def fbnq(x):
a,b=0,1
for i in range(x):
a,b=b,a+b
yield a
c=fbnq(6)
print(next©)
print(next©)
print(next©)
print(next©)
print(next©)
#传送带
a=[1,2,3,4,5,6,7,8,9,10]
for j in range(5):
t=a[0]
for i in range(len(a)-1):
a[i]=a[i+1]
a[-1]=t
print(a)
传送带和斐波那契
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转载自blog.csdn.net/weixin_42218868/article/details/88545365
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