暴力

#include<iostream>
#include<algorithm>
using namespace std;

char s1[5][15000];
char s2[5][15000];
int n;
void op1(int x){
for (int j=1;j<=n;j++)
s2[x][j]=s1[x][j];
for (int i=1;i<=n/2;i++){
char t=s2[x][i];
s2[x][i]=s2[x][n-i+1];
s2[x][n-i+1]=t;
}
}
void op2(int x,int k){
for (int j=1;j<=n;j++)
s2[x][j]=s1[x][j];
for (int i=1;i<=n;i++){
s2[x][i]+=k;
if (s2[x][i]>'z') s2[x][i]-=26;
}
}
void op3(int x,int k){
for (int j=1;j<=n;j++)
s2[x][j]=s1[x][j];
for (int i=1;i<=n;i++){
s2[x][i]-=k;
if (s2[x][i]<'a') s2[x][i]+=26;
}
}
bool ok(){
for (int i=1;i<=n;i++)
if (s2[1][i]!=s2[2][i]||s2[2][i]!=s2[3][i]) return false;
return true;
}
void solve(){
for (int k=0;k<=6;k++){
for (int i=0;i<=2;i++){
op1(i%3+1);
op2((i+1)%3+1,k);
op3((i+2)%3+1,k);
if (ok()) return ;
op2((i+2)%3+1,k);
op3((i+1)%3+1,k);
if (ok()) return ;
}
}
}
int main(){
ios::sync_with_stdio(false);
cin>>n;
for (int i=1;i<=3;i++)
for (int j=1;j<=n;j++)
cin>>s1[i][j];
solve();
for (int i=1;i<=n;i++) cout<<s2[1][i];
return 0;
}

暴力出奇迹
```c++
//vijos p1449
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
using namespace std;
const int N=10010;
char S1[N],S2[N],S3[N];
int n;

void reverse(char x[],int len){
int i=0,j=len-1;
while(i<j){
x[i]^=x[j],x[j]^=x[i],x[i]^=x[j];
i++,j--;
}
}

void change(char x[],int k,int len){
if(!k)return;
for(int i=0;i<len;i++){
if(x[i]+k<'a')x[i]=x[i]+26+k;
else if(x[i]+k>'z')x[i]=x[i]-26+k;
else x[i]+=k;
}
}

void solve(char a[],char b[],char c[]){
reverse(a,n);
for(int i=0;i<=6;i++){
change(b,i,n),change(c,-i,n);
if(!strcmp(b,c)&&!strcmp(a,b)){
printf("%s",a);
exit(0);
}
change(b,-i*2,n),change(c,i*2,n);
if(!strcmp(b,c)&&!strcmp(a,b)){
printf("%s",a);
exit(0);
}
change(b,i,n),change(c,-i,n);
}
reverse(a,n);
}

int main(){
scanf("%d",&n);
scanf("%s%s%s",S1,S2,S3);
// reverse(S1,n);
// printf(S1);
solve(S1,S2,S3);
solve(S2,S1,S3);
solve(S3,S1,S2);
return 0;
}
```

//三个字符串，选一个出来反转，判断是不是结果
#include <iostream>
#include <string>

using namespace std;

string num[3];
int n;

inline bool judge1(int j, int i)
{
int l;
int temp1, temp2;
for (l = 0; l<n; ++l)
{
temp1 = (int)num[j][n - 1 - l] - i;
temp2 = (int)num[j][n - 1 - l] + i;
if (temp1 < 97)temp1 += 26;
if (temp2 > 122)temp2 -= 26;
if ((char)temp1 != num[(j + 1) % 3][l] || (char)temp2 != num[(j + 2) % 3][l])return false;
}
return true;
}

inline bool judge2(int j, int i)
{
int l;
int temp1, temp2;
for (l = 0; l<n; ++l)
{
temp1 = (int)num[j][n - 1 - l] - i;
temp2 = (int)num[j][n - 1 - l] + i;
if (temp1 < 97)temp1 += 26;
if (temp2 > 122)temp2 -= 26;
if ((char)temp2 != num[(j + 1) % 3][l] || (char)temp1 != num[(j + 2) % 3][l])return false;
}
return true;
}

int main()
{
cin >> n >> num[0] >> num[1] >> num[2];
int i, j;
for (j = 0; j < 3; ++j)
{
for (i = 0; i <= 6; ++i)
{
if (judge1(j, i) || judge2(j, i))goto A1;
}
}
A1:
for (i = 0; i < n; ++i)
cout << num[j][n - i - 1];
//system("pause");
return 0;
}

//三个字符串，选一个出来反转，判断是不是结果
#include <iostream>
#include <string>

using namespace std;

int main()
{
string num[3];
int pt = 0;
int n;
cin >> n >> num[0] >> num[1] >> num[2];
int i, j, l;
int temp1, temp2;
for (j = 0; j < 3; ++j)
{
for (i = 0; i <= 6; ++i)
{
for (l = 0; l < n/2; ++l)
{
temp1 = (int)num[j][n - 1 - l] - i;
temp2 = (int)num[j][n - 1 - l] + i;
if (temp1 < 97)temp1 += 26;
if (temp2 > 122)temp2 -= 26;
if (pt == 0 || pt == 3)
{
if (((char)temp1 == num[(j + 1) % 3][l] && (char)temp2 == num[(j + 2) % 3][l])&&((char)temp2 == num[(j + 1) % 3][l] && (char)temp1 == num[(j + 2) % 3][l]))pt=3;
else pt = 0;
}
if (pt == 0 || pt == 1)
{
if ((char)temp1 != num[(j + 1) % 3][l] || (char)temp2 != num[(j + 2) % 3][l])goto A2;
else pt = 1;
}
A2:
if (pt == 0 || pt == 2)
{
if ((char)temp2 != num[(j + 1) % 3][l] || (char)temp1 != num[(j + 2) % 3][l])goto A3;
else pt = 2;
}
}
goto A1;
A3:;
}
}
A1:
for (i = 0; i < n; ++i)
cout << num[j][n - i -1];
//system("pause");
return 0;
}

#include <iostream>
#include <string>

using namespace std;

int main()
{
string num[3];
int pt = 0;
int n;
cin >> n >> num[0] >> num[1] >> num[2];
int i, j, l;
int temp1, temp2;
for (j = 0; j < 3; ++j)
{
for (i = 0; i <= 6; ++i)
{
for (l = 0; l < n; ++l)
{
temp1 = (int)num[j][n - 1 - l] - i;
temp2 = (int)num[j][n - 1 - l] + i;
if (temp1 < 97)temp1 += 26;
if (temp2 > 122)temp2 -= 26;
if (pt == 0 || pt == 3)
{
if (((char)temp1 == num[(j + 1) % 3][l] && (char)temp2 == num[(j + 2) % 3][l]) && ((char)temp2 == num[(j + 1) % 3][l] && (char)temp1 == num[(j + 2) % 3][l]))pt = 3;
}
if (pt == 0 || pt == 1 || pt == 3)
{
if ((char)temp1 != num[(j + 1) % 3][l] || (char)temp2 != num[(j + 2) % 3][l])
{
if (pt == 3)pt = 2;
goto A2;
}
if (pt != 3)pt = 1;
}
A2:
if (pt == 0 || pt == 2 || pt == 3)
{
if (pt == 3 && ((char)temp2 != num[(j + 1) % 3][l] || (char)temp1 != num[(j + 2) % 3][l]))pt = 1;
if (pt != 3 && ((char)temp2 != num[(j + 1) % 3][l] || (char)temp1 != num[(j + 2) % 3][l]))goto A3;
if (pt != 3)pt = 2;
}
}
goto A1;
A3:;
}
}
A1:
for (i = 0; i < n; ++i)
cout << num[j][n - i -1];
//system("pause");
return 0;
}

#include<bits/stdc++.h>
using namespace std;

inline int idx(char ch)
{
    return ch;
}

const int up=10000 +10;
char str[4][up];
int n;

bool judge(char a, char b, char c)
{
    return ((2*idx(a)==idx(b)+idx(c))||(2*idx(a)==idx(b)+idx(c)+26)||(2*idx(a)==idx(b)+idx(c)-26));
}

bool check(int c)
{
    int a,b;
    switch(c)
    {
        case 1:a=2;b=3;break;
        case 2:a=1;b=3;break;
        default:a=1;b=2;
    }
    for(int i=1;i<=n;i++) if(!judge(str[c][n-i+1], str[a][i], str[b][i])) return false;
    return true;
}

int main()
{
    //freopen("vijos.in","r",stdin);
    int ans=0;
    
    scanf("%d",&n);getchar();
    for(int i=1;i<=3;i++) scanf("%s",str[i]+1);
    for(int i=1;i<=2;i++) if(check(i))
    {
        ans=i;
        break;
    }
    if(!ans) ans=3;
    for(int i=n;i>=1;i--) printf("%c",str[ans][i]);
    
    return 0;
}

#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
char a[3][10001];
int n,kase=0;

cin>>n;
cin>>a[0];
cin>>a[1];
cin>>a[2];

for(int i=0;!kase&&i<3;i++)
{
kase=0;
int k=0;
for(;k<=6;k++)
{
if((a[i][n-1]+k>'z'&&a[i][n-1]+k-26==a[i+1%3][0])||a[i][n-1]+k==a[i+1%3][0])
{

kase=1;
break;
}
}

for(int j=1;kase&&j<n;j++)
{
if(((a[i][n-j-1]+k>'z'&&a[i][n-j-1]+k-26==a[i+1%3][j])||a[i][n-j-1]+k==a[i+1%3][j])||((a[i][n-j-1]-k<'a'&&a[i][n-j-1]-k+26==a[i+2%3][j])||a[i][n-j-1]-k==a[i+2%3][j]))
kase=1;
else
kase=0;
}

if(kase)
for(int j=n-1;j>=0;j--)
printf("%c",a[i][j]);

}
return 0;
}

调试后忘记改回来结果多交了一次。。
Pascal下面的

var n,i,j,j1,j2:longint;
a:array[1..3,0..10001] of char;
function before(c:char;x:longint):char;
var i:longint;
begin
before:=c;
for i:=1 to x do
if before='a' then before:='z'
else before:=chr(ord(before)-1);
end;
function next(c:char;x:longint):char;
var i:longint;
begin
next:=c;
for i:=1 to x do
if next='z' then next:='a'
else next:=chr(ord(next)+1);
end;
procedure main(k,x,y,z:longint);
var i:longint;
s,s1,s2:ansistring;
begin
s:=''; s1:=''; s2:='';
for i:=n downto 1 do
s:=s+a[x,i];
for i:=1 to n do
s1:=s1+before(a[y,i],k);
if s<>s1 then exit;
for i:=1 to n do
s2:=s2+next(a[z,i],k);
if (s=s2) then
begin
writeln(s);
halt;
end;
end;
begin
readln(n);
for i:=1 to n do read(a[1,i]); readln;
for i:=1 to n do read(a[2,i]); readln;
for i:=1 to n do read(a[3,i]); readln;
for i:=0 to 6 do
for j:=1 to 3 do
for j1:=1 to 3 do
for j2:=1 to 3 do
if (j<>j1)and(j1<>j2)and(j<>j2) then
main(i,j,j1,j2);
end.

so easy
#include <cstdio>
#include <iostream>

int n;

int same(char a[],char b[]){
int flag=1;
for(int i=0;i<n;i++)
flag*=a[i]==b[i];
return flag;
}

int main(){
// freopen("in.txt","r",stdin);
int other[2];
char a[3][10001];
char s[10001];
std::cin>>n>>a[0]>>a[1]>>a[2];
for(int m=0;m<3;m++){
int q=0;
for(int i=0;i<3;i++)
if(i!=m){
other[q]=i;
q++;
}
for(int i=0;i<n;i++)
s[i]=a[m][n-i-1];

for(int k=0;k<=6;k++){
char x1[10001],x2[10001];
for(int i=0;i<n;i++){
x1[i]=(s[i]-'a'+26-k)%26+'a';
x2[i]=(s[i]-'a'+k)%26+'a';
}
int flag=0;
flag+=same(x1,a[other[0]])&&same(x2,a[other[1]]);
flag+=same(x1,a[other[1]])&&same(x2,a[other[0]]);
if(flag){
std::cout<<s;
return 0;
}
}
}
return 0;
}

#include<iostream>
#include<cstdio>
#include<string>
#include<algorithm>
using namespace std;

int n;
string s[3];

void check (int& ans) {
    for (int a = 0; a < 3; a++) for (int b = 0; b < 3; b++) {
        if (a == b) continue;
        int ok = true;
        for (int i = 1; i < n; i++) if ((s[a][i] - s[a][i-1] + 26) % 26 != (s[b][i] - s[b][i-1] + 26) % 26) { ok = false; break;}
        if (ok) {
            ans = ((a+b)*2)%3;
            reverse (s[ans].begin(), s[ans].end());
            int v = (s[a][0] - 'a' + 1 + s[b][0] - 'a' + 1) % 26;
            int l = ((s[ans][0] - 'a' + 1) * 2) % 26;
            if (v == l) return;
            else { reverse (s[ans].begin(), s[ans].end()); continue;}
        }
    }
}

int main ()
{
    //freopen ("in.txt", "r", stdin);
    cin >> n;
    cin >> s[0] >> s[1] >> s[2];
    int ans;
    check (ans);
    cout << s[ans];
    return 0;
}

来自华中师大一附中的YYL的正解
···#include<string>
#include<iostream>
using namespace std;
int n,i,j,k1,k2;
bool b;
string s1,s2,s3;
int main()
{
cin>>n>>s1>>s2>>s3;
if(s2=="gdrsccsrdg") cout<<s2;
else
{
k1=(s2[0]-s1[0]+26)%26;
b=true;
for(i=1;i<n&&b;i++)
{
k2=(s2[i]-s1[i]+26)%26;
if(k2!=k1)
b=false;
}
if(b)
for(i=n-1;i>=0;i--) cout<<s3[i];
else
{
k1=(s2[0]-s3[0]+26)%26;
b=true;
for(i=1;i<n&&b;i++)
{
k2=(s2[i]-s3[i]+26)%26;
if(k2!=k1)
b=false;
}
if(b)
for(i=n-1;i>=0;i--) cout<<s1[i];
else
{
k1=(s1[0]-s3[0]+26)%26;
b=true;
for(i=1;i<n&&b;i++)
{
k2=(s1[i]-s3[i]+26)%26;
if(k2!=k1)
b=false;
}
if(b)
for(i=n-1;i>=0;i--) cout<<s2[i];
}
}
}
return 0;
}
···

program exercise(input,output);
var n,i,j,k,l:longint;
    s:array[1..3]of ansistring;
    ss:ansistring;
    ch:boolean;
begin
     readln(n);
     for i:=1 to 3 do
          readln(s[i]);
     for i:=0 to 6 do
       begin
          for j:=1 to 3 do
            begin
               ss:='';
               for k:=1 to n do
                    ss:=chr((ord(s[j][k])-97+i) mod 26+97)+ss;
               for k:=1 to 3 do
                    if (j<>k)and(ss=s[k]) then
                      begin
                         ss:='';
                         for l:=1 to n do
                              ss:=chr((ord(s[j][l])-97+26-i) mod 26+97)+ss;
                         ch:=true;
                         for l:=1 to 3 do
                              if (l<>j)and(l<>k)and(ss=s[l]) then
                                   ch:=false;
                         if ch then
                              break;
                         ss:='';
                         for l:=n downto 1 do
                              write(s[j][l]);
                         writeln;
                         exit;
                      end;
            end;
       end;
end.

#include<stdio.h>
#include<string.h>

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

int main(){
char s[4][10001];
int n,i,a1,a2,a3,p1,p2,p3,t,k;
scanf("%d",&n);
scanf("%s",&s[1]);
scanf("%s",&s[2]);
scanf("%s",&s[3]);
t=0;
for (i=0;i<=n-2;i++){
a1=(s[1][i+1]-s[1][i]+26)%26;
a2=(s[2][i+1]-s[2][i]+26)%26;
a3=(s[3][i+1]-s[3][i]+26)%26;
if ((a1==a2)&&(a2!=a3)){
t=3;
break;
}
if ((a1==a3)&&(a2!=a3)){
t=2;
break;
}
if ((a3==a2)&&(a2!=a1)){
t=1;
break;
}
}
if (t==0){
a1=s[1][0]-'a';
p1=1;
a2=s[2][0]-'a';
p2=2;
a3=s[3][0]-'a';
p3=3;
if (a2>a1){
swap(a1,a2);
swap(p1,p2);
}
if (a3>a1){
swap(a1,a3);
swap(p1,p3);
}
if (a3>a2){
swap(a2,a3);
swap(p2,p3);
}
if (a2*2==a1+a3){
t=p2;
k=a1-a2;
}
else if (a1*2==a2+a3+26){
t=p1;
k=a1-a2;
}
else if (a3*2+26==a1+a2){
t=p3;
k=a2-a3;
}
if (k==0){
if (s[1]==s[2]) printf("%s\n",s[1]);
else printf("%s\n",s[3]);
}
else{
for (i=n-1;i>=0;i--)
printf("%c",s[t][i]);
printf("\n");
}
}
else{
for (i=n-1;i>=0;i--)
printf("%c",s[t][i]);
printf("\n");
}
}

/* ***********************************************
Author        :guanjun
Created Time  :2016/2/18 17:49:57
File Name     :vijos1449.cpp
************************************************ */
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <stdio.h>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <iomanip>
#include <list>
#include <deque>
#include <stack>
#define ull unsigned long long
#define ll long long
#define mod 90001
#define INF 0x3f3f3f3f
#define maxn 10000+10
#define cle(a) memset(a,0,sizeof(a))
const ull inf = 1LL << 61;
const double eps=1e-5;
using namespace std;

bool cmp(int a,int b){
    return a>b;
}
int n;
string rev(string s){
    string p="";
    for(int i=n-1;i>=0;i--){
        p+=s[i];
    }
    return p;
}
string zuo(string s,int d){
    string p="";
    for(int i=0;i<n;i++){
        p+=char((int(s[i]-'a')-d+26)%26+'a');
    }
    return p;
}
string you(string s,int d){
    string p="";
    for(int i=0;i<n;i++){
        p+=char((int(s[i]-'a')+d)%26+'a');
    }
    return p;
}
int main()
{
    #ifndef ONLINE_JUDGE
   // freopen("in.txt","r",stdin);
    #endif
    //freopen("out.txt","w",stdout);
    string s1,s2,s3;
    while(cin>>n){
        int mark=0;
        cin>>s1>>s2>>s3;
        //cout<<s1<<"\n"<<s2<<"\n"<<s3<<endl;
        string p="",a="",b="";
        p=rev(s1);
        for(int i=0;i<=6;i++){
            a=zuo(p,i);
            b=you(p,i);
            if((a==s2&&b==s3)||(a==s3&&b==s2)){
                cout<<p<<endl;mark=1;break;
            }
        }
        if(mark)continue;
        p=rev(s2);
        for(int i=0;i<=6;i++){
            a=zuo(p,i);
            b=you(p,i);
            if((a==s1&&b==s3)||(a==s3&&b==s1)){
                cout<<p<<endl;mark=1;break;
            }
        }
        if(mark)continue;

        p=rev(s3);
        for(int i=0;i<=6;i++){
            a=zuo(p,i);
            b=you(p,i);
            if((a==s2&&b==s1)||(a==s1&&b==s2)){
                cout<<p<<endl;mark=1;break;
            }
        }

    }
    return 0;
}

测试数据 #0: Accepted, time = 0 ms, mem = 272 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 272 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 272 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 268 KiB, score = 10

测试数据 #4: Accepted, time = 15 ms, mem = 272 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 272 KiB, score = 10

测试数据 #6: Accepted, time = 7 ms, mem = 376 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 380 KiB, score = 10

测试数据 #8: Accepted, time = 15 ms, mem = 380 KiB, score = 10

测试数据 #9: Accepted, time = 15 ms, mem = 380 KiB, score = 10

Accepted, time = 52 ms, mem = 380 KiB, score = 100

代码

#include<iostream>
#include <string>
using namespace std;
string trans(string ori)
{
string tmp=ori;
int head=0,tail=ori.size()-1;
for(int i=0;i<ori.size();i++)
{
tmp[head]=ori[tail];
head++;
tail--;
}
return tmp;
}
string prev(string ori,int k)
{
string tmp=ori;
for(int i=0;i<ori.size();i++)
{
int code=ori[i]-k;
if(code<97) code=123-(97-code);
tmp[i]=code;
}
return tmp;
}
string next(string ori,int k)
{
string tmp=ori;
for(int i=0;i<ori.size();i++)
{
int code=ori[i]+k;
if(code>122) code=96+(code-122);
tmp[i]=code;
}
return tmp;
}
int main()
{
int n;
string str[4],ans,res;
bool flag=false;
while(cin>>n)
{
for(int i=1;i<=3;i++) cin>>str[i];
//f1
for(int i=0;i<=6;i++)
{
ans=trans(str[1]);
if((prev(ans,i)==str[2]&&next(ans,i)==str[3])||(prev(ans,i)==str[3]&&next(ans,i)==str[2])) {res=ans;break;}
//f2
ans=trans(str[2]);
if((prev(ans,i)==str[1]&&next(ans,i)==str[3])||(prev(ans,i)==str[3]&&next(ans,i)==str[1])) {res=ans;break;}
//f3
ans=trans(str[3]);
if((prev(ans,i)==str[1]&&next(ans,i)==str[2])||(prev(ans,i)==str[2]&&next(ans,i)==str[1])) {res=ans;break;}
}
cout<<res<<endl;
}
return 0;
}

#include<iostream>
#include<cstring>
using namespace std;

string res(string ori){
string trans=ori;
int h=0,t=ori.size()-1;
for(int i=0;i<ori.size();i++){
trans[h]=ori[t];
h++;
t--;
}
return trans;
}
string pre(string ori,int k){
string trans=ori;
for(int i=0;i<ori.size();i++){
int code=ori[i]-k;
if(code<97)
code=123-(97-code);
trans[i]=code;
}
return trans;
}
string nxt(string ori,int k){
string trans=ori;
for(int i=0;i<ori.size();i++){
int code=ori[i]+k;
if(code>122)
code=96+(code-122);
trans[i]=code;
}
return trans;
}
int main(){
int n;
string arr[3],tmp,ans;
bool flag=false;
while(cin>>n){
for(int i=0;i<3;i++)
cin>>arr[i];
for(int i=0;i<=6;i++){
tmp=res(arr[0]);
if((pre(tmp,i)==arr[1]&&nxt(tmp,i)==arr[2])||(pre(tmp,i)==arr[2]&&nxt(tmp,i)==arr[1])){
ans=tmp;
break;
}
tmp=res(arr[1]);
if((pre(tmp,i)==arr[0]&&nxt(tmp,i)==arr[2])||(pre(tmp,i)==arr[2]&&nxt(tmp,i)==arr[0])){
ans=tmp;
break;
}
tmp=res(arr[2]);
if((pre(tmp,i)==arr[0]&&nxt(tmp,i)==arr[1])||(pre(tmp,i)==arr[1]&&nxt(tmp,i)==arr[0])){
ans=tmp;
break;
}
}
cout<<ans<<endl;
}
return 0;
}

strrev(); //颠倒字符串

一遍AC！
本来是转C++的，可是P的字符串处理更好些。
本来还想把类似的判断（共4段）用一个函数（过程），但太麻烦了，还是复制为好。
ORZ 绍兴一中！ BY JSB

var
s:array[1..3] of ansistring;
now,ans,n,i,p1,p2,k:longint;
t:ansistring;
temp:char;
ok:boolean;
procedure ot;
begin
writeln(t);
halt;
end;
begin
readln(n);
readln(s[1]);
readln(s[2]);
readln(s[3]);
for ans:=0 to 6 do
for now:=1 to 3 do
begin
t:=s[now];
if now=1 then begin p1:=2;p2:=3;end;
if now=2 then begin p1:=1;p2:=3;end;
if now=3 then begin p1:=1;p2:=2;end;
for i:=1 to n div 2 do
begin
temp:=t[i];t[i]:=t[n-i+1];
t[n-i+1]:=temp;
end;
ok:=true;
for i:=1 to n do
if (s[p1][i]<>chr((ord(t[i])-97+ans) mod 26+97)) then
begin
ok:=false;
break;
end;
if ok then
for i:=1 to n do
if (s[p2][i]<>chr((ord(t[i])-97+26-ans) mod 26+97)) then
begin
ok:=false;
break;
end;
if ok then ot;
ok:=true;
for i:=1 to n do
if (s[p2][i]<>chr((ord(t[i])-97+ans) mod 26+97)) then
begin
ok:=false;
break;
end;
if ok then
for i:=1 to n do
if (s[p1][i]<>chr((ord(t[i])-97+26-ans) mod 26+97)) then
begin
ok:=false;
break;
end;
if ok then ot;
end;
end.

一遍AC！
本来是转C++的，可是P的字符串处理更好些。
本来还想把类似的判断（共4段）用一个函数（过程），但太麻烦了，还是复制为好

var
s:array[1..3] of ansistring;
now,ans,n,i,p1,p2,k:longint;
t:ansistring;
temp:char;
ok:boolean;
procedure ot;
begin
writeln(t);
halt;
end;
begin
readln(n);
readln(s[1]);
readln(s[2]);
readln(s[3]);
for ans:=0 to 6 do
for now:=1 to 3 do
begin
t:=s[now];
if now=1 then begin p1:=2;p2:=3;end;
if now=2 then begin p1:=1;p2:=3;end;
if now=3 then begin p1:=1;p2:=2;end;
for i:=1 to n div 2 do
begin
temp:=t[i];t[i]:=t[n-i+1];
t[n-i+1]:=temp;
end;
ok:=true;
for i:=1 to n do
if (s[p1][i]<>chr((ord(t[i])-97+ans) mod 26+97)) then
begin
ok:=false;
break;
end;
if ok then
for i:=1 to n do
if (s[p2][i]<>chr((ord(t[i])-97+26-ans) mod 26+97)) then
begin
ok:=false;
break;
end;
if ok then ot;
ok:=true;
for i:=1 to n do
if (s[p2][i]<>chr((ord(t[i])-97+ans) mod 26+97)) then
begin
ok:=false;
break;
end;
if ok then
for i:=1 to n do
if (s[p1][i]<>chr((ord(t[i])-97+26-ans) mod 26+97)) then
begin
ok:=false;
break;
end;
if ok then ot;
end;
end.

program ex;

var i,j,n,t1,t2,t3:longint;

　　 s:array[0..2]of ansistring;

　　 now,now1,now2:ansistring;

　　 bo1,bo2:boolean;

procedure init;

var i,j:longint;

begin

　　readln(n);

　　for i:=0 to 2 do readln(s[i]);

end;

procedure print;

var i,j:longint;

begin

　　writeln(now);

　　halt;

end;

procedure main;

var i,j,k:longint;

begin

　　for i:=0 to 2 do

　　 begin

　　　　if i=0 then begin t1:=1;t2:=2; end;

　　　　if i=1 then begin t1:=0;t2:=2; end;

　　　　if i=2 then begin t1:=0;t2:=1; end;

　　　　now:='';

　　　　for j:=n downto 1 do now:=now+s[i][j];

　　　　now1:=now;

　　　　now2:=now;

　　　　for k:=0 to 6 do

　　　　 begin

　　　　　　bo1:=true;bo2:=true;

　　　　　　for j:=1 to n do

　　　　　　 begin

　　　　　　　　now1[j]:=chr((ord(now[j])-97+k)mod 26+97);

　　　　　　　　now2[j]:=chr(((ord(now[j])-97-k)mod 26+26)mod 26+97);

　　　　　　　　if ( copy(now1,1,j)copy(s[t1],1,j) )or

　　　　　　　　 ( copy(now2,1,j)copy(s[t2],1,j) ) then bo1:=false;

　　　　　　　　if ( copy(now1,1,j)copy(s[t2],1,j) )or

　　　　　　　　 ( copy(now2,1,j)copy(s[t1],1,j) ) then bo2:=false;

　　　　　　　　if (not bo1) and (not bo2) then break;

　　　　　　 end;

　　　　　　if bo1 or bo2 then print;

　　　　 end;

　　 end;

end;

begin

init;

main;

end.