碎碎念
最后两天！两天！
Day12
ll两点才给我发题，只能先摸会python了
摸完第一阶段，题终于来了，康了两眼，简单题！
半个小时胡完，可以继续摸python了嘿嘿嘿
--------碎碎念到此结束，下面是正文（手动分割线）--------
phase1：熟悉语句
# 请仅使用一行语句求出三个数的最小平方和
def two_of_three(x, y, z):
return x**2+y**2+z**2-max(x,y,z)**2
# 下降阶乘
def falling(n, k):
ans=1
for i in range(n-k+1,n+1):
ans=ans*i
return ans
# 判断一个函数是否有两个或者两个连续的8
def double_eights(n):
while n!=0:
if n%100==88:
return True
n/=10
return False
if __name__ == '__main__':
opt=input()
if opt=='1':
x,y,z=map(int,input().split())
print(two_of_three(x,y,z))
elif opt=='2':
x,y=map(int,input().split())
print(falling(x,y))
elif opt=='3':
x=int(input())
print(double_eights(x))phase2：递归操作
# 编写一个递归函数skip_add，它接受一个参数 n 并返回n + n-2 + n-4 + n-6 + ... + 0。假设 n 是非负数。
def skip_add(n):
if n>0:
return n+skip_add(n-2)
else:
return 0
# GCD，给出两个正整数，求出两个正整数的最大公约数
def gcd(a, b):
if b==0:
return a
else:
return gcd(b,a%b)
# 汉诺塔
def print_move(origin, destination):
print("Move the top disk from rod", origin, "to rod", destination)
def move_stack(n, start, end):
assert 1 <= start <= 3 and 1 <= end <= 3 and start != end, "Bad start/end"
if n==1:
print_move(start,end)
else:
mid=6-start-end
move_stack(n-1,start,mid)
print_move(start,end)
move_stack(n-1,mid,end)
if __name__ == '__main__':
opt=input()
if opt=='1':
x=int(input())
print(skip_add(x))
elif opt=='2':
x,y=map(int,input().split())
print(gcd(x,y))
elif opt=='3':
x,y,z=map(int,input().split())
move_stack(x,y,z)phase3：数据抽象
# 99乘法表
def nine_nine_multiplication_table():
return '\n'.join([' '.join([f"{j}*{i}={i*j}" for j in range(1,i+1)]) for i in range(1,10)])
# 实现函数couple，它接受两个列表并返回一个列表，其中包含两个序列的第 i 个元素耦合在一起的列表。您可以假设两个序列的长度相同。
def couple(lst1, lst2):
assert len(lst1) == len(lst2)
return [[lst1[i],lst2[i]] for i in range(len(lst1))]
# 计算两地距离，找出两点中离某点更近的一点
def make_city(name, lat, lon):
return [name, lat, lon]
def get_name(city):
return city[0]
def get_lat(city):
return city[1]
def get_lon(city):
return city[2]
from math import sqrt
def distance(city1, city2):
return sqrt((get_lon(city1)-get_lon(city2))**2+(get_lat(city1)-get_lat(city2))**2)
def closer_city(lat, lon, city1, city2):
x=make_city('x',lat,lon)
if distance(x,city1)>distance(x,city2):
return get_name(city2)
else:
return get_name(city1)
if __name__ == '__main__':
opt=input()
if opt=='1':
print(nine_nine_multiplication_table())
elif opt=='2':
lst1 = ['c', 6]
lst2 = ['s', '1']
print(couple(lst1,lst2))
elif opt=='3':
city1 = make_city('city3', 6.5, 12)
city2 = make_city('city4', 2.5, 15)
print(distance(city1, city2))
elif opt=='4':
berkeley = make_city('Berkeley', 37.87, 112.26)
stanford = make_city('Stanford', 34.05, 118.25)
print(closer_city(38.33, 121.44, berkeley, stanford))
bucharest = make_city('Bucharest', 44.43, 26.10)
vienna = make_city('Vienna', 48.20, 16.37)
print(closer_city(41.29, 174.78, bucharest, vienna))phase4：高阶函数
def count_cond(condition):
return lambda n:sum([condition(n,i) for i in range(1,n+1)])
if __name__ == '__main__':
count_factors = count_cond(lambda n, i: n % i == 0)
print(count_factors(2))
print(count_factors(4))
print(count_factors(12))
is_prime = lambda n, i: count_factors(i) == 2
count_primes = count_cond(is_prime)
print(count_primes(2))
print(count_primes(3))
print(count_primes(4))
print(count_primes(5))
print(count_primes(20))phase5：迭代生成
# 实现count，它接受一个迭代器t并返回该值x出现在 的第一个n元素中的次数t
def count(t,n,x):
c=0
for i in range(n):
p=next(t)
if x==p:
c=c+1
return c
# 实现生成器函数scale(it, multiplier)，它产生给定迭代的元素it，按multiplier.
# 同时也希望你尝试使用yield from语句编写这个函数！
def scale(it, multiplier):
yield from [i*multiplier for i in it]
if __name__ == '__main__':
opt=input()
if opt=='1':
s = iter([10, 9, 10, 9, 9, 10, 8, 8, 8, 7])
print(count(s, 10, 9))
s2 = iter([10, 9, 10, 9, 9, 10, 8, 8, 8, 7])
print(count(s2, 3, 10))
s = iter([3, 2, 2, 2, 1, 2, 1, 4, 4, 5, 5, 5])
print(count(s, 1, 3))
print(count(s, 4, 2))
print(next(s))
s2 = iter([4, 1, 6, 6, 7, 7, 8, 8, 2, 2, 2, 5])
print(count(s2, 6, 6))
elif opt=='2':
m = scale(iter([1, 5, 2]), 5)
print(type(m))
print(list(m))
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