【每日一题】2021.10.21
斐波那契数列F(1)=F(2)=1;n>2时 F(n)=F(n-1)+F(n-2)
用递归和循环两种方法编程求F(50), F(100), F(1000)的值
编程语言:包括但不限于Python
【每日一题】2021.10.21
斐波那契数列F(1)=F(2)=1;n>2时 F(n)=F(n-1)+F(n-2)
用递归和循环两种方法编程求F(50), F(100), F(1000)的值
编程语言:包括但不限于Python
循环
## 将斐波那契额序列
## 获取用户输入数据
# nmber=input()
def fib_loop_while(n):
a, b = 0, 1
for i in range(0, n):
a, b = b, a + b
yield a
#
# for i in fib_loop_while(nmber):
# print(i, end=' ')
for i in fib_loop_while(50):
print(i, end=' ')
for i in fib_loop_while(100):
print(i, end=' ')
for i in fib_loop_while(1000):
print(i, end=' ')
递归
# 递归
def rabbit2(n):
if (n==1 or n==2):
return 1
else:
return rabbit2(n-1)+rabbit2(n-2)
for i in range(1,50):
print(rabbit2(i),end=' ')
for i in range(1,100):
print(rabbit2(i),end=' ')
for i in range(1,1000):
print(rabbit2(i),end=' ')