Gym

题意：给你四串字符，问你当1转化为3时，2能不能同时转化为4，交换规则是，i与j交换则n-i+1与n-j+1交换

思路：把握一条就可以了，不管怎么换，一个字符i对应的那个字符n-i+1是不变的，

ac代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespacestd;
char s1[105],s2[105],t1[105],t2[105];
int vis[105];
int main(){
    freopen("enchanted.in","r",stdin); freopen("enchanted.out","w",stdout);
    while(gets(s1))
    {
        memset(vis,0,sizeof(vis));
        gets(s2);
        gets(t1);
        gets(t2);
        int len=strlen(s1);
        int k=0;
        for(int i=1;i<=len;i++)
        {
            for(int j=1;j<=len;j++)
            {
                if(s1[j-1]==t1[i-1]&&s2[len-j]==t2[len-i]&&vis[j]==0)
                {
                    vis[j]=1;
                    k++;
                    break;
                }
            }
        }
        if(k==len)
        cout<<"Yes"<<endl;
        else
        cout<<"No"<<endl;
    }
    return 0;
}