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=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
return x*f;
}
int n,t,a[maxn],b[maxn],f[50][50];
char s[10];
int main()
{
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
n=read();
for1(i,26)
for1(j,26)
f[i][j]=j>=i?j-i:26-i+j;
for1(i,n)
{
scanf("%s",s);t=0;
for0(j,3)t=t*26+f[s[j]-'a'+1][s[j+1]-'a'+1];
a[t]=i;b[t]=s[0];
}
for1(i,n)
{
scanf("%s",s);t=0;
for0(j,3)t=t*26+f[s[j]-'a'+1][s[j+1]-'a'+1];
printf("%d %d\n",a[t],f[s[0]-'a'+1][b[t]-'a'+1]);
}
return 0;
}

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