Note: this is a harder version of Mirrored string I.
The gorillas have recently discovered that the image on the surface of the water is actually a reflection of themselves. So, the next thing for them to discover is mirrored strings.
A mirrored string is a palindrome string that will not change if you view it on a mirror.
Examples of mirrored strings are "MOM", "IOI" or "HUH". Therefore, mirrored strings must contain only mirrored letters {A, H, I, M, O, T, U, V, W, X, Y} and be a palindrome.
e.g. IWWI, MHHM are mirrored strings, while IWIW, TFC are not.
A palindrome is a string that is read the same forwards and backwards.
Given a string S of length N, help the gorillas by printing the length of the longest mirrored substring that can be made from string S.
A substring is a (possibly empty) string of characters that is contained in another string S. e.g. "Hell" is a substring of "Hello".
The first line of input is T – the number of test cases.
Each test case contains a non-empty string S of maximum length 1000. The string contains only uppercase English letters.
For each test case, output on a line a single integer - the length of the longest mirrored substring that can be made from string S.
3
IOIKIOOI
ROQ
WOWMAN
4
1
3
题解：暴力呀。
1 #include <iostream> 2 #include <stdio.h> 3 #include <algorithm> 4 #include <string.h> 5 #include <math.h> 6 #define PI acos(-1.0) 7 using namespace std; 8 bool judge(char arr) 9 { 10 if(arr=='A'||arr=='H'||arr=='I'||arr=='M'||arr=='O'||arr=='T'||arr=='U'||arr=='V'||arr=='W'||arr=='X'||arr=='Y') 11 return false; 12 return true; 13 } 14 int main() 15 { 16 int i,j,n,k,m,kk; 17 char str[2001]; 18 scanf("%d",&n); 19 while(n--) 20 { 21 int ans=0,pp; 22 scanf(" %s",&str); 23 for(i=0;i<strlen(str);i++) 24 for(j=i;j<strlen(str);j++) 25 { 26 if(judge(str[j])) 27 break; 28 if(str[j]==str[i]) 29 { 30 m=j;int sum=0; 31 for(k=i;k<(i+j+1)/2;k++) 32 { 33 if(str[k]==str[m--]) 34 sum++; 35 } 36 if(sum==(j-i+1)/2) 37 ans=max(ans,j-i+1); 38 } 39 40 } 41 printf("%d ",ans); 42 43 } 44 return 0; 45 }