Day2
描述
大意就是石头剪刀布这个游戏分为对手和你
得分分为 你出手的形状得分 和最终胜利时的得分总和
出手形状得分:
胜利得分:
解法
解法思路：
- 第一部分规则是判断是否要赢或者输或者平局，得出获胜积分,此时选择积分是已经确定的
- 第二部分是规则 需要赢或者输或者平局 出对应的石头、剪刀、布 从而影响选择得分，而获胜积分此时
是确定的
下面我做的是暴力解法,可以优化
第一部分
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| from typing import List
def solution_two(two:List):
game_ans = []
for i in two :
opponent , myself = i
select_dict = {'X': 1,'Y':2,'Z':3}
win_dict = {'W':6,'L':0,'D':3}
opp_dict = {'A':'Rock','B':'Paper','C':'Scissors'}
own_dict = {'X':'Rock','Y':'Paper','Z':'Scissors'}
if own_dict[myself] == 'Rock':
if opp_dict[opponent]=='Paper':
win_score = win_dict['L']
elif opp_dict[opponent]=='Scissors':
win_score = win_dict['W']
else:
win_score = win_dict['D']
if own_dict[myself] == 'Paper':
if opp_dict[opponent]=='Scissors':
win_score = win_dict['L']
elif opp_dict[opponent]=='Rock':
win_score = win_dict['W']
else:
win_score = win_dict['D']
if own_dict[myself] == 'Scissors':
if opp_dict[opponent]=='Rock':
win_score = win_dict['L']
elif opp_dict[opponent]=='Paper':
win_score = win_dict['W']
else:
win_score = win_dict['D']
round_score = select_dict[myself] + win_score
game_ans.append(round_score)
return game_ans
|
第二部分
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| def solution_two_another(two:List):
game_ans = []
for i in two :
opponent , myself = i
select_dict = {'X': 1,'Y':2,'Z':3}
win_dict = {'W':6,'L':0,'D':3}
opp_dict = {'A':'Rock','B':'Paper','C':'Scissors'}
own_dict = {'X':'L','Y':'D','Z':'W'}
if own_dict[myself] == 'L':
if opponent == 'A':
new_type = 'Z'
if opponent == 'B':
new_type = 'X'
if opponent == 'C':
new_type = 'Y'
select_score = select_dict[new_type]
win_score = win_dict['L']
if own_dict[myself] == 'D':
if opponent == 'A':
new_type = 'X'
if opponent == 'B':
new_type = 'Y'
if opponent == 'C':
new_type = 'Z'
select_score = select_dict[new_type]
win_score = win_dict['D']
if own_dict[myself] == 'W':
if opponent == 'A':
new_type = 'Y'
if opponent == 'B':
new_type = 'Z'
if opponent == 'C':
new_type = 'X'
select_score = select_dict[new_type]
win_score = win_dict['W']
round_score = select_score + win_score
game_ans.append(round_score)
return game_ans
|
总结
这个题应该提前计算好每一种情况的值来预处理 简化结果 另外
我看Discord 上有人给出一行的解决方案
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| print(
sum(
(r := (M := "XYZABC".index)(L[2]))
+ [1 + 3 * (r == (o := M(L[0]) - 3)), 7][r == (o + 1) % 3]
+ 1j * [p := [3, 1, 2][o], 4 + o, 11 - p - o][r]
for L in open("input/02.txt")
)
)
|
PreComputed:
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| with open("the.txt",'r') as f:
k = f.read().split('\n')
l = []
for i in k:
l.append(i.split(' '))
m = 0
for i in l:
if ''.join(i) == "AX":
m += 4
if ''.join(i) == "AY":
m += 8
if ''.join(i) == "AZ":
m += 3
if ''.join(i) == "BX":
m += 1
if ''.join(i) == "BY":
m += 5
if ''.join(i) == "BZ":
m += 9
if ''.join(i) == "CX":
m += 7
if ''.join(i) == "CY":
m += 2
if ''.join(i) == "CZ":
m += 6
print(m)
|