Flippy Sequence （思维题）
题目链接：https://cn.vjudge.net/contest/268424#problem/H
参考来源：https://blog.csdn.net/chen_zan_yu_/article/details/83904832
https://blog.csdn.net/BePosit/article/details/83927964
题目描述：
DreamGrid has just found two binary sequences and ( for all ) from his virtual machine! He would like to perform the operation described below exactly twice, so that holds for all after the two operations.
The operation is: Select two integers and (), change to for all .
DreamGrid would like to know the number of ways to do so.
We use the following rules to determine whether two ways are different:
- Let , where , be a valid operation pair denoting that DreamGrid selects integers and for the first operation and integers and for the second operation;
- Let , where , be another valid operation pair denoting that DreamGrid selects integers and for the first operation and integers and for the second operation.
- and are considered different, if there exists an integer () such that .
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 an integer (), indicating the length of two binary sequences.
The second line contains a string () of length , indicating the first binary sequence.
The third line contains a string () of length , indicating the second binary sequence.
It's guaranteed that the sum of in all test cases will not exceed .
Output
For each test case, output an integer denoting the answer.
Sample Input
3
1
1
0
2
00
11
5
01010
00111
Sample Output
0
2
6
Hint
For the second sample test case, there are two valid operation pairs: (1, 1, 2, 2) and (2, 2, 1, 1).
For the third sample test case, there are six valid operation pairs: (2, 3, 5, 5), (5, 5, 2, 3), (2, 5, 4, 4), (4, 4, 2, 5), (2, 4, 4, 5) and (4, 5, 2, 4).
题目大意：
给定两个0,1串，有两次操作机会，第一次操作第一个串，第二次操作第二个串。
选定一个连续区间把这里面的数字翻转，问有多少种方案使得两串相同。
思路：
只要分4种情况即可
1，最初全都相同，那么两次操作必须选择相同位置进行操作才符合要求
2.有一段连续区间不同的，这时两次操作可以把这一段区间让这两次操作分配一下，但是一定要保证
它们两个不重叠且恰好覆盖这段不同的区间，一种分配情况为1——>到区间长度-1因为不能交叉，
它们只能选择分配不同的区间那么情况数就是 ：区间长度-1，还可以它们其中一个全部选择这些不同的
搭配上原来相同的的字符，每一次这种搭配只能是向（不同区间）的一侧，情况又是两端相同字符长度
然后这些情况它们俩都可以交换操作所以情况数*2得出：
sum=2*(两侧相同的个数：(x1+n-1-y2)+不同区间的长度：(y2-x1));
或直接为sum=2*(n-1）
3.有两段连续区间不同，那么他们可以有三种选择，然后两个人不同加倍则为6种选择
4.大于两段连续不同的区间，这样他们没有可选择的余地了。
代码：
#include
#include
#include
using namespace std;
const long long int maxn=1000010;
int main()
{
int T,n;
char a[maxn],b[maxn];
while(scanf("%d",&T)!=EOF)
{
for(int k=0;k1)||(cnt 敲这道题时就在想，写完一定要好好吐槽一下，我天呀，一下午呀，都在弄这道题。看懂了题之后就在那想思路，边敲边完善，不断地调整自己的代码，最后终于对题目有了清晰地理解，却难在了代码上，参考了了大佬的代码，才觉着自己的代码如此麻烦，过程中，自己一直在思考的是，先看相同段与不同段一共有多少段，只有不同段为1和2时才有可能得到解，多于2一定是不对的。然后自己就在尝试如何判断得到的是1个不同段还是两个不同段，结果总是不对，但是上方的那段正确的代码中却巧妙地解决了这个问题。