python, 遍历一遍即可， 与最长公共子串的题一样，只不过输出的是长度而不是子串本身
公共子串计算
http://www.nowcoder.com/questionTerminal/98dc82c094e043ccb7e0570e5342dd1b
while True:
try:
a = input()
b = input()
if len(a) > len(b):
a,b = b,a
max_length = 0
i = 0
while i + max_length < len(a):
while i + max_length < len(a) and a[i:i + max_length + 1] in b:
max_length += 1
i += 1
print(max_length)
except:
break
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