int i=;i<n;i++)
{
if(num[i].b==num[i-].b)
{
temp++;
}
else
{
res+=temp*(temp-)/;
temp=;
}
}
if(temp!=)
{
res+=temp*(temp-)/;
}
printf("%I64d\n",res);
}
return ;
}
Codeforces Round #345 (Div. 2)【A.模拟,B,暴力,C,STL,容斥原理】的更多相关文章
- Codeforces Round #345 (Div. 2) D. Image Preview 暴力 二分
D. Image Preview 题目连接: http://www.codeforces.com/contest/651/problem/D Description Vasya's telephone ...
- Codeforces Round #345 (Div. 2) B. Beautiful Paintings 暴力
B. Beautiful Paintings 题目连接: http://www.codeforces.com/contest/651/problem/B Description There are n ...
- cf之路,1,Codeforces Round #345 (Div. 2)
cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅..... ...
- Codeforces Round #345 (Div. 2)——A. Joysticks(模拟+特判)
A. Joysticks time limit per test 1 second memory limit per test 256 megabytes input standard input o ...
- Codeforces Round #249 (Div. 2) (模拟)
C. Cardiogram time limit per test 1 second memory limit per test 256 megabytes input standard input ...
- Codeforces Round #366 (Div. 2) C 模拟queue
C. Thor time limit per test 2 seconds memory limit per test 256 megabytes input standard input outpu ...
- codeforces Codeforces Round #345 (Div. 1) C. Table Compression 排序+并查集
C. Table Compression Little Petya is now fond of data compression algorithms. He has already studied ...
- Codeforces Round #345 (Div. 1) A. Watchmen 模拟加点
Watchmen 题意:有n (1 ≤ n ≤ 200 000) 个点,问有多少个点的开平方距离与横纵坐标的绝对值之差的和相等: 即 = |xi - xj| + |yi - yj|.(|xi|, |y ...
- 题解——Codeforces Round #508 (Div. 2) T1 (模拟)
依照题意暴力模拟即可A掉 #include <cstdio> #include <algorithm> #include <cstring> #include &l ...
随机推荐
- LINUX:alias命令详解
发现目前安装的g++并没有开启选项 -std=c++11,无法使用c++11的新标准及其中的列表初始化.搜索后得到解决方法:键入:alias g++="g++ -std=c++11&quo ...
- [置顶]
xamarin android 布局尺寸了解
为了使UI界面在不同大小的移动端显示器上能够正常显示,大家可能都知道使用sp作为字体大小的单位,dp作为其他元素长度的单位. 前几天看了一篇文章关于 App设计规范的,文章用心写的非常好,这里是链接 ...
- ES6 字符串的扩展
字符的Unicode表示法 JavaScript允许采用\uXXXX形式表示一公分字符,其中XXXX表示字符的码点. "\u0061" //"a" 但是,这种表 ...
- Android 执行 adb shell 命令
Android 执行Adb shell 命令大多需要root权限,Android自带的Runtime. getRuntime().exec()容易出错,在网上找到了一个执行adb shell命令的类 ...
- Git详解之一:Git起步
起步 本章介绍开始使用 Git 前的相关知识.我们会先了解一些版本控制工具的历史背景,然后试着让 Git 在你的系统上跑起来,直到最后配置好,可以正常开始开发工作.读完本章,你就会明白为什么 Git ...
- pinyin utils
package cn.itcast.bos.utils; import java.util.Arrays; import net.sourceforge.pinyin4j.PinyinHelp ...
- 程序包管理rpm、yum与简单编译安装程序
Linux程序包管理 Linux中软件的安装主要有两种形式:一种是直接下载源代码包自行编译后安装,另一种直接获取rpm软件包进行安装. 程序的组成部分: 二进制程序:程序的主体文件,比如我们运行一个l ...
- VS代码生成工具ReSharper使用手册:配置快捷键(转)
原文:http://blog.csdn.net/fhzh520/article/details/46364603 VS代码生成工具ReSharper提供了丰富的快捷键,可以极大地提高你的开发效率. 配 ...
- Robot Framework学习笔记(八)------ride标签使用
一.edit标签使用 1.导入库 点击 Edit 标签页右侧的"Library"按钮,来添加库.在添加库之前,首先库已经在 Python 下进行了安装.如,添加"Sele ...
- 跟我一起读postgresql源码(八)——Executor(查询执行模块之——可优化语句的执行)
2.可优化语句的执行 可优化语句的共同特点是它们被查询编译器处理后都会生成査询计划树,这一类语句由执行器(Executor)处理.该模块对外提供了三个接口: ExecutorStart.Executo ...