AtCoder Toyota Programming Contest 2023#8（AtCoder Beginner Contest 333）B题解

B - Pentagon?Editorial

?/?

Time Limit: 2 sec / Memory Limit: 1024 MB

Score :?200200?points

Problem Statement

A regular pentagon?�P?is shown in the figure below.

Determine whether the length of the line segment connecting points?�1S1??and?�2S2??of?�P?equals the length of the line segment connecting points?�1T1??and?�2T2?.

Constraints

• Each of?�1S1?,?�2S2?,?�1T1?, and?�2T2??is one of the characters?A,?B,?C,?D, and?E.
• �1≠�2S1?=S2?
• �1≠�2T1?=T2?

---

Input

The input is given from Standard Input in the following format:

�1�2S1?S2?
�1�2T1?T2?

Output

If the length of the line segment connecting points?�1S1??and?�2S2??of?�P?equals the length of the line segment connecting points?�1T1??and?�2T2?, print?Yes; otherwise, print?No.

---

Sample Input 1Copy

Copy

AC
EC

Sample Output 1Copy

Copy

Yes

The length of the line segment connecting point?A?and point?C?of?�P?equals the length of the line segment connecting point?E?and point?C.

---

Sample Input 2Copy

Copy

DA
EA

Sample Output 2Copy

Copy

No

The length of the line segment connecting point?D?and point?A?of?�P?does not equal the length of the line segment connecting point?E?and point?A.

---

Sample Input 3Copy

Copy

BD
BD

Sample Output 3Copy

Copy

Yes

#include<iostream>
using namespace std;
int main()
{
	string a;
	string b;
	cin>>a>>b;
	if(a==b)
		{
			cout<<"Yes";
			return 0;
		}
	if(a[1]==b[1])
		{
			if((a[0]+2==a[1]||a[0]-2==a[1])&&(b[0]+2==b[1]||b[0]-2==b[1]))
				{
					cout<<"Yes";
					return 0;
				}
			if((a[0]+1==a[1]||a[0]-1==a[1])&&(b[0]+1==b[1]||b[0]-1==b[1]))
				{
					cout<<"Yes";
					return 0;
				}
			else
				{
					cout<<"No";
					return 0;
				}
		}
	if((a[0]+1==a[1]||a[0]-1==a[1])&&(b[0]+1==b[1]||b[0]-1==b[1]))
		{
			cout<<"Yes";
			return 0;
		}
	if(a[0]+4==a[1]||a[0]-4==a[1]||b[0]+4==b[1]||b[0]-4==b[1])
	{
		if(b[0]+2==b[1]||b[0]-2==b[1]){
			cout<<"No";
			return 0;
		}
		cout<<"Yes";
		return 0;
	}
	else
		{
			cout<<"No";
			return 0;

		}
	return 0;
}