# a.     獲取內容相同的元素列表

 l1   =  [11,22,33]
 l2   =  [22,33,44]
 for l in l1:
     if l in l2:
         print(l)

# b.    獲取 l1  中有， l2 中沒有的元素列表

 l1   =  [11,22,33]
 l2   =  [22,33,44]
 for l in l1:
     if l not in l2:
         print(l)

# c.    獲取 l2  中有， l1 中沒有的元素列表

 l1   =  [11,22,33]
 l2   
 
=  [22,33,44]
 for i in l2:
     if i not in l1:
         print(i)

# d.    獲取 l1  和 l2   中內容都不同的元素

 l1   =  [11,22,33]
 l2   =  [22,33,44]
 for i in l1:
     if i not in l2:
         print(i)
 for i in l2:
     if i not in l1:
          print(i)

#有12345678 八個數字，如何組成不重複的兩位數，共有多少種？

 count=0
 for i in range(1, 9):
     for i1 in range(1,9):
         if i!=i1:
             count +=1
 print(count)

 count=0
 list1=[1,3]
 for i in range(0,len(list1)):
     for i1 in range(0,len(list1)):
         if i!=i1:
             count +=1

#九九乘法表

 for i in range(1,10):
     
 
for i1 in range(1,i+1):
         print(str(i)+"*"+str(i1)+"="+str(i*i1)+"\t",end="")
     print(end="\n")

    #公雞5文錢一隻，母雞3文錢一隻，小雞3只一文錢，100文錢只能買100只雞，有哪些組合？

 for x in range(1, (100 // 5+1) ):
     for y in range(1, (100 // 3+1) ):
         for z in range(1, 101):
             if x + y + z == 100 and 5 * x + 3 * y + z / 3 == 1000:
                 print(x,y,z)

#li=["dsk","ds","pp"],用下劃線把列表的值連起來

 li=["dsk","ds","pp",12]
 for i in li:
     print(i,end="_")

 li=["dsk","ds","pp",12]
 li[3]=str(li[3])
 v="_".join(li)
 print(v)

#獲取元組tu=("dhik","xjo","xsx")，計算元組的長度並輸出，獲取元組的第二個元素並輸出

 tu=("dhik","xjo","xsx")
 v=len(tu)
 print(v)
 

tu=("dhik","xjo","xsx")
 v=tu[1]
 print(v)

#nums=[1,3,5,2,8,4]
#找出列表中兩個元素相加等於9的結合

nums=[1,3,5,2,8,4]
cound=[]
for i in nums:
    for j in nums:
        if i+j==9:
            cound.append((i,j))

print(cound)