Educational Codeforces Round 33 (Rated for Div. 2) A题
A. Chess For Three
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!
Input
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.
Output
Print YES if the situation described in the log was possible. Otherwise print NO.
input
output
YES
思路:简单的模拟即可。
AC代码:
#include<bits/stdc++.h> using namespace std;
int main(){
int _;
cin>>_;
int flag=;
int arr[_+];
int a,b,c;
a=;b=;c=;
for(int i=;i<=_;i++){
cin>>arr[i];
if(arr[i]==c){
flag=;
}
if(flag){
if(arr[i]==a){
swap(b,c);
}else if(arr[i]==b){
swap(a,c);
}else if(arr[i]==c){
swap(a,b);
}
}
}
if(flag){
puts("YES");
}else{
puts("NO");
}
return ;
}
Educational Codeforces Round 33 (Rated for Div. 2) A题的更多相关文章
- 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 ...
- Educational Codeforces Round 33 (Rated for Div. 2) C题·(并查集变式)
C. Rumor Vova promised himself that he would never play computer games... But recently Firestorm — a ...
- Educational Codeforces Round 33 (Rated for Div. 2) B题
B. Beautiful Divisors Recently Luba learned about a special kind of numbers that she calls beautiful ...
- 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 ...
随机推荐
- CentOS7下使用Harbor搭建Docker私有仓库
相关资料: Harbor官方网站:https://goharbor.io/ Harbor Github地址:https://github.com/goharbor/harbor ⒈安装Docker(必 ...
- spring mvc 简单实现及相关配置实现
配置文件 actio.xml <?xml version="1.0" encoding="UTF-8"?> <controller> & ...
- 【Trie】背单词
参考博客: https://www.luogu.org/problemnew/solution/P3294 https://blog.csdn.net/VictoryCzt/article/detai ...
- springboot加载application.yml文件null
话不多说,直接上代码 本人项目为maven项目 以下是项目结构 pom.xml文件 <?xml version="1.0" encoding="UTF-8" ...
- MySQL 索引的优化
一.MySQL如何使用索引(index) 1.1 索引概述 索引用于快速查找具有特定列值的行. 如果不使用索引,MySQL必须从表的第一行开始,然后扫描整个表来寻找符合条件的行.这种情况下,表越大,扫 ...
- Sql Server 收缩日志文件原理及always on 下的实践
一.准备知识 1.LSN LSN用来标识特定日志在日志文件中位置(详情请见什么是LSN:日志序列号),它由两部分组成:一部分用来标识VLF(虚拟日志文件)的序列号,剩下的用来标识该日志在VLF中的具体 ...
- vue2.0和animate.css的结合使用
animate.css是一款前端动画库,相似的有velocity-animate. 上述是一个完整的结构.其中重要的几个点用箭头表示出来.首先在transition组件内部,需要定义两个基本的clas ...
- 在QT中添加LIB的方法
注意:-L后面跟着的目录不能用空格,否则会出现读不到的情况. 例如win32下添加D:\app\my.lib 就 pro文件中 LIBS += -LD:\app\debug\ -lmy
- ubuntu目录结构(转)
/:根目录,一般根目录下只存放目录,不要存放文件,/etc./bin./dev./lib./sbin应该和根目录放置在一个分区中 /bin:/usr/bin:可执行二进制文件的目录,如常用的命令ls. ...
- VUE实现简单的全选/全不选
<!DOCTYPE html> <html> <head lang="en"> <meta charset="UTF-8&quo ...