【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
2
abcbcdcbd
abcdcbcbd
abc
abc
Sample Output
3
3
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
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#include