POJ2159 Ancient Cipher

Ancient Cipher

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22896		Accepted: 7703

Description

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 contains 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 are equal and do not exceed 100.

Output

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

Sample Output

YES

Source

Northeastern Europe 2004

这道题目是关于字母替换的，开始很快的写出来了，通过了上面的测试用例，但是一提交就是WA。很是郁闷，后来发现是题目没读懂。从A-Z变换到B-A只是题目中举出的一个例子，没说肯定是这样变换。所以A可以映射为B，C却也可以映射为Z。每个字母映射为哪个，我们并不知道。其二，映射后在进行顺序置换，这个置换数组也不知道。所以想通过对给出的字符进行变换后判断和给出的结果是否一致是行不通的。

仔细分析发现，任何一个字母映射的结果是唯一的，如果在明文中x的个数为n，那么在密文中一定有一个字母是x变换后的，所以它出现的概率和明文中x的出现概率必须相等。

代码如下：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>

using namespace std;


int main()
{
    const int MAX = 101;
    char a[MAX];
    char b[MAX];
    char cnta[26],cntb[26];
    memset(cnta,0,26);
    memset(cntb,0,26);

    scanf("%s%s",a,b);
    for(int i=0;i<strlen(a);i++)
        cnta[a[i]-'A']++;

    for(int i=0;i<strlen(b);i++)
        cntb[b[i]-'A']++;

    int flag = 1;
    sort(cnta,cnta+26);
    sort(cntb,cntb+26);

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

    if(flag)
        printf("YES");
    else
        printf("NO");

    return 0;
}