仔细读题，想清楚题意再做。

Pascal Code

var
a,b,c:string;
p:array['A'..'Z'] of char; //下标为密文，值为原文
hash:array['A'..'Z'] of boolean; //下标为原文，值表示这一原文有没有使用过
i:longint;
j:char;
procedure fail;
begin
writeln('Failed');
halt;
end;
begin
readln(a);
readln(b);
readln(c);
fillchar(p,sizeof(p),' ');
fillchar(hash,sizeof(hash),0);
if length(a)<>length(b) then fail;
for i:=1 to length(a) do
begin
if (p[a[i]]<>b[i]) and (p[a[i]]<>' ') then fail;
p[a[i]]:=b[i];
end;
for j:='A' to 'Z' do
begin
if (p[j]=' ') or hash[p[j]] then fail;
hash[p[j]]:=true;
end;
for i:=1 to length(c) do
write(p[c[i]]);
end.

#include<iostream>
#include<string.h>
using namespace std;
int main()
{
string code, text, ncode, ntext;
bool check = true, chr[26];
char match[26];

cin >> code >> text >> ncode;
memset(chr,false,sizeof(chr));

int len = code.length();
int nlen = ncode.length();

int i, j;
//情况2

if(check)
{
for(i = 0; i < len-1; i++)
for(j = i+1; j < len; j++)
{
if((text[i] != text[j]) && (code[i] == code[j]))
check = false;
}

}

//情况3

if(check)
{
for(i = 0; i < len; i++)
chr[(int)text[i]-65] = true;
for(i = 0; i < 26; i++)
if(chr[i] == false) check = false;
}

if(check == false)
{
cout << "Failed" << endl;
}
else
{
for(i = 0; i < len; i++) //建立已知 密文 与 明文对应关系
match[(int)code[i]-65] = text[i];
for(i = 0; i < nlen; i++)
cout << match[(int)ncode[i]-65];
cout << endl;
}

return 0;
}

测试数据 #0: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 512 KiB, score = 10
测试数据 #4: Accepted, time = 15 ms, mem = 504 KiB, score = 10
测试数据 #5: Accepted, time = 15 ms, mem = 508 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #7: Accepted, time = 15 ms, mem = 508 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 508 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 508 KiB, score = 10
Accepted, time = 60 ms, mem = 512 KiB, score = 100
代码
#include <cstdio>
#include <map>
#include <string>
#include <iostream>
using namespace std;
#define M 10000

int ssize;
string origin,sec,trans;
map<char,char> dict;

int main() {
cin>>sec>>origin>>trans;
ssize = sec.size();
if(sec.size()<26) {
printf("Failed\n");
return 0;
}
if(sec.size()==26) {
for(int i = 0; i < ssize; i++) {
for(int j = 0; j < ssize&&i!=j; j++) {
if(sec[i]==sec[j]||origin[i]==origin[j]) {
printf("Failed\n");
return 0;
}
}
}
}
dict[sec[0]] = origin[0];
for(int i = 1; i < ssize; i++) {
if(dict[sec[i]]!=NULL && dict[sec[i]]!=origin[i]) {
printf("Failed\n");
return 0;
}
dict[sec[i]] = origin[i];
}
for(int i = 0; i < trans.size(); i++) {
cout<<dict[trans[i]];
}
cout<<endl;
return 0;
}

评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 848 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 848 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 852 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 848 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 848 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 848 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 848 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 856 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 852 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 848 KiB, score = 10
Accepted, time = 0 ms, mem = 856 KiB, score = 100
代码
var s1,s2,s3:ansistring;
s:array[1..26] of char;
f:array[1..26] of boolean;
i,len,j:longint;
begin
fillchar(s,sizeof(s),'0'); fillchar(f,sizeof(f),false);
readln(s1);
readln(s2);
readln(s3);
len:=length(s1);
for i:=1 to len do begin
f[ord(s1[i])-ord('A')+1]:=true;
if s[ord(s1[i])-ord('A')+1]<>'0' then begin
if s2[i]<>s[ord(s1[i])-ord('A')+1] then begin
writeln('Failed');
halt;
end;
end else s[ord(s1[i])-ord('A')+1]:=s2[i];
end;
for i:=1 to 26 do
if not f[i] then begin
writeln('Failed');
halt;
end;
for i:=1 to 26 do
for j:=i+1 to 26 do
if s[i]=s[j] then begin
writeln('Failed');
halt;
end;
for i:=1 to length(s3) do
write(s[ord(s3[i])-ord('A')+1]);
end.

测试数据 #0: Accepted, time = 0 ms, mem = 532 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 536 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 532 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 532 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 540 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 536 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 532 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 532 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 536 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 532 KiB, score = 10
Accepted, time = 0 ms, mem = 540 KiB, score = 100
###_**C语言精简代码**_
****#include <stdio.h>****
****#include <string.h>****

int main(){
int i,s=0,pan=1;
char s1[101],s2[101],s3[101],a[26]={0},b[26]={0};
scanf("%s %s %s",s1,s2,s3);
for (i=0;i<strlen(s1);i++) {
if (a[s1[i]-'A']==0&&!b[s2[i]-'A'])
{a[s1[i]-'A']=s2[i];s++;b[s2[i]-'A']=1;}
if (a[s1[i]-'A']!=0&&a[s1[i]-'A']!=s2[i]||a[s1[i]-'A']==0&&b[s2[i]-'A'])
pan=0; }
for (i=0;i<strlen(s3);i++)
if (s<26) pan=0;
if (pan==1) {
for (i=0;i<strlen(s3);i++)
printf("%c",a[s3[i]-'A']); }
else printf("Failed");
return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
bool pan=true;
char s1[102],s2[102],key[27],s3[102];
int book[27]={0};
int main()
{
scanf("%s%s",s1,s2);
for(int i=0;i<(int)strlen(s1);i++){key[s1[i]-'A']=s2[i];}
for(int i=0;i<(int)strlen(s2);i++){book[s2[i]-'A']=1;}
for(int i=0;i<26;i++){if(book[i]==0){pan=false;break;}}
scanf("%s",s3);
for(int i=0;i<(int)strlen(s1);i++)
for(int j=0;j<(int)strlen(s1);j++)
if(s1[i]==s1[j]&&s2[i]!=s2[j]){pan=false;break;}
if(!pan)cout<<"Failed"<<endl;

if(pan==true)
{
for(int i=0;i<(int)strlen(s3);i++)
{
cout<<key[s3[i]-'A'];
}
}
//while(1);
return 0;
}

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
bool pan=true;
char s1[102],s2[102],key[27],s3[102];
int book[27]={0};
int main()
{
scanf("%s %s",&s1,&s2);
for(int i=0;i<strlen(s1);i++){key[s1[i]-'A']=s2[i];}
for(int i=0;i<strlen(s2);i++){book[s2[i]-'A']=1;}
for(int i=0;i<26;i++){if(book[i]==0){pan=false;break;}}
scanf("%s",&s3);
for(int i=0;i<strlen(s1);i++)
for(int j=0;j<strlen(s1);j++)
if(s1[i]==s1[j]&&s2[i]!=s2[j]){pan=false;break;}
if(!pan)cout<<"Failed"<<endl;

int len=strlen(s3);
if(pan==true)
{
for(int i=0;i<strlen(s3);i++)
{
cout<<key[s3[i]-'A'];
}
}
return 0;
}//就是瞎搞 注意题设的大多数条件就可以了 逐一判断

测试数据 #0: Accepted, time = 0 ms, mem = 532 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 536 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 532 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 532 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 540 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 536 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 532 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 532 KiB, score = 10

测试数据 #8: Accepted, time = 0 ms, mem = 536 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 532 KiB, score = 10

Accepted, time = 0 ms, mem = 540 KiB, score = 100

C语言代码

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
main()
{
bool pan=true;
int b[26]={0},i,s=0;
char s1[101],s2[101],a[26]={0};
scanf("%s %s",s1,s2);
for (i=0;i<strlen(s1);i++)
{
if (a[(s1[i])-'A']==0&&!b[(s2[i])-'A'])
a[s1[i]-'A']=s2[i],s++,b[s2[i]-'A']=1;
if (a[(s1[i])-'A']!=0&&a[(s1[i])-'A']!=s2[i]||a[(s1[i])-'A']==0&&b[(s2[i])-'A'])
pan=false;
}
scanf("%s",s1);
for (i=0;i<strlen(s1);i++)
if (!a[(s1[i])-'A']||s<26)
pan=false;
if (pan)
{
for (i=0;i<strlen(s1);i++)
printf("%c",a[(s1[i])-'A']);
printf("\n");
}
else puts("Failed");
return 0;
}

#include<cstring>
#include<iostream>
using namespace std;
main()
{
bool b[26]={0};
char a[26]={0};
int i,s=0;
string s1,s2;
cin>>s1>>s2;
for(i=0;i<s1.size();i++)
{
if(a[int(s1[i])-65]==0&&!b[int(s2[i])-65])
{
a[int(s1[i])-65]=s2[i];
s++;
b[int(s2[i])-65]=1;
}
if(a[int(s1[i])-65]!=0&&a[int(s1[i])-65]!=s2[i]||a[int(s1[i])-65]==0&&b[int(s2[i])-65])
{
cout<<"Failed";
return 0;
}
}
cin>>s1;
for(i=0;i<s1.size();i++)
if(!a[int(s1[i])-65]||s<26)
{
cout<<"Failed";
return 0;
}
for(i=0;i<s1.size();i++)
cout<<a[int(s1[i])-65];
}

var
a:array['A'..'Z']of char;
s1,s2,s3:ansistring;
i:longint;c,d:char;
begin
fillchar(a,sizeof(a),'0');
readln(s2);readln(s1);
for i:=1 to length(s1)do
if (a[s2[i]]<>'0')and(a[s2[i]]<>s1[i])then begin writeln('Failed');halt;end
else a[s2[i]]:=s1[i];
readln(s3);
for c:='A'to 'Y'do for d:=chr(ord(c)+1)to 'Z'do if a[c]=a[d] then
begin writeln('Failed');halt;end;
for c:='A' to 'Z' do if a[c]='0'then begin writeln('Failed');halt;end;
for i:=1 to length(S3)do write(a[s3[i]]);
end.
15ms水过

数据略坑。。。题意略吊。。。
var a:array['A'..'Z']of char;
i,j:longint; ch:char;
sa,sb,s,d,fax:ansistring;
procedure kong(st1:ansistring; var st2:ansistring);
var i:longint;
begin
st2:='';
for i:=1 to length(st1) do
if st1[i]<>' ' then st2:=st2+st1[i];
end;

procedure failed;
begin
writeln('Failed');
halt;
end;
begin
readln(fax); kong(fax,sa);
readln(fax); kong(fax,sb);
readln(fax); kong(fax,s);
fillchar(a,sizeof(a),' ');
for i:=1 to length(sb) do
for j:=1 to length(sb) do
if (sb[i]=sb[j])and(sa[i]<>sa[j]) then failed;
for i:=1 to length(sa) do
if (a[sa[i]]=' ') then a[sa[i]]:=sb[i]
else if a[sa[i]]<>sb[i] then failed;
for i:=1 to length(s) do
if a[s[i]]<>' ' then d:=d+a[s[i]] else failed;
for ch:='A' to 'Z' do
if a[ch]=' ' then failed;
writeln(d);

end.

编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 280 KiB, score = 10

测试数据 #1: Accepted, time = 15 ms, mem = 276 KiB, score = 10

测试数据 #2: Accepted, time = 15 ms, mem = 280 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 276 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 276 KiB, score = 10

测试数据 #5: Accepted, time = 15 ms, mem = 280 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 280 KiB, score = 10

测试数据 #7: Accepted, time = 31 ms, mem = 280 KiB, score = 10

测试数据 #8: Accepted, time = 0 ms, mem = 280 KiB, score = 10

测试数据 #9: Accepted, time = 15 ms, mem = 280 KiB, score = 10

Accepted, time = 91 ms, mem = 280 KiB, score = 100

代码

#include<iostream>
#include<cstring>
using namespace std;
string A,B,C;
char d[26]={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
int a[101],b[101],c[101],n,hash1[27],hash2[27];
int main()
{
cin>>A>>B>>C;
char A1[101],B1[101];
n=A.size();
if(n<26) {cout<<"Failed";return 0;}//如果输入文件字母数不到26个，就结束
for(int i=1;i<=n;i++)
{
A1[i]=A[i-1];
B1[i]=B[i-1];
}
for(int i=1;i<=n;i++)//A的值为65
{
a[i]=int(A1[i])-64;
hash1[a[i]]=1;
b[i]=int(B1[i])-64;
hash2[b[i]]=1;
}
//如果没有都出现，就结束（题中说是加密文件，但如果加密文件中都出现了，原文件中没出现，也是不可能的）
for(int i=1;i<=26;i++)
if(hash2[i]==0) {cout<<"Failed";return 0;}
for(int i=1;i<=26;i++)
if(hash2[i]==0) {cout<<"Failed";return 0;}

for(int i=1;i<=n;i++)
{
if(c[a[i]]!=0&&c[a[i]]!=b[i]) {cout<<"Failed";return 0;}//如果有不符合编码规则的，就结束
c[a[i]]=b[i];
}
//如果都不是，就进行把新电报解密
int m=C.size();
for(int i=1;i<=m;i++)
{
int qwe;
char qwe1;
qwe1=C[i-1];
qwe=int(qwe1)-64;
cout<<d[c[qwe]-1];
}
return 0;
}

测试数据 #0: Accepted, time = 0 ms, mem = 500 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 504 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 496 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 504 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 500 KiB, score = 10

测试数据 #5: Accepted, time = 15 ms, mem = 504 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 496 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 500 KiB, score = 10

测试数据 #8: Accepted, time = 0 ms, mem = 500 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 500 KiB, score = 10

Accepted, time = 15 ms, mem = 504 KiB, score = 100

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <climits>
#include <string>
using namespace std;
int i,j;
bool flag[50],flagans=true,used[50];
char trans[50],ans[1001],pd[50];
string a,b,ques;
#define trans(x) trans[x-(int)'A'+1]
#define pd(x) pd[x-(int)'A'+1]
#define flag(x) flag[x-(int)'A'+1]
#define used(x) used[x-(int)'A'+1]
int main()
{
memset(flag,false,sizeof(flag));
memset(used,false,sizeof(used));
cin>>a>>b>>ques;
for(i=0;i<=a.length()-1;i++){
if(((flag((int)a[i]))&&(b[i]!=trans((int)a[i])))||((used((int)b[i]))&&(a[i]!=pd((int)b[i])))){
flagans=false;
break;
}
trans((int)a[i])=b[i];
flag((int)a[i])=true;
pd((int)b[i])=a[i];
used((int)b[i])=true;
}
for(i=1;i<=26;i++) if(!flag[i]) flagans=false;
if(flagans){
for(i=0;i<=ques.length()-1;i++) ans[i]=trans((int)ques[i]);
cout<<ans;
}
else cout<<"Failed";
system("pause");
return 0;
}

测试数据 #0: Accepted, time = 0 ms, mem = 500 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 504 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 496 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 504 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 500 KiB, score = 10

测试数据 #5: Accepted, time = 15 ms, mem = 504 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 496 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 500 KiB, score = 10

测试数据 #8: Accepted, time = 0 ms, mem = 500 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 500 KiB, score = 10

Accepted, time = 15 ms, mem = 504 KiB, score = 100

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <climits>
#include <string>
using namespace std;
int i,j;
bool flag[50],flagans=true,used[50];
char trans[50],ans[1001],pd[50];
string a,b,ques;
#define trans(x) trans[x-(int)'A'+1]
#define pd(x) pd[x-(int)'A'+1]
#define flag(x) flag[x-(int)'A'+1]
#define used(x) used[x-(int)'A'+1]
int main()
{
memset(flag,false,sizeof(flag));
memset(used,false,sizeof(used));
cin>>a>>b>>ques;
for(i=0;i<=a.length()-1;i++){
if(((flag((int)a[i]))&&(b[i]!=trans((int)a[i])))||((used((int)b[i]))&&(a[i]!=pd((int)b[i])))){
flagans=false;
break;
}
trans((int)a[i])=b[i];
flag((int)a[i])=true;
pd((int)b[i])=a[i];
used((int)b[i])=true;
}
for(i=1;i<=26;i++) if(!flag[i]) flagans=false;
if(flagans){
for(i=0;i<=ques.length()-1;i++) ans[i]=trans((int)ques[i]);
cout<<ans;
}
else cout<<"Failed";
system("pause");
return 0;
}

var
ss,st,sd:string;
a,f:array['A'..'Z'] of char;
i:longint;
procedure find;
var
i:longint;
j:char;
begin
for j:='A' to 'Z' do begin a[j]:='0'; f[j]:='0'; end;
for i:=1 to length(ss) do
if a[ss[i]]='0' then
begin
a[ss[i]]:=st[i];
if f[st[i]]='0' then f[st[i]]:=ss[i];
if f[a[ss[i]]]<>ss[i] then begin writeln('Failed');halt;end;
end
else
if (a[ss[i]]<>'0') and (st[i]<>a[ss[i]]) then begin writeln('Failed');halt;end;

for j:='A' to 'Z' do
if a[j]='0' then begin writeln('Failed');halt;end;
end;
begin
readln(ss);
readln(st);
readln(sd);
find;
for i:=1 to length(sd) do
write(a[sd[i]]);
end.

有好几个出错条件，开始第二个点就是过不了了。。 注意可能A-B 又出现 A-C

#include <cstdio>
#include <cstring>

int main(void)
{
char enc[101], src[101], obj[101], key[26];
bool flag[26], isfailed = 0;
int len1, len2;

memset(enc, 0, 101);
memset(src, 0, 101);
memset(obj, 0, 101);
memset(key, 0, 26);
memset(flag, 0, 26);

scanf("%s%s%s", enc, src, obj);

len1 = strlen(enc);
len2 = strlen(obj);
for (int i = 0; i < len1; ++i) {
if (!flag[src[i] - 'A']) {
key[enc[i] - 'A'] = src[i];
flag[src[i] - 'A'] = 1;
}
else {
if(key[enc[i] - 'A'] == src[i]) continue;
else {
isfailed = 1;
break;
}
}
}

for (int i = 0; i < 26; ++i) {
if (!flag[i]) {
isfailed = 1;
break;
}
}

if (isfailed) printf("Failed");
else
for (int i = 0; i < len2; ++i) printf("%c", key[obj[i] - 'A']);
return 0;
}

太水了

本人一开始一直是第三个点过不去，后来才发现没有考虑‘不同字母对应不同密字’，WA了N次...大家一定要注意啊，认真读题，NOIP不过如此！

编译通过...

├ 测试数据 01：答案正确... (0ms, 580KB)

├ 测试数据 02：答案正确... (0ms, 580KB)

├ 测试数据 03：答案正确... (0ms, 580KB)

├ 测试数据 04：答案正确... (0ms, 580KB)

├ 测试数据 05：答案正确... (0ms, 580KB)

├ 测试数据 06：答案正确... (0ms, 580KB)

├ 测试数据 07：答案正确... (0ms, 580KB)

├ 测试数据 08：答案正确... (0ms, 580KB)

├ 测试数据 09：答案正确... (0ms, 580KB)

├ 测试数据 10：答案正确... (0ms, 580KB)

---|---|---|---|---|---|---|---|-

Accepted / 100 / 0ms / 580KB

水题

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <map>

using namespace std;

string a,b,c;
char m[300];
int vis[300];

void fail(){cout<<"Failed"<<endl;exit(0);}

int main(){
cin>>a>>b>>c;
for(int i=0;a.c_str()[i];i++){
if(m[a[i]]==0||m[a[i]]==b[i])
m[a[i]]=b[i],vis[a[i]]=1;
else fail();
}
for(int i='A';i<='Z';i++)if(!vis[i])fail();
for(int i='A';i<='Z';i++)
for(int j='A';j<='Z';j++)
if(i!=j&&m[i]==m[j])fail();
for(int i=0;c.c_str()[i];i++)cout<<m[c[i]];
cout<<endl;
}