凯撒密码 CH Round #57 - Story of the OI Class

题目：http://ch.ezoj.tk/contest/CH%20Round%20%2357%20-%20Story%20of%20the%20OI%20Class/凯撒密码

题解：刚开始想map，结果被出题说的卡map提醒了。

然后直觉告诉我可以hash相邻字母的距离，然后就这样做了。。。

代码：

 #include<cstdio>
 #include<cstdlib>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #include<iostream>
 #include<vector>
 #include<map>
 #include<set>
 #include<queue>
 #include<string>
 #define inf 1000000000
 #define maxn 1000000
 #define maxm 500+100
 #define eps 1e-10
 #define ll long long
 #define pa pair<int,int>
 #define for0(i,n) for(int i=0;i<=(n);i++)
 #define for1(i,n) for(int i=1;i<=(n);i++)
 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
 #define mod 1000000007
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}
     return x*f;
 }
 int n,t,a[maxn],b[maxn],f[][];
 char s[];
 int main()
 {
     freopen("input.txt","r",stdin);
     freopen("output.txt","w",stdout);
     n=read();
     for1(i,)
      for1(j,)
       f[i][j]=j>=i?j-i:-i+j;
     for1(i,n)
     {
         scanf("%s",s);t=;
         for0(j,)t=t*+f[s[j]-'a'+][s[j+]-'a'+];
         a[t]=i;b[t]=s[];
     }
     for1(i,n)
     {
         scanf("%s",s);t=;
         for0(j,)t=t*+f[s[j]-'a'+][s[j+]-'a'+];
         printf("%d %d\n",a[t],f[s[]-'a'+][b[t]-'a'+]);
     }
     return ;
 }