题目：
The Fibonacci numbers are the numbers in the following integer sequence (Fn):
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, ...
such as
F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.
Given a number, say prod (for product), we search two Fibonacci numbers F(n) and F(n+1) verifying
F(n) * F(n+1) = prod.
Your function productFib takes an integer (prod) and returns an array:
[F(n), F(n+1), true] or {F(n), F(n+1), 1} or (F(n), F(n+1), True)
depending on the language if F(n) * F(n+1) = prod.
If you don't find two consecutive F(m) verifying F(m) * F(m+1) = prodyou will return
[F(m), F(m+1), false] or {F(n), F(n+1), 0} or (F(n), F(n+1), False)
F(m) being the smallest one such as F(m) * F(m+1) > prod.
Examples
productFib(714) # should return [21, 34, true],
# since F(8) = 21, F(9) = 34 and 714 = 21 * 34
productFib(800) # should return [34, 55, false],
# since F(8) = 21, F(9) = 34, F(10) = 55 and 21 * 34 < 800 < 34 * 55
Notes: Not useful here but we can tell how to choose the number n up to which to go: we can use the "golden ratio" phi which is (1 + sqrt(5))/2 knowing that F(n) is asymptotic to: phi^n / sqrt(5). That gives a possible upper bound to n.
You can see examples in "Example test".
References
http://en.wikipedia.org/wiki/Fibonacci_number
题意：
题意就是输入一个数，找出斐波那契序列中连续两个数相乘等于此数，则输出这两个数和True；否则输出这刚好大于这个数的两个数的乘积和False解题思路：
思路：
我的思路是先循环遍历；从0、1、1、2、3、5……两两相乘，直到等于输入的数，则输出[两个数，True],否则输出刚好大于这个数乘积的两个数[两个数，False]
代码：
def productFib(prod):
f=[]
f.append(0)
f.append(1)
list1=[]
for i in range(2,prod):
f.append(f[i-1]+f[i-2])
if f[i]*f[i-1] == prod:
list1.append(f[i-1])
list1.append(f[i])
list1.append(True)
return list1
if f[i]*f[i-1] > prod:
list1.append(f[i-1])
list1.append(f[i])
list1.append(False)
return list1
return None我的思路很简单，接下来看看大佬的精简代码：
def productFib(prod):
a, b = 0, 1
while prod > a * b:
a, b = b, a + b
return [a, b, prod == a * b]相当精炼：真的棒！只用了两个变量就解决了这个问题。
题目二：
这道题没写出来，写了可以但是timeout
Longest Palindrome
Find the length of the longest substring in the given string s that is the same in reverse.
As an example, if the input was “I like racecars that go fast”, the substring (racecar) length would be 7.
If the length of the input string is 0, the return value must be 0.
Example:
"a" -> 1
"aab" -> 2
"abcde" -> 1
"zzbaabcd" -> 4
"" -> 0题意：
题意就是找到一串连续字符串左边和右边对称并且要是最长的；
如："zzbaabcd" -> 4 其中baab就是这个字符串
代码如下：
我的解法很简单就是把字符串每个子串部分放在一个列表中，然后把字符串颠倒，将其子串放在一个列表中，然后匹配，计算最长匹配值返回。
然而超时了：
def longest_palindrome(s):
#s[::-1]颠倒字符串
a=s
b=s[::-1]
#print(b[-1])
length=len(s)
ls=[]
ls1=[]
maxlength=0
for i in range(0,length):
for j in range(i+1,length+1):
c = a[i:j]
ls.append(c)
#print(ls)
for i in range(0,length):
for j in range(i+1,length+1):
d = b[i:j]
ls1.append(d)
#print(ls1)
for j in ls:
if j in ls1:
num=len(j)
maxlength=max(num,maxlength)
return maxlength#AC代码：大神精简代码
def longest_palindrome(s):
for i in range(len(s), 0, -1):
for j in range(len(s)-i+1):
#print(i,j)
sub = s[j:j+i]
#print(sub)
if sub == sub[::-1]:
return i
return 0
#print(longest_palindrome("abcdefghba"), 1)加了两个print，理解此代码的输出过程：我大概理解了是固定长度，变区间；比如一开始i就代表字符串长度10，然后区间就只有一个0-10；接下来i减少1，则区间可以是0-9，1-10；i=8时，就有三个区间：0-8、1-9、2-10；以此类推……
10 0
abcdefghba
9 0
abcdefghb
9 1
bcdefghba
8 0
abcdefgh
8 1
bcdefghb
8 2
cdefghba
7 0
abcdefg
7 1
bcdefgh
7 2
cdefghb
7 3
defghba
6 0
abcdef
6 1
bcdefg
6 2
cdefgh
6 3
defghb
6 4
efghba
5 0
abcde
5 1
bcdef
5 2
cdefg
5 3
defgh
5 4
efghb
5 5
fghba
4 0
abcd
4 1
bcde
4 2
cdef
4 3
defg
4 4
efgh
4 5
fghb
4 6
ghba
3 0
abc
3 1
bcd
3 2
cde
3 3
def
3 4
efg
3 5
fgh
3 6
ghb
3 7
hba
2 0
ab
2 1
bc
2 2
cd
2 3
de
2 4
ef
2 5
fg
2 6
gh
2 7
hb
2 8
ba
1 0
a
1 1