流程控制练习题

Posted 2020-09-30 海带123

 

使用尽可能多的方法实现list去重

方法1：

a=[1,1,2,2,3,3,4,5]

print set(a)

方法2：

a=[1,1,2,2,3,3,4,5]

b=[]

for i in a:

    if i in b:

        continue

    else:

        b.append(i)

print b

方法3:

a=[1,1,2,2,3,3,4,5]

d={}

for i in a:

    d[i]=""



print d.keys()

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

while 1:

    grade=raw_input("please input your grade:")

    if grade==".":

        break

    else:

        grade=int(grade)

        if grade>=90:

            print "your level is A"

        elif grade >=60 and grade <90:

            print "your level is B"

        else:

            print "your level is C"

 

实现数学中多项式求和公式的打印

a6x^6 + a5x^5 + a4x^4 + a3x^3 + a2x^2 + a1x^1 + a0

exp=""

for i in range(6,0,-1):

    exp+=(‘a%s^x%s+‘ %(i,i))

exp+=‘a0‘

print exp

 

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_list=["jane","fosterwu","janet","jordan","flore",‘michel‘]

d={}



for name in name_list:

    if d.has_key(name[0]):

        d[name[0]]+=1

    else:

        d[name[0]] = 1



for letter,times in d.items():

    print "%s occurs %s times" %(letter,times)

 

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

a=int(raw_input("please input side length:"))

b=int(raw_input("please input side length:"))

c=int(raw_input("please input side length:"))



if a>0 and b>0 and c>0 and a+b>c and a+c>b and b+c>a and a-b<c and a-c<b and b-c<a:

    print "it is a triangle"

else:

    print "it is not a triangle"

实现字典的fromkeys方法

def from_keys(lista,value):

    d={}

    for i in lista:

        d[i]=value

    return d



print from_keys([‘a‘,"b","c"],10)

键盘读入一字符串，逆序输出：

line=raw_input("input a sentence:")

print line[::-1]

读入一个整数n，输出n的阶乘

def factorial(n):

    result=1

    for i in range(1,n+1):

        result*=i

    return result



print factorial(5)

 

打印1/2, 1/3, 1/4,….1/10

result=[]

for i in range(2,11):

    result.append("1/"+str(i))



print ",".join(result)

 

写一个函数实现一个数学公式

import math

def area(radius):

    return math.pi*radius**2



print area(10)

输入数字a，n，如a，4，则打印a+aa+aaa+aaaa之和

def print_exp(letter,times):

    result=[]

    for i in range(1,times+1):

        result.append(letter*i)

    return result



print "+".join(print_exp("a",4))

 

求100个随机数之和，随机数要求为0—9的整数

import random

result=0

for i in range(100):

    result+=random.randint(0,9)



print result

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_number_sum=0

oven_number_sum=0

for i in range(1,101):

    if i%2==0:

        oven_number_sum+=i

    else:

        odd_number_sum+=i



print odd_number_sum

print oven_number_sum

 

输入10个数，并显示最大的数与最小的数

biggest_number=None

least_number=None

for i in range(5):

    number=float(raw_input("please input a number:"))

    if i ==0:

        biggest_number,least_number=i,i

    if number>biggest_number:

        biggest_number=number

    if number<least_number:

        least_number=number



print biggest_number

print least_number

 

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。

number=raw_input("please input a number:")

print "number is %s length" %(len(number))

print str(number)[::-1]

 

求1000以内的所有水仙花数

for i in range(100,1000):

    length=len(str(i))

    if int(str(i)[0])**length+int(str(i)[1])**length+int(str(i)[2])**length==i:

        print i

 

编程求s=1!+2!+3!+…..+n!

def factorial(n):

    result=1

    for i in range(1,n+1):

        result*=i

    return result



def compute_factorial_sum(n):

    sum=0

    for i in range(1,n+1):

        sum+=factorial(i)

    return sum



print compute_factorial_sum(3)

 

钞票换硬币

result=[]

for i in range(0,101):

    for j in range(0,51):

        for m in range(0,21):

            if i+j*2+m*5==100:

                result.append("%s,%s,%s" %(i,j,m))



print len(result)

print result

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

sentence="i am a boy!"

search_word="i"

def find_word(word,target_sentence):

    word_length=len(word)

    for i in range(len(target_sentence)):

        if sentence[i:i+word_length]==word:

            return i

    return False



print find_word(search_word,sentence)

 

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

请见讲义

将一个矩阵转置

a=[[1,2,3],[4,5,6]]
def covert_array(array):
    b=[]
    for i in range(len(a[0])):
        x=[]
        for j in range(len(a)):
            x.append(a[j][i])
        b.append(x)
    return b

print covert_array(a)