In [14]:

    

#写函数，返回某个元素/对象在一个list中的位置，如果不在，则返回-1.

def place():
    
    numbers = ['3','ert','34','3','f','3','5','r345','d','rtr','we','26']
    i = 0
    n = char(input("please enter an element,end with a tab: ")
    

    for n in numbers : 
        if numbers[i] == n :
             print("the place is: ",i+1)
             i = i+1
        else :
             i = i+1
    else :
        i = -1
             
    return i
             
print(place())


    

    




  File "<ipython-input-14-83ae6fbb92da>", line 10
    for n in numbers :
                     ^
SyntaxError: invalid syntax

In [18]:

    

#写函数，可求两个向量的夹角余弦值，向量可放在list中。主程序调用该函数。
import math

def cosine():
    vector1 = []
    vector2 = []
    i = 0
    x1 = 0
    x2 = 0
    y1 = 0
    y2 = 0
    cos = 0
    
    for i < 2:
        vector1.append(i)
        vector2.append(i)
        i += 1
        
    x1 = vector1[0]
    y1 = vector1[1]
    x2 = vector2[0]
    y2 = vector2[1]
        
    m1 = sqrt(x1*x1+y1*y1)
    m2 = sqrt(x2*x2+y2*y2)
    cos = (x1*x2 + y1*y2)/(m1*m2)
    
    return cos

print(cosine())


    

    




  File "<ipython-input-18-7424f6ed3415>", line 14
    for i < 2:
          ^
SyntaxError: invalid syntax

In [5]:

    

#挑战性习题：python语言老师为了激励学生学python，自费买了100个完全相同的Macbook Pro，分给三个班级，每个班级至少分5个
#用穷举法计算共有多少种分法？

def method():
    n1 = 5
    n2 = 5
    n3 = 5
    i = 0
    j = 85
    k = (85-i-j)
    
    for (i >= 0) and (i <= 85):
        for (j >= 0） and (j <= 85）:
            for （k >= 0） and (k <= 85）:
                n1 = n1 + i
                n2 = n2 + j
                n3 = n3 + k
                i = i+1
                j = j-1
                print("class1: ",n1)
                print("class2: ",n2)
                print("class3: ",n3)

method()


    

    




  File "<ipython-input-5-bd0c1eede77c>", line 12
    for i >= 0 and i <= 85:
           ^
SyntaxError: invalid syntax