Educational Codeforces Round 33 (Rated for Div. 2)
1 second
256 megabytes
standard input
standard output
Alex, Bob and Carl will soon participate in a team chess tournament. Since they are all in the same team, they have decided to practise really hard before the tournament. But it's a bit difficult for them because chess is a game for two players, not three.
So they play with each other according to following rules:
- Alex and Bob play the first game, and Carl is spectating;
- When the game ends, the one who lost the game becomes the spectator in the next game, and the one who was spectating plays against the winner.
Alex, Bob and Carl play in such a way that there are no draws.
Today they have played n games, and for each of these games they remember who was the winner. They decided to make up a log of games describing who won each game. But now they doubt if the information in the log is correct, and they want to know if the situation described in the log they made up was possible (that is, no game is won by someone who is spectating if Alex, Bob and Carl play according to the rules). Help them to check it!
The first line contains one integer n (1 ≤ n ≤ 100) — the number of games Alex, Bob and Carl played.
Then n lines follow, describing the game log. i-th line contains one integer ai (1 ≤ ai ≤ 3) which is equal to 1 if Alex won i-th game, to 2 if Bob won i-th game and 3 if Carl won i-th game.
Print YES if the situation described in the log was possible. Otherwise print NO.
3
1
1
2
YES
2
1
2
NO
In the first example the possible situation is:
- Alex wins, Carl starts playing instead of Bob;
- Alex wins, Bob replaces Carl;
- Bob wins.
The situation in the second example is impossible because Bob loses the first game, so he cannot win the second one.
三个人 比赛,两个人下棋 一个人旁观,下棋之后输的人去旁观,胜的人和之前旁观的人比赛下棋 。模拟下就可以了
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin>>n;
int ans=;
for(int i=; i<n; ++i)
{
int x;
cin>>x;
if(x==ans)
{
cout<<"NO\n";
return ;
}
ans=-x-ans;
}
cout<<"YES\n";
return ;
}
2 seconds
256 megabytes
standard input
standard output
Recently Luba learned about a special kind of numbers that she calls beautiful numbers. The number is called beautiful iff its binary representation consists of k + 1 consecutive ones, and then k consecutive zeroes.
Some examples of beautiful numbers:
- 12 (110);
- 1102 (610);
- 11110002 (12010);
- 1111100002 (49610).
More formally, the number is beautiful iff there exists some positive integer k such that the number is equal to (2k - 1) * (2k - 1).
Luba has got an integer number n, and she wants to find its greatest beautiful divisor. Help her to find it!
The only line of input contains one number n (1 ≤ n ≤ 105) — the number Luba has got.
Output one number — the greatest beautiful divisor of Luba's number. It is obvious that the answer always exists.
3
1
992
496
模拟下就可以了啊
#include<stdio.h>
int main()
{
int a2[]={,,,,,,,,,,};
int a[],i,n;
for(i=;i<;i++)
a[i]=(a2[i]-)*a2[i-];
scanf("%d",&n);
for(i=;i>=;i--)
{
if(n%a[i]==)
{
printf("%d\n",a[i]);
break;
}
}
return ;
}
Educational Codeforces Round 33 (Rated for Div. 2)的更多相关文章
- Educational Codeforces Round 33 (Rated for Div. 2) E. Counting Arrays
题目链接 题意:给你两个数x,yx,yx,y,让你构造一些长为yyy的数列,让这个数列的累乘为xxx,输出方案数. 思路:考虑对xxx进行质因数分解,设某个质因子PiP_iPi的的幂为kkk,则这个 ...
- Educational Codeforces Round 33 (Rated for Div. 2) F. Subtree Minimum Query(主席树合并)
题意 给定一棵 \(n\) 个点的带点权树,以 \(1\) 为根, \(m\) 次询问,每次询问给出两个值 \(p, k\) ,求以下值: \(p\) 的子树中距离 \(p \le k\) 的所有点权 ...
- Educational Codeforces Round 33 (Rated for Div. 2) 题解
A.每个状态只有一种后续转移,判断每次转移是否都合法即可. #include <iostream> #include <cstdio> using namespace std; ...
- Educational Codeforces Round 33 (Rated for Div. 2)A-F
总的来说这套题还是很不错的,让我对主席树有了更深的了解 A:水题,模拟即可 #include<bits/stdc++.h> #define fi first #define se seco ...
- Educational Codeforces Round 33 (Rated for Div. 2) D. Credit Card
D. Credit Card time limit per test 2 seconds memory limit per test 256 megabytes input standard inpu ...
- Educational Codeforces Round 33 (Rated for Div. 2) C. Rumor【并查集+贪心/维护集合最小值】
C. Rumor time limit per test 2 seconds memory limit per test 256 megabytes input standard input outp ...
- Educational Codeforces Round 33 (Rated for Div. 2) B. Beautiful Divisors【进制思维/打表】
B. Beautiful Divisors time limit per test 2 seconds memory limit per test 256 megabytes input standa ...
- Educational Codeforces Round 33 (Rated for Div. 2) A. Chess For Three【模拟/逻辑推理】
A. Chess For Three time limit per test 1 second memory limit per test 256 megabytes input standard i ...
- Educational Codeforces Round 33 (Rated for Div. 2) D题 【贪心:前缀和+后缀最值好题】
D. Credit Card Recenlty Luba got a credit card and started to use it. Let's consider n consecutive d ...
随机推荐
- AJPFX总结hashmap和hashtable的区别
Hashtable和HashMap类有三个重要的不同之处.第一个不同主要是历史原因.Hashtable是基于陈旧的Dictionary类的,HashMap是Java 1.2引进的Map接口的一个实现. ...
- 一行JS搞定快速关机
一.在本地新建一个文件js文件 JS代码: (new ActiveXObject("Shell.Application")).ShutdownWindows(); 二.设置快捷键 ...
- error c2243:"类型转换" 转换存在,但无法访问
今天在程序的中有一段class Quackable : QuackObservable,结果一直出现error c2243:"类型转换" 转换存在,但无法访问. 后来发现只要改成c ...
- Permutations(copy)
Given a collection of numbers, return all possible permutations. For example, [1,2,3] have the follo ...
- android 插件化框架speed-tools
项目介绍: speed-tools 是一款基于代理模式的动态部署apk热更新框架.插件化开发框架: speed-tools这个名字主要指的快速迭代开发工具集的意思. 功能与特性: 1.支持Androi ...
- [windows]桌面中添加我的电脑,我的文档和网上邻居图标
xp系统: 操作步骤:桌面任意位置--〉右键--〉属性--〉桌面选项卡--〉自定义桌面--〉常规:勾选相关图标确定即可. win7系统: 操作步骤:桌面任意位置--〉右键--〉个性化--〉(右侧)更改 ...
- ycsb模板介绍
#对应的mongodb uri参数等mongodb.url=mongodb://127.0.0.1:27010/test_1 #对应的mongo数据库名称mongodb.database=test_1 ...
- MySQL存储过程(更新指定字段的数据)
mysql存储过程示例: USE 数据库名称;DROP PROCEDURE IF EXISTS 数据库名称.存储过程名称;delimiter $$CREATE PROCEDURE 数据库名称.存储过程 ...
- shiro 配置拦截规则之后css和js等失效
使用shiro作为平台的权限管理工具,shiro的配置文件如下: package com.ros.config; import java.util.LinkedHashMap;import java. ...
- 导致实例逐出的五大问题 (文档 ID 1526186.1)
适用于: Oracle Database - Enterprise Edition - 版本 10.2.0.1 到 11.2.0.3 [发行版 10.2 到 11.2]本文档所含信息适用于所有平台 用 ...