【2019浙江省赛 - K 】Strings in the Pocket（马拉车，思维）

【2019浙江省赛 - K 】Strings in the Pocket（马拉车，思维）

题干：

BaoBao has just found two strings  and  in his left pocket, where  indicates the -th character in string , and  indicates the -th character in string .

As BaoBao is bored, he decides to select a substring of  and reverse it. Formally speaking, he can select two integers  and  such that  and change the string to .

In how many ways can BaoBao change  to  using the above operation exactly once? Let  be an operation which reverses the substring , and  be an operation which reverses the substring . These two operations are considered different, if  or .

Input

There are multiple test cases. The first line of the input contains an integer , indicating the number of test cases. For each test case:

The first line contains a string  (), while the second line contains another string  (). Both strings are composed of lower-cased English letters.

It's guaranteed that the sum of  of all test cases will not exceed .

Output

For each test case output one line containing one integer, indicating the answer.

Sample Input

2abcbcdcbdabcdcbcbdabcabc

Sample Output

33

Hint

For the first sample test case, BaoBao can do one of the following three operations: (2, 8), (3, 7) or (4, 6).

For the second sample test case, BaoBao can do one of the following three operations: (1, 1), (2, 2) or (3, 3).

题目大意：

给你两个串s1和s2,可以翻转s1串的一个区间，只能翻转一次，,问有多少对l,r使得翻转后的s1串等于s2串

解题报告：

当两串完全相同的时候就是马拉车，不同的时候就先两边往里扫到两串第一个不相同的位置，在第一个串往外扩，看能扩几次，答案就是多少。

注意无数细节，，，那个初始化（见注释）还有那个while(1)里面必须是if(l>=ed)就可以break了一开始写成了l>ed，，，WA了一上午难受难受。。。

其实不是细节多，，而是代码姿势不太对，，，还是太菜，，不过这么简单的一题比赛的时候没有开有点小亏。做网络同步赛打了三小时比赛水了7题，就去准备晚上给学弟讲课的PPT去了。。。这成绩对比了下，好像7题放现场赛也只是个铜（不过罚时较少可能是银？）。。不过这题是真不难。。

AC代码：

#include#include#include#include#include

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