這題越推越覺得應該用程序寫，所以就用python寫了一個，為什麽用python，因為真的很方便。

先看看理論上是否可行，10道題，每題4個選項，也就是4的10次冪，2的20次冪，也就是20bit，2.5Byte的數據要遍歷，這個數量級一般計算機妥妥夠用，4Byte以內通常都可以，要是20題就是5Byte就比較吃力了，6Byte就要幾周時間了，8Byte就別想了，當然是遍歷完，要是在前面就出結果了例外。

下面是python源碼：

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),

結果：

2018年刑偵科推理試題 python實現