Educational Codeforces Round 33 (Rated for Div. 2) A. Chess For Three【模拟/逻辑推理】
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.
//刚开始一直想找找有什么规律可以直接得到结论···还是模拟大法好= =
【题意】：A，B，C三人参加比赛，两两对局，初始AB比赛，C围观，接下来输者围观。给你胜利者日志，求该日志是否正确。
【分析】：可以模拟游戏：有3个布尔变量 / 3 bools数组，然后根据谁的胜利进行更新。
初始球员1,2。观众：3。 结果必须在玩家之间，所以结果必须是1/2。如果结果=1，观众=2，玩家=3。在失败者和观众之间交换。
如果结果=2，观众=1，玩家=3。同理现在玩家：1，3，观众：2···一直这样判断直到发现悖论(谁获胜了却是观众)或直到循环结束。
（也可以位运算的1表示比赛 0表示旁观。每次赢得那个人保持1不变。剩下两个人变成相反 。www
【代码】：
#include <bits/stdc++.h> using namespace std; const int N = 200; int w[N]; int main() { int n; scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&p[i]); int a=1,b=2,c=3; for(int i=1;i<=n;i++) { if(w[i]==a) swap(b,c);//如果结果=1，观众=2，玩家=3。在失败者2和观众3之间交换。 else if(w[i]==b) swap(a,c);//如果结果=2，观众=1，玩家=3。在失败者1和观众3之间交换。 else return 0*puts("NO");//则不可能的局输出NO } puts("YES"); return 0; }
