Commit 2db0ac81 by BellCodeEditor

auto save

parent f14d92f5
'''
#猴子吃桃子
def pench(n):
if n==10:
return 1
num = (pench(n+1)+1)*2
return num
peach_num = pench(7)
print(peach_num)
#请用递归实现1+2+3....+num的累加和
#出口:1
def sum_num(number):
if number == 1:
return 1
sum_temp = sum_num(number-1)
return number+sum_temp
res = sum_num(3)
print(res)
#求阶乘 5! 5!= 5*4*3*2*1
#兔子数量
def f(n):
if n<=2:
return 1
v = f(n-1)+f(n-2)
return v
# elif n>2:
# v = f(n-1)+f(n-2)
# return v
res = f(6)
print(res)
#汉诺塔
def hlt(n):
if n==1:
return 1
count = 2*hlt(n-1)+1
return count
res = hlt(3)
print(res)
'''
#求阶乘 5! 5!= 5*4*3*2*1
def f(x):
if x==0:
return 0
elif x==1:
return 1
else:
y = x*f(x-1)
return y
print(f(5))
def Hanoi(n):
if n==1:
return 1
count = 2*Hanoi(n-1)+1
return count
res = Hanoi(3)
print(res)
#请用递归的知识实现1+2+3+...+num的累加和
def sum_numbers(num):
# 递归的出口
if num == 1:
return 1
# 数字的累加
temp = sum_numbers(num - 1)
return num + temp
result = sum_numbers(100)
print(result)
\ No newline at end of file
def f(n):
if n<=2:
return 1
values = f(n-1)+f(n-2)
return values
result = f(6)
print(result)
\ No newline at end of file
#利用递归方法求5!。5! 5*4*3*2*1
def f(x):
if x == 0:
return 0
elif x == 1:
return 1
else:
return (x * f(x-1))
print(f(5))
#第二种方式
f = 1
for i in range(1,6):
f = f * i
print(f)
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