【CF1027A】Palindromic Twist（模拟）

题意：输入T组字符串，每个字符串都必须改变一次，每个字母改变的规则是变成相邻的字母，字母a只能变b，z只能变y，判断改变后的字符依旧是否能够变成回文串

n<=1e2

思路：

 #include<cstdio>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<iostream>
 #include<algorithm>
 #include<map>
 #include<set>
 #include<queue>
 #include<vector>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef vector<int> VI;
 #define fi first
 #define se second
 #define MP make_pair
 #define N   410000
 #define MOD 1000000007
 #define eps 1e-8
 #define pi acos(-1)
 #define oo 110000000000000

 char c[N][],a[N];
 int b[N];

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 void swap(int &x,int &y)
 {
     int t=x;x=y;y=t;
 }

 int main()
 {
     //freopen("1.in","r",stdin);
     //freopen("1.out","w",stdout);
     int cas;
     scanf("%d",&cas);
     while(cas--)
     {
         int n;
         scanf("%d",&n);
         scanf("%s",a+);
         for(int i=;i<=n;i++)
         {
             if(a[i]=='a') {b[i]=; c[i][]='b'; continue;}
             if(a[i]=='z') {b[i]=; c[i][]='y'; continue;}
             b[i]=; c[i][]=a[i]+; c[i][]=a[i]-;
         }
     //    for(int i=1;i<=n;i++) printf("%d ",b[i]);

         int flag=;
         for(int i=;i<=n/;i++)
         {
             int t=;
             for(int j=;j<=b[i];j++)
              for(int k=;k<=b[n-i+];k++)
               if(c[i][j]==c[n-i+][k]) t=;
             if(t==) flag=;
         }
         if(flag) printf("YES\n");
          else printf("NO\n");
     }
     return ;
 }