백준: 15025번 Judging Moose (Python3)

백준: 15025번 Judging Moose (Python3)

Judging Moose 성공다국어

 

시간 제한메모리 제한제출정답맞힌 사람정답 비율

1 초

512 MB

1363

838

780

62.251%

문제

When determining the age of a bull moose, the number of tines (sharp points), extending from the main antlers, can be used. An older bull moose tends to have more tines than a younger moose. However, just counting the number of tines can be misleading, as a moose can break off the tines, for example when fighting with other moose. Therefore, a point system is used when describing the antlers of a bull moose.

The point system works like this: If the number of tines on the left side and the right side match, the moose is said to have the even sum of the number of points. So, “an even 6-point moose”, would have three tines on each side. If the moose has a different number of tines on the left and right side, the moose is said to have twice the highest number of tines, but it is odd. So “an odd 10-point moose” would have 5 tines on one side, and 4 or less tines on the other side.

Can you figure out how many points a moose has, given the number of tines on the left and right side?

입력

The input contains a single line with two integers ℓ and r, where 0 ≤ ℓ ≤ 20 is the number of tines on the left, and 0 ≤ r ≤ 20 is the number of tines on the right.

출력

Output a single line describing the moose. For even pointed moose, output “Even x” where x is the points of the moose. For odd pointed moose, output “Odd x” where x is the points of the moose. If the moose has no tines, output “Not a moose”

예제 입력 1 

2 3

예제 출력 1 

Odd 6

예제 입력 2 

3 3

예제 출력 2 

Even 6

예제 입력 3 

0 0

예제 출력 3 

Not a moose

답안

left, right = map(int, input().split())
 
if left==0 and right==0:
    print("Not a moose")
elif left==right:
    fmt = f"Even {2*left}"
    print(fmt)
else:
    fmt = f"Odd {max(left, right)*2}"
    print(fmt)