题解 | #公共子串计算#
a = input()
b = input()
s = []
if len(a) > len(b):
a, b = b, a # a存短的字符串
n = len(a)
for i in range(0, n):
for j in range(0,i + 1):
if a[j:n- i + j] in b:
s.append(len(a[j:n - i + j]))
s.append(0)
print(max(s))
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