poj2159 Ancient Cipher

Ancient Cipher

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

题意：给你两个字符串，问第一个能不能通过凯撒加密的逆过程得到得到第二个

解题思路：所谓的凯撒加密先是将先通过一个置换的过程将字符串重组，然后再让每个字符加上一个数变成另一个字符，

但是我们无法模拟这个过程，因为字符串的大小最大100，如果模拟大概有100！种情况，运行时间超过1s，

我们可以发现一个字符串相同的字符对应另一个字符串的字符也是相同的字符，因此，

如果一个字符串中某个字符的频率是n，那个在另一个字符串中某个字符的频率也是n，则这两个很有可能是相对应的，

所以我们可以找出字符中所有字符的频率，然后让字符的频率按顺序排列，同理另一个字符串做同样的操作，

如果他们排列的顺序是一样的，那么他们可以通过凯撒加密的逆过程进行变换。

参考代码：

#include <iostream>
#include <string.h>
#include <algorithm>
using namespace std;
int main(){
	char x[102],y[102];
	int a[26],b[26];
	while (cin>>x>>y){
		int len_x=strlen(x),len_y=strlen(y),flag=0;
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		for (int i=0;i<len_x;i++)
			a[x[i]-'A']++;
		for (int i=0;i<len_y;i++)
			b[y[i]-'A']++;
		sort(a,a+26);
		sort(b,b+26);
		for (int i=0;i<26;i++){
			if (a[i]!=b[i]){
				cout<<"NO"<<endl;
				flag=1;
				break;
			}
		}
		if (flag==0)
			cout<<"YES"<<endl;
	}
	return 0;
}