New Ket Network 2018 Beijing University of Information Technology tenth session of the Program Design Competition and ACM tryouts-play Basketball

Links: Https://www.nowcoder.com/acm/contest/118/D

Source: Niu Ke Net

Topic Link: Click to open the link
Time limit: C/s 1 sec, other languages 2 seconds
Space limitations: C/C + + 32768K, other languages 65536K
64bit IO Format:%lld
Title Description
1, 2, 3 in playing basketball, but the two dozen one is always unfair, so they decided to a one-on, another person on the edge of watching, who lost who and the end and the side of the person watching the exchange. Now give you a sequence of victors (each round of winning people) and ask this sequence to be incompatible. (First singled out, 3 watched below)
Input Description:

The first number is the number n (1<=n<=100), which represents the size of the input winner sequence, and the next n lines describe the winner sequence. Line I contains a positive integer a[i], (1<=a[i]<=3), representing A[i] win the game

Output Description:

Output Yes if the winner sequence is legal, otherwise no

Example 1 input

3
1
1
2
2
1
2

Output

YES
NO

Description

In the first example, 1 won 2, 3 instead of 2, 1 won 3, 2 instead of 3, 2 won the
second example, 1 won 2, 3 instead of 2, when 2 was obviously already on the field, so it was impossible for the victor

Analysis:

Water problem

#include <stdio.h>

struct book{
	int win,x1,x2;
} S[105];

int main ()
{
	int n,a;
	while (~SCANF ("%d", &n))
	{
		int bj=0;
		s[0].x1=1;///initial state 
		s[0].x2=2;
		for (int i=1;i<=n;i++)
		{
			scanf ("%d", &a),/////Winner//////////// 
			for the time we enter, so if the value of BJ changes, it does not need to be judged. Because the wrong sequence has been found 
			if (!BJ)
			{
				if (a==s[i-1].x1| | A==S[I-1].X2)///If the winner is the contestant stating that it is eligible for the winner 
				{
					s[i].x1=a;///one of the next contestants is the winner of this round if 
					(s[i-1].x1+s[i-1].x2==3)// /Find another contestant in the next round 
						s[i].x2=3;
					else if (s[i-1].x1+s[i-1].x2==5)
						s[i].x2=1;
					else if (s[i-1].x1+s[i-1].x2==4)
						s[i].x2=2;
				}
				else
					bj=1;
			}
		}
		if (BJ)
			printf ("no\n");
		else
			printf ("yes\n");
	}
	return 0;
}