[py]简易pick lucky num程序

程序功能:

1,用户输入数字,当用户输入指定数字时候,输出他输入的循环那次

2,第二次询问是否还要输

3,如果no 则 终止

4,如果yes则继续输入

  判断输入是否大于首次输入的

    如果大于则开始循环输出他第二次的lucky num

    否则循环让输入,直到它输入为大于首次

程序1:较为繁琐

#!/usr/bin/env python
####coding:utf8

lucky_num=input('lucky_num:')
count=1
while count < 100000:
    # count+=1
    print 'loop:',count
    if lucky_num==count:
        print 'you have got your lucky num:',count
        op=raw_input("if you want to continue(y/n):")
        if op=="n":break
        else:
            while 1:
                lucky_num_new=input('lucky_num:')
                if lucky_num>lucky_num_new:
                    print "too small,you should bigger than %s pls reinput your lucky num:"% lucky_num
                else:
                    lucky_num=lucky_num_new
                    break
        # print 'loop:',count
    count+=1
else:
    print 'wow. that fantastic',count

 

执行结果:

py@lanny:~/t$ python t.py 
lucky_num:5
loop: 1
loop: 2
loop: 3
loop: 4
loop: 5
you have got your lucky num: 5
if you want to continue(y/n):y
lucky_num:3
too small,you should bigger than 5 pls reinput your lucky num:
lucky_num:2
too small,you should bigger than 5 pls reinput your lucky num:
lucky_num:4
too small,you should bigger than 5 pls reinput your lucky num:
lucky_num:9
loop: 6
loop: 7
loop: 8
loop: 9
you have got your lucky num: 9
if you want to continue(y/n):

 

程序2:

#!/usr/bin/env python
#coding:utf8

lucky_num=input('pls input your lucky_num:')
count=0

while count<10000:
    print 'loop',count
    if lucky_num==count:
        print 'ok,now is loop:',count
        op=raw_input('again?(y/n)')
        if op=='n':break
        else:
            while 1:
                if lucky_num<=count:
                    lucky_num=input('pls input your lucky_num:')
                else:
                    break
    count+=1

 

转载于:https://www.cnblogs.com/iiiiher/p/5870140.html