《2018年刑侦科推理试题》Python版求解⁄(⁄ ⁄•⁄ω⁄•⁄ ⁄)⁄
Eajack Lau
3月 07, 2018
昨天刷知乎想法，看到Milo Yip的深夜暴力編程，感觉有点意思，所有尝试用py写了下…思路没什么，就是暴力遍历而已，纯粹一时兴起~先抛结果：BCACACDABA
Python代码如下：
# !/usr/bin/env python3
# -*- coding : utf-8 -*-
# ==================================================================================================================
# Author: Eajack
# date:2018/3/7
# ==================================================================================================================
# Function：
# 2018年刑侦科推理试题，暴力遍历……
# ==================================================================================================================
# Results：
# ['B', 'C', 'A', 'C', 'A', 'C', 'D', 'A', 'B', 'A']
# [Finished in 5.7s]
# 经过检验，答案都是对的（如果没眼花的话……）
# ==================================================================================================================
def preparation(answers):
global maxCount,maxOption,minCount,minOption
numCountsDict = {i:answers.count(i) for i in set(answers)}
numCountsList = []
for key,value in numCountsDict.items():
numCountsList.append([value,key])
numCountsList = sorted(numCountsList,reverse = True)
maxCount = numCountsList[0][0]
maxOption = numCountsList[0][1]
minCount = numCountsList[-1][0]
minOption = numCountsList[-1][1]
def Q3Test(answers):
if(answers[2] == 1):
return (answers[5]==answers[1]==answers[3] and answers[2]!=answers[5])
else:
if(answers[2]==2):
return (answers[2]==answers[1]==answers[3] and answers[5]!=answers[2])
else:
if(answers[2]==3):
return (answers[2]==answers[5]==answers[3] and answers[1]!=answers[2])
else:
return (answers[2]==answers[5]==answers[1] and answers[3]!=answers[2])
def Q4Test(answers):
trueList = [int(answers[0]==answers[4]),int(answers[1]==answers[6]),int(answers[0]==answers[8]),int(answers[5]==answers[9])]
if(sum(trueList)==1 and trueList[answers[3]-1]==1):
return True
else:
return False
def Q6Test(answers):
if(answers[5] == 1):
return (answers[1]==answers[3]==answers[7])
else:
if(answers[5]==2):
return (answers[0]==answers[5]==answers[7])
else:
if(answers[5]==3):
return (answers[2]==answers[9]==answers[7])
else:
return (answers[4]==answers[8]==answers[7])
def Q8Test(answers):
if(answers[7] == 1):
return (abs(answers[6]-answers[0])!=1)
else:
if(answers[7]==2):
return (abs(answers[4]-answers[0])!=1)
else:
if(answers[7]==3):
return (abs(answers[1]-answers[0])!=1)
else:
return (abs(answers[9]-answers[0])!=1)
if __name__ == '__main__':
#
# globals
Q2toQ5 = {1:3,2:4,3:1,4:2}
Q4toQDouble = {1:[1,5],2:[2,7],3:[1,9],4:[6,10]}
Q5toQX = {1:8,2:4,3:9,4:7}
Q7toQY = {1:3,2:2,3:1,4:4}
Q9toQZ = {1:6,2:10,3:2,4:9}
Q10toQ = {1:3,2:2,3:4,4:1}
answers2ADs = {1:'A',2:'B',3:'C',4:'D'}
maxCount = []
maxOption = []
minCount = []
minOption = []
#
for Q1 in range(1,5):
for Q2 in range(1,5):
for Q3 in range(1,5):
for Q4 in range(1,5):
for Q5 in range(1,5):
for Q6 in range(1,5):
for Q7 in range(1,5):
for Q8 in range(1,5):
for Q9 in range(1,5):
for Q10 in range(1,5):
answers = [Q1,Q2,Q3,Q4,Q5,Q6,Q7,Q8,Q9,Q10]
preparation(answers)
# 2题 检验
if(Q2toQ5[Q2] != Q5):
continue;
else:
# 3题 检验
if(Q3Test(answers) != True):
continue;
else:
# 4题 检验
if(Q4Test(answers) != True):
continue;
else:
# 5题 检验
if(Q5 != answers[Q5toQX[Q5]-1]):
continue;
else:
# 6题 检验
if(Q6Test(answers) != True):
continue;
else:
# 7题 检验
if(minOption != Q7toQY[Q7]):
continue;
else:
# 8题 检验
if(Q8Test(answers) != True):
continue;
else:
# 9题 检验
if((Q1==Q6) == (answers[Q9toQZ[Q9]-1] == Q5)):
continue;
else:
# 10题 检验
if(maxCount - minCount != Q10toQ[Q10]):
continue;
else:
answersADs = [answers2ADs[answer] for answer in answers ]
print(answersADs)