第一種:
L = range(100,1000)
print([m for m in L if str(m)[0] == str(m)[2]])
第二種:
print([m*100+n*10+m for m in range(1,10) for n in range(0,10)])
無需if哦。
L = range(100,1000)
print([m for m in L if str(m)[0] == str(m)[2]])
第二種:
print([m*100+n*10+m for m in range(1,10) for n in range(0,10)])
無需if哦。
2020-06-04
先刪除index(2)的話,則下一次python執行操作的時候bart的index即為(2)因此如果運營pop(3)的話則超出list的元素個數界限。將兩個順序顛倒刪除即可解決。
2020-06-03
x1 = 1
d = 3
n=100
sum = 0
for x in range(1,n+1):
# print(x,x1)
sum += x1
# print(x,x1,sum)
x1 += d
x100 = x1-d
# print(x100)
s = "sum = {}".format(sum)
print(s)
d = 3
n=100
sum = 0
for x in range(1,n+1):
# print(x,x1)
sum += x1
# print(x,x1,sum)
x1 += d
x100 = x1-d
# print(x100)
s = "sum = {}".format(sum)
print(s)
2020-06-01
x1 = 1
d = 3
n = 100
for x in range(1,n+1):
# print(x,x1)
x1 += d
x100 = x1 - d
s = "x100 = {}".format(x100)
print(s)
d = 3
n = 100
for x in range(1,n+1):
# print(x,x1)
x1 += d
x100 = x1 - d
s = "x100 = {}".format(x100)
print(s)
2020-06-01
import math
def quadratic_equation(a, b, c):
t = math.sqrt(b*b - 4*a*c)
x1 = (-b - t)/(2*a)
x2 = (-b + t)/(2*a)
return x2,x1
print quadratic_equation(2, 3, 0)
print quadratic_equation(1, -6, 5)
注意分母2*a,一定得括號括起來,不然程序會認為先÷2,然后×a,結果就錯咯
def quadratic_equation(a, b, c):
t = math.sqrt(b*b - 4*a*c)
x1 = (-b - t)/(2*a)
x2 = (-b + t)/(2*a)
return x2,x1
print quadratic_equation(2, 3, 0)
print quadratic_equation(1, -6, 5)
注意分母2*a,一定得括號括起來,不然程序會認為先÷2,然后×a,結果就錯咯
2020-05-28
def square_of_sum(L):
sum=0
for a in L:
sum+=a**2
return sum
print square_of_sum([1, 2, 3, 4, 5])
print square_of_sum([-5, 0, 5, 15, 25])
sum=0
for a in L:
sum+=a**2
return sum
print square_of_sum([1, 2, 3, 4, 5])
print square_of_sum([-5, 0, 5, 15, 25])
2020-05-25
我覺得這個網特別好,我們那個年代根本沒有這么好的東西。大家免費學習,要有感恩的心。計算機這東西說復雜還是很復雜的,細節問題和版本有很大關系。
2020-05-25