1、一行解决杨辉三角
def triangles():
L = [1]
yield L
while True:
L = [1] + [L[i-1] + L[i] for i in range(1, len(L))] + [1]
yield L
n = 0
results = []
for t in triangles():
results.append(t)
n += 1
if n == 10:
break
for x in results:
print(x)
>>>
[1]
[1, 1]
[1, 2, 1]
[1, 3, 3, 1]
[1, 4, 6, 4, 1]
[1, 5, 10, 10, 5, 1]
[1, 6, 15, 20, 15, 6, 1]
[1, 7, 21, 35, 35, 21, 7, 1]
[1, 8, 28, 56, 70, 56, 28, 8, 1]
[1, 9, 36, 84, 126, 126, 84, 36, 9, 1]
2、求最大质数值
给定一个n值,求小于等于n的最大的一个质数
from math import sqrt, floor
# 判断是否是质数值,是True,否则False
def is_prime(n):
if n > 1:
if n == 2:
return True
if n % 2 == 0:
return False
for i in range(3, floor(sqrt(n))+1, 2):
if n % i ==0:
return False
return True
return False
n = int(input())
# 因为求给定值范围内的最大质数,反着依次判断是否为质数,取第一个符合的值即可
while 1:
if is_prime(n):
print(n)
break
else:
n -= 1
3、假设没有 float() 方法,把 str 转换成 float 形式
from functools import reduce
d = [str(i) for i in range(10)]
digits = {k:int(k) for k in d}
def prod(num_str):
# 求出小数部分的长度
length = len(num_str.split('.')[1])
# 去除'.'
num_str = num_str.replace('.', '')
num_integer = reduce(lambda x, y: x*10 + y,map(lambda x:digits[x], num_str))
return num_integer*(10**(-length))
print(prod('789456.123'))
>>>
789456.123
4、输入1000以内的回文(即正反读相同)
def is_palindrome(n):
if str(n) == str(n)[::-1]:
return n
L = list(filter(is_palindrome, range(1001)))
print(L)
>>>
[1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, 191, 202, 212, 222, 232, 242, 252, 262, 272, 282, 292, 303, 313, 323, 333, 343, 353, 363, 373, 383, 393, 404, 414, 424, 434, 444, 454, 464, 474, 484, 494, 505, 515, 525, 535, 545, 555, 565, 575, 585, 595, 606, 616, 626, 636, 646, 656, 666, 676, 686, 696, 707, 717, 727, 737, 747, 757, 767, 777, 787, 797, 808, 818, 828, 838, 848, 858, 868, 878, 888, 898, 909, 919, 929, 939, 949, 959, 969, 979, 989, 999]
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