Python编程36题练习

其中1-30题采用Python2.7,31-36题采用Python3.7

(1)有1、2、3、4四个数字,能组成哪些互不相同且无重复数字的三位数?

# -*- coding: utf-8 -*-
for i in range(1,5):
    for j in range(1,5):
        for k in range(1,5):
            num = i*100+j*10+k;
            if(i!=j & j!=k):
                print num

 

(2)企业发放的奖金根据利润提成。利润(i)低于或等于10万元时,奖金可提10%;利润高于10万元,低于20万元时,低于10万元的部分按10%提成,高于10万元的部分,可提成7.5%;20万到40万之间时,高于20万元的部分,可提成5%;40万到60万之间时高于40万元的部分,可提成3%;60万到100万之间时,高于60万元的部分,可提成1.5%,高于100万元时,超过100万元的部分按1%提成,求当利润为i时应发放奖金总数。

# -*- coding: utf-8 -*-
num = input()
sum = 0
if(num <= 100000):
    sum = sum + num * 0.1
if(num > 100000 & num <= 200000):
    sum = sum + 100000 * 0.1 + (num-100000) * 0.075
if(num > 200000 & num <= 400000):
    sum = sum + 100000 * 0.1 + 100000 * 0.075 + (num - 200000) * 0.05
if(num >400000 & num <= 600000):
    sum = sum + 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + (num - 400000) * 0.03
if(num > 600000 & num <=1000000):
    sum = sum + 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + 200000 * 0.03 + (num - 600000) * 0.015
if(num > 1000000):
    sum = sum + 100000 * 0.1 + 100000 * 0.075 + 200000 * 0.05 + 200000 * 0.03 + 400000 * 0.015 + (num - 1000000) * 0.01
print sum

 

(3)输入某年某月某日,判断这一天是这一年的第几天?

# -*- coding: utf-8 -*-
 
def leapyear(year):
    return ((year % 400 == 0) | (year % 4 == 0 & year % 100 != 0))
 
sum = 0
year = input("请输入年:")
month = input("请输入月:")
day = input("请输入日:")
if(month > 1):
    sum = sum + 31
if(month > 2):
    if(leapyear(year)):
        sum = sum + 29
    else:
        sum = sum + 28
if(month > 3):
    sum = sum + 31
if(month > 4):
    sum = sum + 30
if(month > 5):
    sum = sum + 31
if(month > 6):
    sum = sum + 30
if(month > 7):
    sum = sum + 31
if(month > 8):
    sum = sum + 31
if(month > 9):
    sum = sum + 30
if(month > 10):
    sum = sum + 31
if(month > 11):
    sum = sum + 30
 
sum = sum + day
print ("第%d天" %(sum))

 

(4)使用□■输出国际象棋棋盘(8*8)。

# -*- coding: utf-8 -*-
 
for i in range(1,9):
    for j in range(1,9):
        if((i+j)%2 == 0):
            print("□"),
        else:
            print("■"),
    print("")

 

(5)有一对兔子,从出生后第3个月起每个月都生一对兔子,小兔子长到第三个月后每个月又生一对兔子,假如兔子都不死,问第N个月的兔子总数为多少?

# -*- coding: utf-8 -*-
num = input()
a = 1
b = 1
 
def fun(num):
    if(num == 1):
        return 1
    elif(num == 2):
        return 1
    else:
        return fun(num - 1) + fun(num - 2)
print 2*fun(num)

 

(6)将一个正整数分解质因数。例如:输入90,打印出90=2*3*3*5。

# -*- coding: utf-8 -*-
num = input()
print(num),
print("="),
for i in range(2,num):
    if(num == 1):
        break;
    while(num % i == 0):
        print(i),
        num = num / i;
        if(num != 1):
            print("*"),

 

(7)输入一个字符串,分别统计出其中英文字母、空格、数字和其它字符的个数。

# -*- coding: utf-8 -*-
str = raw_input()
alpha = 0
space = 0
number = 0
other = 0
for i in str:
    if(i.isalpha()):
        alpha=alpha + 1
    elif(i.isdigit()):
        number = number + 1
    elif(i.isspace()):
        space = space + 1
    else:
        other = other + 1
print("英文字母数=%d"%alpha)
print("数字数=%d"%number)
print("空格数=%d"%space)
print("其他字符数=%d"%other)

 

(8)输入a,n(0<a<10, 0<n<10),求s=a+aa+aaa+aaaa+aa…a(n个a)的值。例如a=2,n=5时为:2+22+222+2222+22222。

# -*- coding: utf-8 -*-
a = input()
n = input()
 
sum = 0
for i in range(n,0,-1):
    sum = sum + i * a
    a = a * 10
print sum

 

(9) 一球从100米高度自由落下,每次落地后反跳回原高度的一半;再落下,求它在第10次落地时,共经过多少米?第10次反弹多高?

# -*- coding: utf-8 -*-
num = 100.0
sum = 100.0
for i in range(0,9):
    num = num / 2.0
    sum = sum + num * 2.0
print ("第十次:%f"%(num/2.0))
print ("总距离:%f"%sum)

 

(10) 猴子吃桃问题:猴子第一天摘下若干个桃子,当即吃了一半,还不过瘾,又多吃了一个。第二天早上又将剩下的桃子吃掉一半,又多吃了一个。以后每天早上都吃了前一天剩下的一半零一个。到第10天早上想再吃时,见只剩下一个桃子了。求第一天共摘了多少?

# -*- coding: utf-8 -*-
num= 1
for i in range(0,9):
    num= (num + 1)*2
print num

 

(11) 两个乒乓球队进行比赛,各出三人。甲队为a,b,c三人,乙队为x,y,z三人。已抽签决定比赛名单。有人向队员打听比赛的名单。a说他不和x比,c说他不和x,z比,请编程序找出三队赛手的名单。

# -*- coding: utf-8 -*-
def fun(num):
    if num == 0:
        return "X"
    if num == 1:
        return "Y"
    if num == 2:
        return "Z"
 
for i in range(0,3):
    for j in range(0,3):
        if(i != j):
            for k in range(0,3):
                if(i != k and j != k):
                    if(k != 0 and i != 0 and k != 2):
                        print "A vs",
                        print fun(i)
                        print "B vs",
                        print fun(j)
                        print "C vs",
                        print fun(k)

 

(12) 求1+2!+3!+…+20!的和。

# -*- coding: utf-8 -*-
def fun(num):
    if(num == 1):
        return 1
    else:
        return num * fun(num-1)
 
sum = 0
for i in range(1,21):
    sum = sum + fun(i)
 
print sum

 

(13) 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。

# -*- coding: utf-8 -*-
 
def printConverse(num):
    count = 0
    while(num != 0):
        temp = num % 10
        print temp,
        num = num/10
        count = count + 1
    return count
 
num = input()
print num,
print "逆序输出:",
a = printConverse(num)
print "位数为:",
print a

 

(14)有一分数序列:2/1,3/2,5/3,8/5,13/8,21/13…求出这个数列的前20项之和,用分数形式表现出来。

(1)使用fractions库函数实现

# -*- coding: utf-8 -*-
import fractions
from fractions import  Fraction
a = 2
b = 1
sum = 0
for i in range(0,20):
    sum = sum + Fraction(a,b)
    temp = a
    a = a + b
    b = temp
 
print sum

(2)不用fractions库函数实现

# -*- coding: utf-8 -*-
a = 1
b = 2
denominator = 1
denominator = denominator * a * b
arrayDen = [1,2]
arrayNum = [2]
for i in range(0,18):
    sum = a + b
    arrayDen.append(sum)
    a = b
    b = sum
    denominator = denominator * sum
    arrayNum.append(sum)
 
arrayNum.append(a+b)
 
sumDen = denominator
sumNum = 0
for i in range(0,20):
    sumNum = sumNum + sumDen/arrayDen[i]*arrayNum[i]
 
for i in range(0,20):
    if(sumDen % arrayDen[i] == 0 and sumNum % arrayDen[i] == 0):
        sumDen = sumDen / arrayDen[i]
        sumNum = sumNum / arrayDen[i]
    if(sumDen % (i+1) == 0 and sumNum % (i+1) ==0):
        sumDen = sumDen/(i+1)
        sumNum = sumNum/(i+1)
print sumNum,
print "/",
print sumDen

 

(15)求0—7所能组成的数字中,不重复的8位奇数的个数。

# -*- coding: utf-8 -*-
sum = 1
#从1,3,5,7中选择奇数作为末尾
sum = sum * 4
#选择非0首位
sum = sum * 7
for i in range(6,0,-1):
    sum = sum * i
print sum

 

(16)已有结论“一个偶数总能表示为两个素数之和”,试输入一个偶数,用代码求出它对应的两个素数,若不止一组,尝试求相差最小的一组,比如16=13+3=11+5,则11+5更符合要求。

# -*- coding: utf-8 -*-
def isPrime(num):
    for i in range(2,num):
        if num % i == 0:
            return False
    return True
 
num = input()
temp = num / 2
for i in range(0,temp):
    if( isPrime( temp + i ) and isPrime( temp - i )):
        print temp+i,temp-i
        break;

 

(17) 对一个列表元素进行排序,要求:非冒泡排序法,非内置排序函数。

# -*- coding: utf-8 -*-
 
array = [1,5,5,6,3,6,2,7,9,1,0]
def selectionSort(arr):
    for i in range(0,len(arr)):
        min = i;
        for j in range(i+1,len(arr)):
            if(arr[j] < arr[min]):
                min = j
        if i != min:
            arr[min],arr[i] = arr[i],arr[min]
    return arr
 
print selectionSort(array)

 

(18) 有一个已经排好序的列表。现输入一个数,要求按原来的规律将它插入数组中。(注意,并不知列表原来是什么顺序)

# -*- coding: utf-8 -*-
 
array = [9,5,3,2,1]
 
num = input()
def insertArray(array , num):
    if(len(array) == 0 or len(array) == 1):
        array.append(num)
    else:
        if(array[0] < array[len(array)-1]):
            #升序
            for i in range(0,len(array)-1):
                if(num <= array[0]):
                    array.insert(0,num)
                    break
                elif(num >= array[len(array)-1]):
                    array.append(num)
                    break
                elif(array[i]<num and array[i+1]>=num):
                    array.insert(i+1,num)
        else:
            #降序
            for i in range(0,len(array)-1):
                if (num >= array[0]):
                    array.insert(0, num)
                    break
                elif (num <= array[len(array) - 1]):
                    array.append(num)
                    break
                elif (array[i] > num and array[i + 1] <= num):
                    array.insert(i + 1, num)
 
insertArray(array,num)
print array

 

(19) 一个列表逆序输出。(注意仅逆序输出,不要改变原列表元素)

# -*- coding: utf-8 -*-
 
array = [9,5,3,2,1]
 
for i in range(len(array)-1,-1,-1):
    print array[i],

 

(20) 将一个整数依次按 十六进制、八进制、二进制输出。

# -*- coding: utf-8 -*-
num = input()
print hex(num)
print oct(num)
print bin(num)

 

(21)  写如下几个进制转换函数,实现八进制、十进制、十六进制字符串的互转,不能使用内建转换函数,dec2oct,dec2hex,oct2dec,oct2hex,hex2oct,hex2dec,比如hex2oct(“FF”)==dec2oct(“255”)==”377″。

 

(22)输入整数V和整数m,n(0<=m<n<=31,0是低位),取V的二进制的第m到n位数字。例如:输入52,二进制为:0011 0100,取第3到5位的结果为:011。

# -*- coding: utf-8 -*-
V = input()
m = input()
n = input()
 
a =  str(bin(V))
print a
for i in range(m,n+1):
    print a[len(a)-1-i],

 

(23) 输入一个列表,将列表中最大的数与第一个元素交换,最小的与最后一个元素交换,输出之。

# -*- coding: utf-8 -*-
array = [7,1,5,6,6,2,1,4,8,9,5]
max = 0
min = 0
for i in range(0,len(array)):
    if(array[i] > array[max]):
        max = i
    if(array[i] < array[min]):
        min = i
 
array[max],array[0] = array[0],array[max]
array[min],array[len(array)-1] = array[len(array)-1],array[min]
 
print array

 

(24) 有n个人围成一圈,顺序排号。从第一个人开始报数(从1到3报数),凡报到3的人退出圈子,问最后留下的是原来第几号的那位。

# -*- coding: utf-8 -*-
n = input()
array = []
count = 0
for i in range(1,n+1):
    array.append(i)
 
while(len(array) > 1):
    q = []
    for i in array:
        count = count + 1
        if(count % 3 != 0):
            q.append(i)
    array = q
 
print  array[0]

 

(25) 创建一个列表,依次输入整形元素并插入列表尾;

(1)反向输出之,

(2)输出其单数列的元素,再输出其双数列的元素

(3)让单数列元素按升序排列,双数列元素按降序排列

# -*- coding: utf-8 -*-
n = input()
array = []
for i in range(0,n):
    temp = input()
    array.append(temp)
 
for i in range(n-1,-1,-1):
    print array[i],
 
print
 
for i in range(0,n):
    if(i % 2 == 0):
        print array[i],
print
 
for i in range(0,n):
    if(i % 2 == 1):
        print array[i],
print
 
for i in range(2,n,2):
    for j in range(i+1,n,2):
        if(array[j] > array[j-2]):
            array[j],array[j-2] = array[j-2],array[j]
 
for i in range(1,n,2):
    for j in range(i+1,n,2):
        if(array[j] < array[j-2]):
            array[j],array[j-2] = array[j-2],array[j]
 
print array

 

(26)创建一个字典,依次输入整形键、值并放入字典中。

1)依次打印键、值

2)以键的升序打印值

3)以值的升序打印键,同值则以键序

4)删除字典中键为1的元素,删除字典中值为1的元素

5)遍历字典,将字典中键为单数的元素都删去

# -*- coding: utf-8 -*-
dict = {}
a = []
b = []
for i in range(0,3):
    a = input()
    b = input()
    dict[a]=b
 
for key in dict.keys():
    print key,dict[key]
 
#以键排序
a = sorted(dict.items(),key = lambda  e:e[0],reverse=False)
#以值排序
b = sorted(dict.items(),key = lambda  e:e[1],reverse=False)
print a
print b
 
#删除字典中键为1的元素
for key in dict.keys():
    if(key == 1):
        dict.pop(1)
 
#删除字典中值为1的元素
for key in dict.keys():
    if(dict[key] == 1):
        dict.pop(key)
 
#普通遍历将字典中键为单数的元素都删去
count = 0
for key in dict.keys():
    if(count % 2 == 0):
        dict.pop(key)
    count = count + 1
 
print dict

 

(27) 编写一个类CPeople,保存着人的姓名str、年龄int、性别bool、工号int;依次从键盘或者文件输入每个人的如上信息。

1)遍历输出所有人的信息,格式为:“姓名:xxx;年龄:xx;性别:男/女;工号:xxxxxx”

2)遍历输出所有男性/女性员工信息,注意列对齐

3)按年龄升序输出所有员工姓名,同年龄则以工号升序为序

4)存储people类对象的数据结构使用list/dict,重新实现上述需求

# -*- coding: utf-8 -*-
class CPeople(object):
 
    def __init__(self , name , age , gender ,number):
        self.name = name
        self.age = age
        self.gender = gender
        self.number = number
 
liu = CPeople("liu",20,True,101)
chen = CPeople("chen",20,True,102)
deng = CPeople("deng",21,True,104)
yang = CPeople("yang",21,True,103)
tian = CPeople("tian",20,True,105)
xia = CPeople("xia",20,False,106)
 
arr = [liu,chen,yang,deng,tian,xia]
for i in arr:
    print i.name,i.age,i.gender,i.number
 
#输出男性
print "BOY:"
for i in arr:
    if(i.gender):
        print i.name,i.age,i.gender,i.number
 
#输出女性
print "GIRL:"
for i in arr:
    if(i.gender == False):
        print i.name,i.age,i.gender,i.number
 
#按年龄升序输出所有员工姓名,同年龄则以工号升序为序
try:
    import  operator
except ImportError:
    cmpfun = lambda CPeople:CPeople.age
else:
    cmpfun = operator.attrgetter("age","number")
 
    arr.sort(key = cmpfun , reverse = False)
    for i in arr:
        print  i.name

 

(28) 某个公司采用公用电话传递数据,数据是四位的整数,在传递过程中是加密的,加密规则如下:

每位数字都加上5,然后用和除以10的余数代替该数字,再将第一位和第四位交换,第二位和第三位交换。

写出其加密、解密函数。

# -*- coding: utf-8 -*-
data = 1007
#加密函数
def encryptionFun(num):
    num1 = num/1000
    num2 = num/100 % 10
    num3 = num/10 %10
    num4 = num % 10
    num1 = (num1 + 5) % 10
    num2 = (num2 + 5) % 10
    num3 = (num3 + 5) % 10
    num4 = (num4 + 5) % 10
    num1,num4 = num4,num1
    num2,num3 = num3,num2
    return num1*1000+num2*100+num3*10+num4
 
#解密函数
def decodeFun(num):
    num1 = num / 1000
    num2 = num / 100 % 10
    num3 = num / 10 % 10
    num4 = num % 10
    num1, num4 = num4, num1
    num2, num3 = num3, num2
    num1 = (num1 + 10 - 5)%10
    num2 = (num2 + 10 - 5)%10
    num3 = (num3 + 10 - 5)%10
    num4 = (num4 + 10 - 5)%10
    return num1*1000+num2*100+num3*10+num4
 
 
print encryptionFun(data)
print decodeFun(encryptionFun(data))

 

(29) 输入一个字符串,一个子串,计算字符串中子串出现的次数,已匹配过的字符不能再算做下一次匹配字符中。例如”abababab”, “abab” -> 2。

# -*- coding: utf-8 -*-
str= "abababab"
substr = "abab"
 
count = 0
i = 0
while(i <= len(str)-1):
    if(str.find(substr, i, len(str) ) != -1):
        count = count + 1
        i = i + len(substr)
    else:
        i = i + 1
print count

 

(30) 输入一个字符串,出现频率最高的3个字。

# -*- coding: utf-8 -*-
text ="系统初始为Win7.64bit系统,每个项目有特定的操作系统环境需求,具体请咨询所属项目同事;如需调整默认启动系统,请在Win7系统下进行如下操作(必须在Win7系统下才能设置):计算机(右键)-属性-高级系统设置-高级-启动和故障恢复(设置)-默认操作系统"
dict = {}
for word in text:
    if word not in dict:
        dict[word] = 1
    else:
        dict[word] = dict[word] + 1
b = sorted(dict.items(),key = lambda  e:e[1],reverse=True)
print(b)

 

(31) lst = [8,5,0,2,4,6,9,1,3,7],输入一个由0~9组成的列表,将其排序,要求lst位置靠前的元素,其排序后位置同样靠前。

# -*- coding: utf-8 -*-
lst = [8,5,0,2,4,6,9,1,3,7]
list = [0,1,2,3,4,5,6,7,8,9]
dict = {}
count = 0
for i in lst:
    dict[i] = count
    count = count + 1
 
for i in range(0,len(list)):
    for j in range(1,len(list)):
        if(dict[list[j]]<dict[list[j-1]]):
            list[j],list[j-1] = list[j-1],list[j]
 
print (list)

 

(32)在键盘上,左手每隔2秒按下一个字母键,依次循环按下A-Z,右手每隔5秒按下一个数字键,依次循环按下0-9,写代码模拟求按下的第100个键,是那个手?按下的哪个键?

# -*- coding: utf-8 -*-
a = 'A'
b = '1'
count = 0
second = 0
press = ''
 
while(True):
    if(second % 2 == 0):
        press = a
        a = chr(ord(a)+1)
        count = count + 1
 
    if(second % 5 == 0):
        press = b
        b = chr(ord(b)+1)
        count = count + 1
 
    second = second + 1
 
    if(count == 100):
        break
 
print(press,end="")
print("键")
if(press > 'A' and press < 'Z'):
    print("左手")
else:
    print("右手")

 

(33)有如下字典,记录每个地图场景能去往哪个地图场景

DATA = {

“中州城”   : (“师道殿”, “云梦泽”, “逐风原”, “清水湾”, “千魂塔一层”, “帮派地图”, ),

“北漠城”   : (“鸣沙洲”, “百花谷”, ),

“月牙湾”   : (“云梦泽”, “伏波港”, ),

“伏波港”   : (“月牙湾”, “百花谷”, ),

“鸣沙洲”   : (“北漠城”, “苍茫山”, ),

“逐风原”   : (“平安镇”, “中州城”, ),

“百花谷”   : (“北漠城”, “伏波港”, ),

“千魂塔一层”  : (“千魂塔二层”, “中州城”, ),

“云梦泽”   : (“月牙湾”, “中州城”, ),

“苍茫山”   : (“鸣沙洲”, ),

“清水湾”   : (“中州城”, “平安镇”, ),

“师道殿”   : (“绝情谷”, “真武门”, “蛮王殿”, “中州城”, “药王宗”, “天策府”, “拜火教”, “清虚观”, ),

“清虚观”   : (“师道殿”, ),

“天策府”   : (“师道殿”, ),

“真武门”   : (“师道殿”, ),

“拜火教”   : (“师道殿”, ),

“绝情谷”   : (“师道殿”, ),

“蛮王殿”   : (“师道殿”, ),

“药王宗”   : (“师道殿”, ),

“千魂塔二层”  : (“千魂塔一层”, ),

“平安镇”   : (“清水湾”, “逐风原”, ),

“帮派地图” : (“中州城”, ),

}

 

写一个函数FindPath(source, target),返回一个list,元素为从source到target所经过的各个地图名的一条即可。

# -*- coding: utf-8 -*-
DATA = {
   "中州城"  : ("师道殿", "云梦泽", "逐风原", "清水湾", "千魂塔一层", "帮派地图", ),
   "北漠城"  : ("鸣沙洲", "百花谷", ),
   "月牙湾"  : ("云梦泽", "伏波港", ),
   "伏波港"  : ("月牙湾", "百花谷", ),
   "鸣沙洲"  : ("北漠城", "苍茫山", ),
   "逐风原"  : ("平安镇", "中州城", ),
   "百花谷"  : ("北漠城", "伏波港", ),
   "千魂塔一层"    : ("千魂塔二层", "中州城", ),
   "云梦泽"  : ("月牙湾", "中州城", ),
   "苍茫山"  : ("鸣沙洲", ),
   "清水湾"  : ("中州城", "平安镇", ),
   "师道殿"  : ("绝情谷", "真武门", "蛮王殿", "中州城", "药王宗", "天策府", "拜火教", "清虚观", ),
   "清虚观"  : ("师道殿", ),
   "天策府"  : ("师道殿", ),
   "真武门"  : ("师道殿", ),
   "拜火教"  : ("师道殿", ),
   "绝情谷"  : ("师道殿", ),
   "蛮王殿"  : ("师道殿", ),
   "药王宗"  : ("师道殿", ),
   "千魂塔二层"    : ("千魂塔一层", ),
   "平安镇"  : ("清水湾", "逐风原", ),
   "帮派地图" : ("中州城", ),
}
 
lst = []
def FindPath(source ,target):
    if(source == target):
        return lst
    elif(target in lst):
        flag = True
        return lst
    for i in DATA[source]:
        if(i in lst):
            break
        else:
            lst.append(i)
            if(target in FindPath(i,target)):
                return lst
 
    return lst
 
list = FindPath("平安镇","中州城")
print(list)

 

(34) 有如下字符串

s = “””师门任务20次,0,80

除魔任务20次,0,100

闹事妖怪10次,0,50

运镖任务5次,0,40

挖宝5次,0,40

角色升级1次,0,50

完成4次江湖悬赏令,10,40

任务链整百环3次,0,90

获得修炼经验400点,0,50

野外暗雷战斗60次,0,50″””

将如上字符串解析为如下格式,注意对齐方式,以及第一项“今天总计”并不出现在s中,而是各行数字计算值。

# -*- coding: utf-8 -*-
import  re
s = """
   师门任务20次,0,80
   除魔任务20次,0,100
   闹事妖怪10次,0,50
   运镖任务5次,0,40
   挖宝5次,0,40
   角色升级1次,0,50
   完成4次江湖悬赏令,10,40
   任务链整百环3次,0,90
   获得修炼经验400点,0,50
   野外暗雷战斗60次,0,50"""
lst =  (re.findall(r"\d+",s))
 
#计算总数
count = 0
sum1 = 0
lst1 = []
sum2 = 0
lst2 = []
for i in lst:
    if (count % 3 == 1):
        sum1 = sum1 + int(i)
        lst1.append(i)
    elif (count % 3 == 2):
        sum2 = sum2 + int(i)
        lst2.append(i)
    count = count + 1
lst1.insert(0,str(sum1))
lst2.insert(0,str(sum2))
 
s1 = list(s)
count = 0
for i in range(0,len(s1)):
    if(s1[i] is ","):
        if(count % 2 == 0):
            s1[i] = ":"
            count = count + 1
        else:
            s1[i] = "/"
            count = count + 1
s = ''.join(s1)
s1 = re.split("\n\t|:",s)
 
count = 0
res = []
res.append("今天总计:")
for i in range(0,len(s1)):
    if(count % 2 == 0):
        count = count + 1
    else:
        res.append(s1[i])
        count = count + 1
 
 
print(" 每天完成以下事项可得活跃度")
tplt = "{0:{1}>10}"
for i in range(0,len(res)):
    print(tplt.format(res[i],chr(12288)),end="")
    print(":",end="")
    print(lst1[i],end="")
    print("/",end="")
    print(lst2[i])

 

(35)举例说明python中可变类型与不可变类型都有哪些?

不可变类型:数字、字符串、元组

可变类型:列表、字典

 

(36)编写函数,检测给定的两维列表中是否有重复数据(已知该列表中保存的是正整数):check_data(lList),如果没有重复,则函数返回真。要求算法的时间复杂度不大于O(n)。

# -*- coding: utf-8 -*-
lst1 = [['apple','banana','orange'],
        ['rabbit','tiger','forg'],
        ['microsoft','apple','amazon'],
        ['intel','amd']]
 
def check_data(List):
    map = {}
    for i in List:
        for j in i:
            if(j not in map):
                map[j] = 1
            else:
                return False
    return True
 
print(check_data(lst1))

 

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