for item in L" 循环中的语法错误
我感觉我在这里漏掉了一些很简单的东西,但在这个函数里:
def triplets(perimeter):
triplets, n, a, b, c = 0 #number of triplets, a, b, c, sides of a triangle, n is used to calculate a triple
L = primes(int(math.sqrt(perimeter)) #list of primes to divide the perimeter
for item in L: #iterate through the list of primes
if perimeter % item == 0: #check if a prime divides the perimeter
n = perimeter / item
a = n**2 - (n+1)**2 #http://en.wikipedia.org/wiki/Pythagorean_triple
b = 2n*(n+1)
c = n**2 + n**2
if a+b+c == perimeter: #check if it adds up to the perimeter of the triangle
triplets = triplets + 1
return triplets
我遇到了这个错误:
for item in L:
^
SyntaxError: invalid syntax
为了完整起见,我的整个程序是这样的:
import math
def primes(n): #get a list of primes below a number
if n==2: return [2]
elif n<2: return []
s=range(3,n+1,2)
mroot = n ** 0.5
half=(n+1)/2-1
i=0
m=3
while m <= mroot:
if s[i]:
j=(m*m-3)/2
s[j]=0
while j<half:
s[j]=0
j+=m
i=i+1
m=2*i+3
return [2]+[x for x in s if x]
def triplets(perimeter):
triplets, n, a, b, c = 0 #number of triplets, a, b, c, sides of a triangle, n is used to calculate a triple
L = primes(int(math.sqrt(perimeter)) #list of primes to divide the perimeter
for item in L: #iterate through the list of primes
if perimeter % item == 0: #check if a prime divides the perimeter
n = perimeter / item
a = n**2 - (n+1)**2 #http://en.wikipedia.org/wiki/Pythagorean_triple
b = 2n*(n+1)
c = n**2 + n**2
if a+b+c == perimeter: #check if it adds up to the perimeter of the triangle
triplets = triplets + 1
return triplets
def solve():
best = 0
perimeter = 0
for i in range(1, 1000):
if triplets(i) > best:
best = triplets(i)
perimeter = i
return perimeter
print solve()
我使用的是Python 2.7.1。在for循环后面有一个分号,primes(n)
这个函数是可以正常工作的,我觉得可能是一些很简单的错误,但我就是搞不清楚是什么导致了这个无效的语法。
4 个回答
0
在它之前的那行代码有个错误:
L = primes(int(math.sqrt(perimeter))
你有三个左括号,但只有两个右括号。
1
你少了一个括号:
L = primes(int(math.sqrt(perimeter)))
^
|
this one
我经常遇到这种情况,你只需要看看前面的那一行就行了。
18
你在前面的一行代码中缺少一个右括号:
L = primes(int(math.sqrt(perimeter)) #list of primes to divide the perimeter
# ^ ^ ^ ^^
#nesting count 1 2 3 21
看看我们在下面的“嵌套计数”中为什么没有达到0?