2018年刑侦科推理试题 python实现

这题越推越觉得应该用程序写，所以就用python写了一个，为什么用python，因为真的很方便。

先看看理论上是否可行，10道题，每题4个选项，也就是4的10次幂，2的20次幂，也就是20bit，2.5Byte的数据要遍历，这个数量级一般计算机妥妥够用，4Byte以内通常都可以，要是20题就是5Byte就比较吃力了，6Byte就要几周时间了，8Byte就别想了，当然是遍历完，要是在前面就出结果了例外。

下面是python源码：

ask_all = 0;

ask = [0,0,0,0,0,0,0,0,0,0];

while ask_all < 2**20:

for i in range(10):

ask[i] = ((ask_all >> (2*i)) & 0x03);

ask_all = ask_all + 1;

#ASK2

select = [2,3,0,1]

if select[ask[1]] != ask[4]:

continue

#ASK3

flag = 0;

select = [2,5,1,3]

for i in select:

if (ask[i] == ask[select[ask[2]]]) & (i != select[ask[2]]):

flag = 1;

break;

if flag == 1:

continue

#ASK4

select = [[0,4],[1,6],[0,8],[5,9]]

if ask[select[ask[3]][0]] != ask[select[ask[3]][1]]:

continue

#ASK5

select = [7,3,8,6]

if ask[select[ask[4]]] != ask[4]:

continue

#ASK6

select = [[1,3],[0,5],[2,9],[4,8]]

if (ask[select[ask[5]][0]] != ask[7]) | (ask[select[ask[5]][1]] != ask[7]):

continue

#ASK7

select = [2,1,0,3]

count = [ask.count(0),ask.count(1),ask.count(2),ask.count(3)]

if select[ask[6]] != count.index(min(count)):

continue

#ASK8

select = [6,4,1,9]

if (ask[select[ask[7]]] == (ask[0]+1)) | (ask[select[ask[7]]] == (ask[0]-1)):

continue

#ASK9

select = [5,9,1,8]

if ask[0] == ask[5]:

if ask[select[ask[8]]] == ask[4]:

continue

else:

if ask[select[ask[8]]] != ask[4]:

continue

#ASK10

select = [3,2,4,1]

if select[ask[9]] != (max(count) - min(count)):

continue

for i in range(10):

print i+1,

print(‘‘);

for i in ask:

print chr(ord(‘A‘)+i),

结果：

原文地址：https://www.cnblogs.com/dracowangs/p/8495878.html