CF1896H1.Cyclic Hamming (Easy Version)
普及/提高-
通过率:0%
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题目描述
This is the easy version of the problem. The only difference between the two versions is the constraint on k . You can make hacks only if all versions of the problem are solved.
In this statement, all strings are 0 -indexed.
For two strings a , b of the same length p , we define the following definitions:
- The hamming distance between a and b , denoted as h(a,b) , is defined as the number of positions i such that 0≤i<p and ai=bi .
- b is a cyclic shift of a if there exists some 0≤k<p such that b(i+k)modp=ai for all 0≤i<p . Here xmody denotes the remainder from dividing x by y .
You are given two binary strings s and t of length 2k+1 each. Both strings may contain missing characters (denoted by the character '?'). Your task is to count the number of ways to replace the missing characters in both strings with the characters '0' or '1' such that:
- Each string s and t contains exactly 2k occurrences of each character '0' and '1'
- h(s,c)≥2k for all strings c that is a cyclic shift of t .
As the result can be very large, you should print the value modulo 998244353 .
输入格式
The first line of the input contains a single integer k ( 1≤k≤7 ).
The second line of the input contains string s of size 2k+1 , consisting of the characters '0', '1' and '?'.
The third line of the input contains string t of size 2k+1 , consisting of the characters '0', '1' and '?'.
It is guaranteed that both strings s and t contains no more than 2k character '0' or '1'.
输出格式
Print a single integer — the answer to the problem modulo 998244353 .
输入输出样例
输入#1
1 0011 0101
输出#1
1
输入#2
1 0011 0110
输出#2
0
输入#3
1 0??1 01??
输出#3
2
输入#4
2 000????? 01010101
输出#4
3
输入#5
2 0??????? 1???????
输出#5
68
输入#6
5 0101010101010101010101010101010101010101010101010101010101010101 ????????????????????????????????????????????????????????????????
输出#6
935297567
说明/提示
In the first example, we can check that the condition h(s,c)≥2k for all cyclic shift c of t is satisfied. In particular:
- for c=0101 , h(s,c)=h(0110,0101)=2≥21 ;
- for c=1010 , h(s,c)=h(0110,1010)=2≥21 .
In the second example, there exists a cycle shift c of t such that h(s,c)<2k (in particular, c=0011 , and h(s,c)=h(0011,0011)=0 ).
In the third example, there are 2 possible ways to recover the missing characters:
- s=0101 , t=0110 ;
- s=0011 , t=0101 .
In the fourth example, there are 3 possible ways to recover the missing characters:
- s=00011110 , t=01010101 ;
- s=00011011 , t=01010101 ;
- s=00001111 , t=01010101 .