A. Palindromic Twist(문자열 변경 + 메모의 판단)

제목 링크 A. Palindromic Twist time limit per test 2 seconds memory limit per test 256 megabytes inputstandard input output output You are given a string s consisting of n lowercase Latin letter.n is even.
For each position i ( 1 ≤ i ≤ n ) in string s you are required to change the letter on this position either to the previous letter in alphabetic order or to the next one (letters ‘a’ and ‘z’ have only one of these options). Letter in every position must be changed exactly once.
For example, letter ‘p’ should be changed either to ‘o’ or to ‘q’, letter ‘a’ should be changed to ‘b’ and letter ‘z’ should be changed to ‘y’.
That way string “codeforces”, for example, can be changed to “dpedepqbft” (‘c’ → ‘d’, ‘o’ → ‘p’, ‘d’ → ‘e’, ‘e’ → ‘d’, ‘f’ → ‘e’, ‘o’ → ‘p’, ‘r’ → ‘q’, ‘c’ → ‘b’, ‘e’ → ‘f’, ‘s’ → ‘t’).
String s is called a palindrome if it reads the same from left to right and from right to left. For example, strings “abba” and “zz” are palindromes and strings “abca” and “zy” are not.
Your goal is to check if it’s possible to make string s a palindrome by applying the aforementioned changes to every position. Print “YES” if string s can be transformed to a palindrome and “NO” otherwise.
Each testcase contains several strings, for each of them you are required to solve the problem separately.
Input The first line contains a single integer T ( 1 ≤ T ≤ 50 ) — the number of strings in a testcase.
Then 2 T lines follow — lines ( 2 i − 1 ) and 2 i of them describe the i -th string. The first line of the pair contains a single integer n ( 2 ≤ n ≤ 100 , n is even) — the length of the corresponding string. The second line of the pair contains a string s , consisting of n lowercase Latin letters.
Output Print T lines. The i -th line should contain the answer to the i -th string of the input. Print “YES” if it’s possible to make the i -th string a palindrome by applying the aforementioned changes to every position. Print “NO” otherwise.
Example inputCopy 5 6 abccba 2 cf 4 adfa 8 abaazaba 2 ml outputCopy YES NO YES NO NO Note The first string of the example can be changed to “bcbbcb”, two leftmost letters and two rightmost letters got changed to the next letters, two middle letters got changed to the previous letters.
The second string can be changed to “be”, “bg”, “de”, “dg”, but none of these resulting strings are palindromes.
The third string can be changed to “beeb” which is a palindrome.
The fifth string can be changed to “lk”, “lm”, “nk”, “nm”, but none of these resulting strings are palindromes. Also note that no letter can remain the same, so you can’t obtain strings “ll” or “mm”. 제목: 제목은 문자열의 문자를 조작하여 앞의 문자가 되고 뒤의 문자가 된다는 뜻이다. 변화가 없으면 안 된다. a와z처럼 두 가지 변화만 있을 뿐이다.그리고 변화된 문자열이 회문열의 사고방식인지 살펴보자. 두 가지 변수를 정의한다. 하나는 앞에서 뛰고, 하나는 뒤에서 뛰고, 중간으로 뛰고, 하나는 모든 문자를 들어 올리는 상황을 보고 회문의 요구를 만족시키는지 살펴보자.

#include
#include
#include
char s[101];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        scanf("%s",s);
        int f=0;
        for(int i=0,j=n-1;i<=n/2-1&&j>=n/2;i++,j--)
        {
            if((s[i]+1==s[j]+1)||(s[i]+1==s[j]-1)
               ||(s[i]-1==s[j]+1)||(s[i]-1==s[j]-1))//   4   
            {
                f=1;
            }
            else
            {
                f=0;
                break;
            }
        }
        if(f==0)printf("NO
");
        else printf("YES
");
    }
    return 0;
}