题面：

Problem A. Super-palindrome

You are given a string that is consistedof lowercase English alphabet. You are supposed to change it into asuper-palindrome string in minimum steps. You can change one character instring to another letter per step.

A string is called asuper-palindrome string if all its substrings with an odd length are palindromestrings. That is, for a string s, if its substring s_i···j satisfies j −i + 1 is oddthen s_i+k = s_j−k fork= 0,1,··· ,j− i + 1.

Input

The first line contains an integer T (1 ≤ T ≤100) representing the number of test cases.

For each test case, the only line containsa string, which consists of only lowercase letters. It is guaranteed that thelength of string satisfies 1 ≤ |s| ≤ 100.

Output

For each test case, print one line with aninteger refers to the minimum steps to take.

Example

standard input		standard output
3 ncncn aaaaba aaaabb	0 1 2

Explanation

For second testcase aaaaba, just change letter bto a in one step.

    

    题目描述：给你一个字符串，问你需要更改里面多少个字符，使得原字符串变为一个超级回文串。超级回文串是指字串为奇数长度的都为回文串的串。

    题目分析：分析样例之后容易得出，对于这样一个超级回文串，只有两种情况可以满足。

    

一为串内所有的字符都是相同的。

    二是该串只有两种字符组成，且这两种字符是交替出现的。

     能够发现这个规律的话，这道题就可以进行模拟。因为数据范围很小（100而已），因此我们可以直接进行暴力的匹配。

    我们首先先将串中不相同的字符储存起来，枚举两种字符的种类，再跟原串进行匹配，最后去最小的更改量即可。

   

总结：最近做字符串算法做得有些多了，上来拿到题直接就往Manache上面去想了（还好队友很给力），对于这些题还得拓宽思维才行！！

    

#include 
    
     #define maxn 1005using namespace std;string str,tmp;bool vis[maxn];int main(){    int t;    cin>>t;    while(t--){        cin>>str;        memset(vis,0,sizeof(vis));        for(int i=0;i