poj 2159 D - Ancient Cipher 文件加密

Posted 2020-09-29 木子丘

　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　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
本题的意思是字符串的加密方法有两种，一种是将字母的顺序打乱，还有一种是将每一个字母都替换成另一个字母（这里千万不要被题目的例子迷惑了，这边要保证的是字母出现的频率相同，巨坑）
如果后面的单词符合两种加密方式的结合，输出YES否则输出NO
附上代码（有待改进）

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<algorithm>
 4 #include<iostream>
 5 using namespace std;
 6 char str1[110],str2[110];
 7 int a[30],b[30];
 8 int main(){
 9     freopen("in.txt","r",stdin);
10     scanf("%s",str1);
11     scanf("%s",str2);
12     int flag=0;
13     int len=strlen(str2);
14     int len1=strlen(str1);
15     if(len1!=len){
16         printf("NO\n");
17     }
18     else{
19         sort(str1,str1+len1);
20         sort(str2,str2+len);
21         for(int i=0;i<len1;i++){
22             if(str1[i]>=‘a‘) a[str1[i]-‘a‘]++;
23             else a[str1[i]-‘A‘]++;
24         }
25         for(int i=0;i<len;i++){
26             if(str2[i]>=‘a‘) b[str2[i]-‘a‘]++;
27             else b[str2[i]-‘A‘]++;
28         }
29         sort(a,a+26);
30         sort(b,b+26);
31         int c,d;
32         for(int i=0;i<26;i++){
33             if(a[i]!=0){
34                 c=i;
35                 break;
36             }
37         }
38         for(int i=0;i<26;i++){
39             if(b[i]!=0){
40                 d=i;
41                 break;
42             }
43         }
44         int len3=max(26-c,26-d);
45         //cout<<len3;
46         for(int i=0;i<len3;i++){
47             if(a[c+i-1]!=b[d+i-1]){
48                 printf("NO\n");
49                 flag=1;
50                 break;
51             }
52         }
53         if(flag==0){
54             printf("YES\n");
55         }
56         return 0;
57     }
58 
59    /* for(int j=-25;j<=26;j++){
60         for(int i=0;i<len;i++){
61             str2[i]=str2[i]+j;
62             if(str2[i]<‘Z‘){
63                 str2[i]=str2[i]+26;
64             }
65             if(str2[i]>‘z‘){
66                 str2[i]=str2[i]-26;
67             }
68         }
69         if(strcmp(str1,str2)==0){
70             printf("YES\n");
71             flag=1;
72             break;
73         }
74   }*/
75 }