紫书第四章训练1A

原文：http://blog.csdn.net/Bob__Huang/article/details/70665281

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 file contains several test cases. Each of them consists of 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 file are equal and do not exceed 100. 
Output 
For each test case, print one output line. 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 
MAMA 
ROME 
HAHA 
HEHE 
AAA 
AAA 
NEERCISTHEBEST 
SECRETMESSAGES 
Sample Output 
YES 
NO 
YES 
YES 
NO

这个题像个阅读理解，读英语是关键啊“For example, applying the permutation （2,1,5,4,3,7,6,10,9,8） to the message “VICTORIOUS” one gets the message “IVOTCIRSUO”. ”这句话可能误导了某些人因为其变换后的字符串是前字符串的混乱顺序串，但是一看输入输出就可以知道不是这个意思。其实他的真实意思是加密，每个字母均对应一个字母，也就是变成了字符串长度一样的字符出现频率相等的就输出YES的题，sort排下序，比较就结束了。

crq注：

题意是判断两个字符串之间是否能映射，如HAHA->HEHE（可以将A→E）

这里面有个隐含条件：如果A→B，那么A在第一个序列中出现的次数和B在第二个序列中出现的次数一定是一样的！

所以可以直接统计出两个字符串中各个字符出现的次数，计入数组，在进行排序，判断以下是否次数都相等即可。

AC代码：

#include <stdio.h>
#include <algorithm>
using namespace std;
int main()
{char s[105];
while(~scanf("%s",s)){
    int a[30]={0},b[30]={0};
//26个英文字母
    for(int i=0;s[i];i++)
    a[s[i]-'A']++;
    sort(a,a+26);
    scanf("%s",s);
    for(int i=0;s[i];i++)
    b[s[i]-'A']++;
    sort(b,b+26);
    int f=0;
    for(int i=0;i<26;i++)
    if(a[i]!=b[i]){
        puts("NO");
        f=1;
        break;
    }
    if(!f)puts("YES");
}

    return 0;
}