POJ 2159 Ancient Cipher(내 워터프루프 - 이중 암호화)

Ancient Cipher
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 20020
 
Accepted: 6786
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
두 가지 암호화 형식이 있다. 1. 26개의 알파벳을 각각 다른 알파벳에 대응하여 암호화한다.2. 문자열의 각 자모의 위치 순서를 순서대로 정렬합니다.현재 두 개의 문자열을 주었는데, 상기 두 가지 암호화 형식을 통해 두 번째 문자열을 첫 번째 문자열로 암호화할 수 있는지 물었다.가능하면 "YES"를 출력하고, 그렇지 않으면 "NO"를 출력합니다.
먼저 이 두 가지 암호화 형식을 분석하면 첫 번째는 모든 자모가 서로 다른 자모를 통해 대체될 수 있음을 보장할 수 있다.두 번째는 문자열의 순서를 마음대로 바꿀 수 있다는 것이다.그래서 우리는 두 문자열의 서로 다른 주파수의 자모 수량이 같다는 것을 보증하기만 하면 된다. 즉, 문자열에서 한 번의 자모 개수만 나타나는 것은 밀문에서 한 번의 자모 개수만 나타나는 것과 같다.
문자열에서 두 번만 나타나는 알파벳 개수는 밀문에서 두 번만 나타나는 알파벳 개수와 같다.그래서 먼저 각 자모의 개수를 통계한 다음에 정렬을 하고 마지막에 한 분을 비교하면 된다.
코드(1AC):

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>

using namespace std;

int first[30];
int second[30];
char str1[110];
char str2[110];

int main(void){
    int i, j;
    int len1, len2, flag;

    while (scanf("%s", str1) != EOF){
        scanf("%s", str2);
        len1 = strlen(str1);
        len2 = strlen(str2);
        memset(first, 0, sizeof(first));
        memset(second, 0, sizeof(second));
        for (i = 0; i < len1; i++){
            first[str1[i] - 'A']++;
        }
        for (i = 0; i < len2; i++){
            second[str2[i] - 'A']++;
        }
        sort(first, first + 26);
        sort(second, second + 26);
        for (flag = 1, i = 0; i < 26; i++){
            if (first[i] != second[i]){
                flag = 0;
                break;
            }
        }
        if (len1 == len2 && flag){
            printf("YES
");
        }
        else{
            printf("NO
");
        }
    }
    return 0;
}