生成器函数的第三种方式send（）一道例题

# .生成器函数的几种使用方法：①生成器._next()②next(生成器)③send()配合上一个yield
# send的原理及使用方法。例：
# def chibaozi(name):
#     print("等待中。。。。")
#     baozi=yield
#     print("%s开心的吃起了%s" %(name, baozi))
# def zaobaozi():
#     qi=chibaozi("佩奇")
#     qi.__next__()
#     qi.send("大包子")#这里send返回的值传给了chibaozi函数中的yield,而yield在赋值给baozi
# zaobaozi()
所有用递归能解决的问题都可以用for来替代解决，比如下面例子，能同时被3和7整除的数的和# def zc(ks,js):

#     fh=[]
#     while ks <= js:
#         if ks % 3 ==0 and ks % 7==0:
#             fh.append(ks)
#         ks +=1
#     z=sum(fh)
#     print("符合被3和被7整除的数有%s,它们的和是%s" % (fh, z))
#
# zc(1,100)

# def fuc(ks ,js ,a=0,b=0):
#     if ks ==js:
#         return a ,b
#     if ks%3==0 and ks%7==0:
#         a+=1
#         b +=ks
#     jg=fuc(ks+1,js,a,b )
#     return jg
# zz=fuc(1,100)
# print(zz)

# def js(ks,js):
#     a=0
#     for i in range(ks,js):
#         if i%3==0 and i%7==0:
#             a +=i
#     return a
# z=js(1,100)
# print(z)