本文提供了一系列Python编程练习题，包括列表去重、成绩等级判断、多项式求和公式打印、统计名字首字母出现次数、判断是否能构成三角形等基础算法和数据处理问题，适合初学者巩固编程基础。
1使用尽可能多的方法实现list去重。
方法一：利用内置函数set去重
old_list = [1,2,2,3,3,4,5]
set(old_list)
方法二：定义一个新的列表存放结果，然后遍历旧列表，如果不在新列表中，就把元素插入到新列表中
old_list = [1,2,2,3,3,4,5]
new_list = []
for i in old_list:
if i not in new_list:
new_list.append(i)
print(new_list)
方法三：遍历列表中的元素，如果count某个元素的个数大于1，就把该元素删除。
old_list = [1,2,2,3,3,4,5]
for i in old_list:
if old_list.count(i) > 1:
old_list.remove(i)
print(old_list)
方法四：利用字典的key值唯一去重
old_list = [1,2,2,3,3,4,5]
j = 1
new_list = []
d = {}
for i in old_list:
d[i] = j
for k in d.keys():
new_list.append(k)
print(new_list)
方法五：先把list重新排序，然后从list的最后开始扫描
>>> a = [1,2,3,4,4,5]
>>> a.sort()
>>> a
[1, 2, 3, 4, 4, 5]
>>> last = a[-1]
>>> for i in range(len(a)-2,-1,-1):
... if last == a[i]:
... del a[i]
... else:
... last = a[i]
...
>>> print(a)
[1, 2, 3, 4, 5]
2.成绩等级判断：利用条件运算符的嵌套来完成此题：学习成绩>=90分的同学用A表示，60-89分之间的用B表示，60分以下的用C表示。
grade=float(input("请输入成绩："))
if grade>=90:
print("A")
elif grade>=60 and grade<=89:
print("B")
else:
print("C")
3.实现数学中多项式求和公式的打印
比如：a6x^6 + a5x^5 + a4x^4 + a3x^3 + a2x^2 + a1x^1 + a0
#r = a6x^6 + a5x^5 + a4x^4 + a3x^3 + a2x^2 + a1x^1 + a0
r = ""
n = 7
for i in list(range(n))[::-1]:
if i == 0:
r+="a0"
else:
r+="a"+str(i)+"x^"+str(i)+"+"
print(r)
4.统计名字列表中，各名字的首字母在名字列表中出现的次数。
name=['foster','foe','lily','mickel','live','moon','ruby','cindy','miya']
d = {}
for i in name:
if i[0] in d:
d[i[0]]+=1
else:
d[i[0]]=1
print(d)
for k,v in d.items():
print("姓名以字母 %s 开头的有 : %d人" %(k,v))
5.输入三个数，判断是否能构成三角形
能构成三角形三边关系：
三边都大于零
两边之和大于第三边，两边之差小于第三边
def check_triangle(a,b,c):
if not (a and b and c):
return False
flag = a + b > c and a + c > b and b + c > a
if not flag:
return False
return True
print(check_triangle(3, 4, 5))
print(check_triangle(1, 2, 3))
print(check_triangle(3, 6, 9))
print(check_triangle(2, 8, 5))
6.实现字典的fromkeys方法
例如：
seq = (‘name’, ‘age’, ‘sex’)
dict = dict.fromkeys(seq, 10)
print “New Dictionary : %s” % str(dict)
结果：New Dictionary : {‘age’: 10, ‘name’: 10, ‘sex’: 10}
>>> def myfromkeys(seq,value):
... seq_dict = {}
... for i in seq:
... seq_dict[i] = value
... return seq_dict
...
>>> print(myfromkeys(('name', 'age', 'sex'),10))
{'name': 10, 'age': 10, 'sex': 10}
7.键盘读入一字符串，逆序输出
>>> s = input("请输入字符串：")
请输入字符串：I am a good boy!
>>> for i in s[::-1]:
... print(i,end=" ")
...
! y o b d o o g a m a I >>>
8.读入一个整数n，输出n的阶乘
>>> n = int(input("please input a integer:"))
please input a integer:5
>>> result = 1
>>> for i in range(2,n+1):
... if i == 2:
... result *=2
... else:
... result*=i
...
>>> print(result)
120
9.打印1/2, 1/3, 1/4,….1/10
>>> for i in range(2,11):
... if i==10:
... print("1/%s" % i)
... else:
... print("1/%s" % i,end=",")
...
1/2,1/3,1/4,1/5,1/6,1/7,1/8,1/9,1/10
10.输入数字a，n，如a，4，则打印a+aa+aaa+aaaa之和
>>> def sum(a,n):
... result = 1
... for i in range(n):
... if i == n-1:
... result*=a
... else:
... result=result*a+1
...
... return result
...
>>> print(sum(3,4))
120
times = input("please input the times:")
digit = input("please input the digit:")
result = 0
for i in range(1,int(times)+1):
result += int(digit*i)
print(result)
times = input("please input the times:")
digit = input("please input the digit:")
temp_value = int(digit)
sum = int(digit)
for i in range(int(times)-1):
sum = temp_value*10+int(digit)+sum
temp_value = temp_value*10 + int(digit)
print(sum)
11.求100个随机数之和，随机数要求为0—9的整数
>>> import random
>>> sum = 0
>>> for i in range(100):
... random_num = random.randint(0,9)
... sum+=random_num
...
>>> print(sum)
409
13.要求在屏幕上分别显求1到100之间奇数之和与偶数之和
>>> odd_num_sum = 0
>>> even_num_sum = 0
>>> for i in range(1,101,2):
... print(i,end=" ")
... odd_num_sum+=i
...
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 5
7 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 >>> for j in ra
nge(2,101,2):
... print(j,end=",")
... even_num_sum+=j
...
2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40,42,44,46,48,50,52,54,56,
58,60,62,64,66,68,70,72,74,76,78,80,82,84,86,88,90,92,94,96,98,100,>>> print(odd
_num_sum)
2500
>>> print(even_num_sum)
2550
14.输入10个数，并显示最大的数与最小的数
>>> a = []
>>> for i in range(10):
... a.append(float(input("请输入一个数字：")))
...
请输入一个数字：4
请输入一个数字：2.4
请输入一个数字：9.8
请输入一个数字：2.3
请输入一个数字：1.9
请输入一个数字：3.45
请输入一个数字：6.42
请输入一个数字：0.56
请输入一个数字：6
请输入一个数字：7.98
>>> print(max(a),min(a))
9.8 0.56
l = list(range(10))
def func(l):
max_item = l[0]
min_item = l[0]
for val in l:
if val > max_item:
max_item = val
elif val < min_item:
min_item = val
return max_item,min_item
max_number,min_number = func(l)
print("最大数: ",max_number)
print("最小数: ",min_number)
15.给一个不多于5位的正整数，要求：一、求它是几位数，二、逆序打印出
各位数字。
>>> num = input("请输入一个不多于5位的整数：")
请输入一个不多于5位的整数：86324
>>> print(len(num))
5
>>> print(num[::-1])
42368
16.求1000以内的所有水仙花数（水仙花数：它的每个位上的数字的 n 次幂
之和等于它本身，例如：1^3 + 5^3+ 3^3 = 153）
>>> num_list = []
>>> for num in range(1,1000):
... n = len(str(num))
... if n == 2:
... if int(str(num)[0])**2 + int(str(num)[1])**2 == num:
... num_list.append(num)
... elif n == 3:
... if int(str(num)[0])**3 + int(str(num)[1])**3 +int(str(num)[2])**3 ==
num:
... num_list.append(num)
...
>>> print(num_list)
[153, 370, 371, 407]
17.编程求s=1!+2!+3!+……+n!
>>> def factorial_sum(n):
... r = 1
... s = 0
... for i in range(1,n+1):
... r*=i
... s+=r
... return s
...
>>>
>>> print(factorial_sum(3))
9
18.钞票换硬币 把一元钞票换成一分、二分、五分硬币（每种至少一枚），有多种换法，分 别有哪些？
remain = 100-1-2-5
result = []
count = 0
for i in range(int(remain/1)):
for j in range(int(remain/2)):
for k in range(int(remain/5)):
if (i + 2*j + 5*k) == remain:
count += 1
result.append((i,j,k))
print(count)
print(result)
19.自己实现在一句话中查找某个单词的算法，存在返回索引号，否则返回False
>>> s = "you are a beautiful girl good"
>>>
>>> def find_word(s,tar_word):
... length = len(tar_word)
... for i in range(len(s)):
... if s[i:i+length] == tar_word:
... return i
... return False
...
>>> print(find_word(s,"girl"))
20
20.读入一个十进制整数，实现十进制转二进制算法将其转成二进制数 要求：不能使用现成进制转换函数，自己写代码实现
def dec_to_bin(num):
remainder_list = []
while True:
temp = num%2
num = num//2
remainder_list.append(str(temp))
if num == 0:
break
return remainder_list[::-1]
print("".join(dec_to_bin(345)))
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