chess for three A

 A类题   Chess for three 

在 cf上看到这个题，觉得这个题很有意思，所以决定把这个记录下来。

题目链接：点击打开链接

这个题是三人下棋。题目如下：

A. Chess For Three

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

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!

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 2if 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.

Examples

input

3112

output

YES

input

212

output

NO

题目大意：A、B和C三个人进行1V1的下棋，输者换第三个人上，第一场为A和B下。判断有没有这种胜利的情况，如果有输出YES，没有输出NO。

题目陷阱：有可能第一场胜利情况不存在。

我的思路：直接暴力过。记录每一场比赛后剩下来的那个人，每次都分情况进行一次判断，如果有相悖就结束循环。

C语言代码如下:

#include<stdio.h>
int main()
{
int a[1000];
int n,k,flag=1;
while(~scanf("%d",&n))
{
flag=1;
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
k=3;//k为不在场上的那个人 
for(int i=0;i<n;i++)
 {
  if(i==0)
  {
  if(a[i]==1)k=2;
  if(a[i]==2)k=1;
  if(a[i]==3)
  {
  flag=0;
  break;
 }
}
else 
{
if(a[i-1]==1&&k==2)
{
 if(a[i]==k)
 {
  flag=0;
  break;
 }
 if(a[i]==1)k=3;
 if(a[i]==3)k=1;
}
else if(a[i-1]==2&&k==1)
{
 if(a[i]==k)
 {
  flag=0;
  break;
 }
 if(a[i]==2)k=3;
 if(a[i]==3)k=2;
}
else if(a[i-1]==1&&k==3)
{
 if(a[i]==k)
 {
  flag=0;
  break;
 }
 if(a[i]==1)k=2;
 if(a[i]==2)k=1;
}
else if(a[i-1]==2&&k==3)
{
 if(a[i]==k)
 {
  flag=0;
  break;
 }
 if(a[i]==1)k=2;
 if(a[i]==2)k=1;
}
else if(a[i-1]==3&&k==1)
{
 if(a[i]==k)
 {
  flag=0;
  break;
 }
 if(a[i]==3)k=2;
 if(a[i]==2)k=3;
}
else if(a[i-1]==3&&k==2)
{
 if(a[i]==k)
 {
  flag=0;
  break;
 }
 if(a[i]==3)k=1;
 if(a[i]==1)k=3;
}
}
 }
        if(flag==0)printf("NO\n");
else printf("YES\n");
}
return 0;
}

本人是刚入门的小菜鸟，我知道我的方法一定不是最优的，也存在许多小问题，望多多包涵。