poj2159 Ancient Cipher 水题

Ancient Cipher

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 33600 Accepted: 10945

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

JWPUDJSTVPVICTORIOUS

Sample Output

YES

Source

Northeastern Europe 2004

我这个八嘎。。。

题意读错wa三回就算了。。。后来想想就算题意读错了照样也能按最重提交的这个做法做啊

果然几天不写题就傻掉了

题意：

几个考古学家在石板上发现密文，秘闻中的26个字母分别有唯一对映的另一个字母，并且密文是打乱顺序的。

先给一条密文一条原文，问是否能有原文变幻为密文

代码：

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#include <vector>#include <cmath>using namespace std;char change[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 num1[26],num2[26];int judge[110];int main(){    string str1,str2;    int i,j;    cin>>str1>>str2;    if(str1==str2){        printf("YES\n");        return 0;    }    memset(num1, 0, sizeof(num1));    memset(num2, 0, sizeof(num2));    int len = (int)str1.length();    for(i=0;i<len;i++){        num1[str1[i]-'A']++;        num2[str2[i]-'A']++;    }    sort(num1, num1+26);    sort(num2, num2+26);    bool flag=true;    for(i=0;i<26;i++){        if(num1[i]!=num2[i]){            flag=false;        }    }    if(flag){        printf("YES\n");    }else{        printf("NO\n");    }    return 0;}