23980번 - Common Anagrams 서브태스크다국어

Ayla has two strings A and B, each of length L, and each of which is made of uppercase English alphabet letters. She would like to know how many different substrings of A appear as anagrammatic substrings of B. More formally, she wants the number of different ordered tuples (i, j), with 0 ≤ i ≤ j < L, such that the i-th through j-th characters of A (inclusive) are the same multiset of characters as at least one contiguous substring of length (j - i + 1) in B.

The first line of the input gives the number of test cases, T. T test cases follow. Each test case starts with one line, containing L: the length of the string. The next two lines contain one string of L characters each: these are strings A and B, in that order.

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the answer Ayla wants, as described above.

• 1 ≤ T ≤ 100.
• 1 ≤ L ≤ 50.

• The two strings A and B will consist only of the characters A and B.

No additional constraints.

In Sample Case #1, L = 3, A = ABB, and B = BAB There are 6 substrings of A:

• A. The substring A in B is (trivially) an anagram.
• B. The substring B in B is (trivially) an anagram.
• B. The substring B in B is (trivially) an anagram.
• AB. The substring AB in B is (trivially) an anagram.
• BB. There is no corresponding anagrammatic substring in B.
• ABB. The substring BAB in B is an anagram.

In total, there are 5 substrings with a corresponding anagrammatic substring in B, so the answer is 5.

In Sample Case #2, note that it is the same as Sample Case #1, except that A and B are swapped. This changes the answer to 6!

In Sample Case #3, note that the substring CAT in A has the corresponding substring TAC in B which is an anagram. This still counts, even though the strings are at different indices in their respective strings.

In Sample Case #4, note that although the substring SUB in A has several corresponding substrings in B which are anagrams, it only counts once.

In Sample Case #5, note that every substring of A has a corresponding anagrammatic substring in B, so the answer is 10.