• Python 练习


    (1)有四个数字:1、2、3、4,能组成多少个互不相同且无重复数字的三位数?各是多少?

    #!python2
    #-*- coding:utf-8 -*-
    
    result=[]
    for i in xrange(1,5):
        for j in xrange(1,5):
            for k in xrange(1,5):
                if i != j and i != k and j != k:
                    result.append((i,j,k))
                    
    result
    len(result)

    (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,从键盘输入当月利润I,求应发放奖金总数?

    #!python2
    #-*- coding:utf-8 -*-
    
    revenue=float(raw_input("Please input your revenue?"))
    
    boundary=[1000000,600000,400000,200000,100000,0]
    rate=[0.01,0.015,0.03,0.05,0.075,0.1]
    bonus=0
    
    for idx in xrange(6):
        if revenue>boundary[idx]:
            print "Over %d: %d" % (boundary[idx],(revenue-boundary[idx])*rate[idx])
            bonus+=(revenue-boundary[idx])*rate[idx]
            revenue=boundary[idx]
    print "Total bonus is %d" % bonus

    (3)一个整数,它加上100和加上268后都是一个完全平方数,请问该数是多少?(假定范围是10000以内)

    #!python2
    #-*- coding:utf-8 -*-
    
    import math
    
    for i in xrange(10000):
        x=int(math.sqrt((i+100)))
        y=int(math.sqrt(i+268))
        
        if (x*x==i+100) and (y*y==i+268) :
            print i

    (4)输入某年某月某日,判断这一天是这一年的第几天?

    #!python2
    #-*- coding:utf-8 -*-
    
    year=int(raw_input("please input year: "))
    month=int(raw_input("please input month: "))
    day=int(raw_input("please input day: "))
    
    months=(0,31,59,90,120,151,181,212,243,273,304,334)
    
    sum=0
    
    if 0<month<=12:
         sum=months[month-1]
         
    else:
        print "Please input a valid month!"
        
    
    sum+=day
    
    leap=0
    
    if (year %400 ==0) or ((year%4==0) and (year%100!=0)):
        
        leap=1
        
        print "%d year is a leap year!" %year
        
    if leap==1 and month >2:
        sum+=1
        
    print "It is %dth day of year %d" %(sum,year)

    (5)输入三个整数x,y,z,请把这三个数由小到大输出

    #!python2
    #-*- coding:utf-8 -*-
    
    l1=[]
    
    for i in xrange(3):
        x=int(raw_input("please input a int number: "))
        l1.append(x)
        
    l1.sort()
    print l1

    (6)9*9 乘法口诀表

    #!python2
    #-*- coding:utf-8 -*-
    
    
    for i in xrange(1,10):
        print 
        for j in xrange(1,i+1):
            print " %d * %d = %d " %(i,j,i*j),

    (7)暂停一秒输出

    #!python2
    #-*- coding:utf-8 -*-
    
    
    import time
    
    d1={1:"a",2:"b"}
    
    for k,v in d1.items():
        print k,v
        time.sleep(5)

    (8) 暂停一秒输出,并格式化当前时间

    #!python2
    #-*- coding:utf-8 -*-
    
    
    import time
    
    print time.strftime("%Y-%m-%d %H:%M:%S",time.localtime(time.time()))
    
    time.sleep(10)
    
    print time.strftime("%y-%m-%d %H:%M:%S",time.localtime(time.time()))

    (9)古典问题: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,问每个月的兔子总数为多少?

    #!python2
    #-*- coding:utf-8 -*-
    
    h = 0
    leap = 1
    from math import sqrt
    from sys import stdout
    
    for m in range(101,201):
        k = int(sqrt(m + 1))
        for i in range(2,k + 1):
            if m % i == 0:
                leap = 0
                break
        if leap == 1:
            print '%-4d' % m
            h += 1
            if h % 10 == 0:
                print ''
        leap = 1
    print 'The total is %d' % h

    (10)打印出所有的"水仙花数",所谓"水仙花数"是指一个三位数,其各位数字立方和等于该数本身

    #!python2
    #-*- coding:utf-8 -*-
    
    for n in xrange(100,1000):
        i = n/100
        j=n/10%10
        z=n%10
        
        if n==i**3+j**3+z**3:
            print n

    (11)将一个正整数分解质因数

    #!python2
    #-*- coding:utf-8 -*-
    
    def func(n):
        print "{} =".format(n),
        if not isinstance(n,int) or n<=0:
            print "Please input a valid integer."
            exit(0)
        elif n==1:
            print "{}".format(n)
            
        while n!=1:
            
            for index in xrange(2,n+1):
                
                if n%index==0:
                    
                    n/=index
                    
                    if n==1:
                        print index
                    else :
                        print "{} *".format(index),
                    break
                                  
    func(100)

    (12)一个排序

    #!python2   
    #-*- coding:utf-8 -*-
    
    data_list=[]
    
    data_list.append({"s":1,"v":1.1})
    data_list.append({"s":10,"v":1.9})
    data_list.append({"s":3,"v":1.6})
    data_list.append({"s":2,"v":1.2})
    data_list.append({"s":8,"v":1.4})
    
    data_list
    
    data_list.sort(key=lambda v:v.get("v"), reverse=False)

    (13) 时间相关

    #!python2   
    #-*- coding:utf-8 -*-
    
    import datetime
    
    #输出今日日期,格式为 dd/mm/yyyy
    
    today=datetime.datetime.today().strftime("%d/%m/%Y")
    
    print today
    
    #创建日期对象
    
    birthday=datetime.date(2000,1,1)
    
    print birthday.strftime("%d/%m/%Y")
    
    #日期算术运算
    
    tmp_date=birthday+datetime.timedelta(days=1)
    
    print tmp_date.strftime("%d/%m/%Y")
    
    #日期替换
    
    replace_date=birthday.replace(year=2012)
    
    print replace_date.strftime("%d/%m/%Y")

    (14)输入一行字符,分别统计出其中英文字母、空格、数字和其它字符的个数

    #!python2   
    #-*- coding:utf-8 -*-
    
    import string
    
    s=raw_input("Please input a word: 
    ")
    letters=0
    space=0
    digit=0
    others=0
    for c in s:
    
        if c.isalpha():
            letters+=1
        elif c.isspace():
            space+=1
        elif c.isdigit():
            digit+=1
        else:
            others+=1
    print "letters=%d, space =%d, digit=%d,others=%d" %(letters,space,digit,others)
        
        

    (15)求s=a+aa+aaa+aaaa+aa...a的值,其中a是一个数字。例如2+22+222+2222+22222(此时共有5个数相加),几个数相加有键盘控制

    #!python2   
    #-*- coding:utf-8 -*-
    
    a=int(raw_input("Please input number: 
    "))
    n=int(raw_input("Please input loop times: 
    "))
    l=[]
    tn=0
    for i in xrange(n):
        tn=tn+a
        a=a*10
        l.append(tn)
        
    
    sum_l=reduce(lambda x,y:x+y,l)    
    print l
    print sum_l

    (16)一个数如果恰好等于它的因子之和,这个数就称为"完数"。例如6=1+2+3.编程找出1000以内的所有完数。

    #!python2
    #-*- coding:utf-8 -*-
    
    for i in xrange(2,1001):
        k=[]
        n=-1
        s=i
        for j in xrange(1,s):
            if i%j==0:
                s-=j
                k.append(j)
                
        if s==0:
            print i
            print k

    (17)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?

    #!python2
    #-*- coding:utf-8 -*-
    
    t=[]
    h=[]
    h2=100.0
    for i in xrange(1,11):
        t.append(h2)
        h2/=2.0
        h.append(h2)
        
    print sum(t)
    print h[9]

    (18)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

    #!python2
    #-*- coding:utf-8 -*-
    
    p=1
    for i in xrange(9,0,-1):
        p=(p+1)*2
        
    print p

    (19)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。

    #!python2
    #-*- coding:utf-8 -*-
    
    for i in range(ord('x'),ord('z') + 1):
        for j in range(ord('x'),ord('z') + 1):
            if i != j:
                for k in range(ord('x'),ord('z') + 1):
                    if (i != k) and (j != k):
                        if (i != ord('x')) and (k != ord('x')) and (k != ord('z')):
                            print 'order is a -- %s	 b -- %s	c--%s' % (chr(i),chr(j),chr(k))

    (20)

    #!python2
    #-*- coding:utf-8 -*-
    
    from sys import stdout
    
    
    for i in xrange(4):
        for j in xrange(3-i):
            stdout.write(" ")
        for k in xrange(2*i+1):
            stdout.write("*")
        print 
    
    for i in xrange(3):
        for j in xrange(i+1):
            stdout.write(" ")
        for k in xrange(5-2*i):
            stdout.write("*")
        print
            

    (21) 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。

    #!python2
    #-*- coding:utf-8 -*-
    
    #methon 1
    a=1.0
    b=2.0
    s=0.0
    
    for n in xrange(1,21):
        s+=b/a
        a,b=b,a+b
        
    print s
    
    
    
    a=1.0
    b=2.0
    s=0.0
    
    
    #methon 2
    l=[]
    for n in xrange(1,21):
        l.append(b/a)
        a,b=b,a+b
    
    s=reduce(lambda x,y:x+y,l)
    print s

    (22)求1+2!+3!+...+20!的和

    #!python2
    #-*- coding:utf-8 -*-
    
    #methon 1
    
    s=0
    for n in xrange(1,21):
        j=1
        for i in xrange(1,n+1):
            j*=i
        s+=j
        
    print s
    
    #method 2
    
    s=0
    i=1
    for n in xrange(1,21):
        i*=n
        s+=i
      
    print s
    
    #method 3
    
    l=xrange(1,21)
    s=0
    
    def op(n):
        j=1
        for i in xrange(1,n+1):
            j*=i
        return j
    
    
    s=sum(map(op,l))
    print s

    (23)利用递归方法求5!

    #!python2
    #-*- coding:utf-8 -*-
    
    def fn_1(n):
        if n==0:
            sum=1
        else:
            sum=fn_1(n-1)*n
        return sum
    
    for i in xrange(6):
        print fn_1(i)

    (24)利用递归函数调用方式,将所输入的5个字符,以相反顺序打印出来

    #!python2
    #-*- coding:utf-8 -*-
    
    def fn_1(s,l):
        if l==0:
            return
        print s[l-1],
        fn_1(s,l-1)
        
    s=raw_input("Please input a string:")
    l=len(s)
    
    fn_1(s,l)

    (25)有5个人坐在一起,问第五个人多少岁?他说比第4个人大2岁。问第4个人岁数,他说比第3个人大2岁。问第三个人,又说比第2人大两岁。问第2个人,说比第一个人大两岁。最后问第一个人,他说是10岁。请问第五个人多大?

    #!python2
    #-*- coding:utf-8 -*-
    
    def fn_1(n):
        if n==1:
            c=10
            print c
            return
        else:
            c=fn_1(n-1)+2
            print c
        return c
        
    fn_1(5)
        

    (26)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。

    #!python2
    #-*- coding:utf-8 -*-
    
    n=int(raw_input("Please input a number:"))
    
    a=n/10000
    b=n%10000/1000
    c=n%1000/100
    d=n%100/10
    e=n%10
    
    if a!=0:
        print "5 位数:",e,d,c,b,a
    elif b!=0:
        print "4 位数:",e,d,c,b
    elif c!=0:
        print "3 位数:",e,d,c
    elif d!=0:
        print "2 位数:",e,d
    else:
        print "1 位数:",e
        
        
        
    
    
    
    
    
        

    (27)一个5位数,判断它是不是回文数。即12321是回文数,个位与万位相同,十位与千位相同。

    #!python2
    #-*- coding:utf-8 -*-
    
    #method 1
    
    n=int(raw_input("Please input a number:"))
    
    a=n/10000
    b=n%10000/1000
    d=n%100/10
    e=n%10
    
    if a==e and b==d:
        print "%d 是个回文数" %(n)
    else:
        print "%d 不是个回文数"%(n)
    
    
    #method 2
    
    n=int(raw_input("Please input a number:"))
    x=str(n)
    flag=True
    
    for i in xrange(len(x)/2):
        if x[i]!=x[-i-1]:
            flag=False
            break
            
    if flag:
        print "%d 是个回文数" %(n)
    else:
        print "%d 不是个回文数"%(n)

    (28)请输入星期几的第一个字母来判断一下是星期几,如果第一个字母一样,则继续判断第二个字母

    #!/usr/bin/python
    # -*- coding: UTF-8 -*-
    
    first_letter = raw_input("please input first letter(input Y abort):")
    
    if first_letter.upper() == 'S':
        print ('please input second letter:')
        second_letter = raw_input("please input:")
        if second_letter.upper() == 'A':
            print ('Saturday')
        elif second_letter.upper()  == 'U':
            print ('Sunday')
        else:
            print ('data error')
        
    elif first_letter.upper() == 'F':
        print ('Friday')
        
    elif first_letter.upper() == 'M':
        print ('Monday')
        
    elif first_letter.upper() == 'T':
        print ('please input second letter')
        second_letter = raw_input("please input:")
     
        if second_letter.upper()  == 'U':
            print ('Tuesday')
        elif second_letter.upper()  == 'H':
            print ('Thursday')
        else:
            print ('data error')
            
    elif first_letter.upper() == 'W':
        print ('Wednesday')
    else:
        print ('data error')
  • 相关阅读:
    怎么看到数据库以前做过的日志?
    感觉很好的网站
    uview 滑动切换
    Flyweight享元(结构型模式)
    悟空,真的是空?
    Interpreter解释器(行为型模式)
    Proxy代理(结构型模式)
    [转]有一种爱叫索取
    Command命令(行为型模式)
    Composite组合(结构型模式)
  • 原文地址:https://www.cnblogs.com/dadadechengzi/p/6514453.html
Copyright © 2020-2023  润新知