Ancient Cipher 字符串加密 重排

Ancient Roman empire had a strong government system with various departments, including a secret

service department. Important documents were sent between provinces and the capital in encrypted

form to prevent eavesdropping. The most popular ciphers in those times were so called
substitution

cipher and
permutation cipher.

Substitution cipher changes all occurrences of each letter to some other letter. Substitutes for all

letters must be different. For some letters substitute letter may coincide with the original letter. For

example, applying substitution cipher that changes all letters from ‘A’ to ‘Y’ to the next ones in the

alphabet, and changes ‘Z’ to ‘A’, to the message “VICTORIOUS”
one gets the message “WJDUPSJPVT”.

Permutation cipher applies some permutation to the letters of the message. For example, applying the permutation
⟨2,
1,
5, 4,
3,
7, 6,
10,
9, 8⟩
to the message “VICTORIOUS” one gets the message

“IVOTCIRSUO”.

It was quickly noticed that being applied separately, both substitution cipher and permutation

cipher were rather weak. But when being combined, they were strong enough for those times. Thus,

the most important messages were first encrypted using substitution cipher, and then the result was

encrypted using permutation cipher. Encrypting the message “VICTORIOUS” with the combination of

the ciphers described above one gets the message “JWPUDJSTVP”.

Archeologists have recently found the message engraved on a stone plate. At the first glance it

seemed completely meaningless, so it was suggested that the message was encrypted with some substitution and permutation ciphers. They have conjectured the possible text of the original message that

was encrypted, and now they want to check their conjecture. They need a computer program to do it,

so you have to write one.
Input
Input file contains several test cases. Each of them consists of two lines. The first line contains the

message engraved on the plate. Before encrypting, all spaces and punctuation marks were removed, so

the encrypted message contains only capital letters of the English alphabet. The second line contains

the original message that is conjectured to be encrypted in the message on the first line. It also contains

only capital letters of the English alphabet.

The lengths of both lines of the input file are equal and do not exceed 100.
Output
For each test case, print one output line. Output ‘YES’ if the message on the first line of the input file

could be the result of encrypting the message on the second line, or ‘NO’ in the other case.
Sample Input
JWPUDJSTVP

VICTORIOUS

MAMA

ROME

HAHA

HEHE

AAA

AAA

NEERCISTHEBEST

SECRETMESSAGES
Sample Output
YES

NO

YES

YES

NO

题目大意：通过两种方式加密一个字符串：

1.将两个字母表建立映射关系，如A->B,B->C...

2.按照某种排列将1得到的序列重排

由上面两条规则，实际上还是建立一种映射关系，但是我们不容易确定这种映射，但是可以判断出是否具有可行方案。

1.统计加密前的字符串的26个字母出现的次数

2.统计加密后的字符串的26个字母出现的次数

3.对统计得到的两组数据排列并且比较

（1）如果相同，则是有可行方案

（2）如果不相同，则没有可行方案

这个很好理解，对于同一个字母，如果加密前后出现的次数一样，那么他们就可以互相替代，如果每一个字母都满足这种性质，就有解决方案。

#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdio.h>
#define MAX 110
using namespace std;

bool judge(int *orgi,int *guess,int len)
{
for(int i=0;i<len;i++)
{
if(orgi[i]!=guess[i])
return false;
}
return true;
}
int main()
{
char orig[MAX],guess[MAX];
int orig_c[26],guess_c[26];
int len,i;
while(~scanf("%s",orig))
{
len=strlen(orig);
memset(orig_c,0,sizeof(orig_c));
memset(guess_c,0,sizeof(guess_c));
scanf("%s",guess);
for(i=0;i<len;i++)
{
orig_c[orig[i]-'A']++;
guess_c[guess[i]-'A']++;
}
sort(orig_c,orig_c+26);
sort(guess_c,guess_c+26);

if(judge(orig_c,guess_c,26))
printf("YES\n");
else
printf("NO\n");

}
return 0;
}