zoukankan      html  css  js  c++  java
  • uva 1339

    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 $ langle$2, 1, 5, 4, 3, 7, 6, 10, 9, 8$ 
angle$ 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

    将字符串输入进去后,用sort排序,再用两个数组将各个字母的个数存进数组里。再对比两个数组的数字是否完全一样。。。

    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    
    const int MAXN = 110;
    
    int main()
    {
        char str1[MAXN], str2[MAXN];
        int a[26], b[26];
        while(scanf("%s%*c", str1) != EOF)
        {
            scanf("%s%*c", str2);
    
            memset(a, 0, sizeof(a));
            memset(b, 0, sizeof(b));
    
            int len = strlen(str1);
            sort(str1, str1+len);
            sort(str2, str2+len);
    
            for(int i = 0; i < len; ++i)
            {
                a[str1[i] - 'A']++;
                b[str2[i] - 'A']++;
            }
    
            sort(a, a+26);
            sort(b, b+26);
    
            int mark = 0;
            for(int i = 0; i < 26; ++i)
            {
                if(a[i] == b[i])
                {
                    mark++;
                    //printf("%d %d %d", a[i], b[i], mark);
                }
            }
            //printf("%d
    ", mark);
            if(mark == 26)
                printf("YES
    ");
            else
                printf("NO
    ");
        }
        return 0;
    }
  • 相关阅读:
    A* Pathfinding for Beginners
    OpenGL Draw Mesh
    CentOS6.5下安装、配置SSH
    Ubuntu18.04下搭建LAMP环境
    滚动合集
    关闭页面触发事件
    在table中tr的display:block在firefox下显示布局错乱问题
    sharepoint添加模板及删除模板
    常用正则表达式
    javascript对象的property和prototype是这样一种关系
  • 原文地址:https://www.cnblogs.com/ya-cpp/p/3978147.html
Copyright © 2011-2022 走看看