POJ1107 ZOJ1042 UVALive2291 W's Cipher【密码+模拟】

W’s Cipher
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 5811 Accepted: 2903

Description

Weird Wally’s Wireless Widgets, Inc. manufactures an eclectic assortment of small, wireless, network capable devices, ranging from dog collars, to pencils, to fishing bobbers. All these devices have very small memories. Encryption algorithms like Rijndael, the candidate for the Advanced Encryption Standard (AES) are demonstrably secure but they don’t fit in such a tiny memory. In order to provide some security for transmissions to and from the devices, WWWW uses the following algorithm, which you are to implement.

Encrypting a message requires three integer keys, k1, k2, and k3. The letters [a-i] form one group, [j-r] a second group, and everything else ([s-z] and underscore) the third group. Within each group the letters are rotated left by ki positions in the message. Each group is rotated independently of the other two. Decrypting the message means doing a right rotation by ki positions within each group.

Consider the message the_quick_brown_fox encrypted with ki values of 2, 3 and 1. The encrypted string is _icuo_bfnwhoq_kxert. The figure below shows the decrypting right rotations for one character in each of the three character groups.

Looking at all the letters in the group [a-i] we see {i,c,b,f,h,e} appear at positions {2,3,7,8,11,17} within the encrypted message. After a right rotation of k1=2, these positions contain the letters {h,e,i,c,b,f}. The table below shows the intermediate strings that come from doing all the rotations in the first group, then all rotations in the second group, then all the rotations in the third group. Rotating letters in one group will not change any letters in any of the other groups.

All input strings contain only lowercase letters and underscores(_). Each string will be at most 80 characters long. The ki are all positive integers in the range 1-100.

Input

Input consists of information for one or more encrypted messages. Each problem begins with one line containing k1, k2, and k3 followed by a line containing the encrypted message. The end of the input is signalled by a line with all key values of 0.

Output

For each encrypted message, the output is a single line containing the decrypted string.

Sample Input

2 3 1
_icuo_bfnwhoq_kxert
1 1 1
bcalmkyzx
3 7 4
wcb_mxfep_dorul_eov_qtkrhe_ozany_dgtoh_u_eji
2 4 3
cjvdksaltbmu
0 0 0

Sample Output

the_quick_brown_fox
abcklmxyz
the_quick_brown_fox_jumped_over_the_lazy_dog
ajsbktcludmv

Source

Mid-Central USA 2001

Regionals 2001 >> North America - Mid-Central USA

问题链接：POJ1107 ZOJ1042 UVALive2291 W’s Cipher
问题简述：（略）
问题分析：
这个问题是一个密码问题，用模拟法来实现，按照题意模拟即可。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C语言程序如下：

/* POJ1107 ZOJ1042 UVALive2291 W's Cipher */#include <stdio.h>
#include <string.h>#define N 80
char s[N + 1], t1[N + 1], t2[N + 1], t3[N + 1], s1[N + 1], s2[N + 1];void fix(char t[],int k,int len)
{int i, j;k = len - k % len;for(i = 0; i < k; i++)s1[i] = t[i];s1[i]='\0';for(i = k, j = 0; i < len; i++)s2[j++] = t[i];s2[j]='\0';strcat(s2, s1);strcpy(t, s2);
}int main(void)
{int k1, k2, k3, len, i1, i2, i3, i;while(~scanf("%d%d%d", &k1, &k2, &k3) && (k1 || k2 || k3)) {memset(t1, 0, sizeof(t1));memset(t2, 0, sizeof(t2));memset(t3, 0, sizeof(t3));scanf("%s", s);len = strlen(s);i1 = i2 = i3 = 0;for(i = 0; i < len; i++)if(s[i] >= 'a' && s[i] <= 'i')  t1[i1++] = s[i];else if(s[i] >= 'j' && s[i] <= 'r') t2[i2++] = s[i];else if( (s[i] >= 's' && s[i] <= 'z' ) || s[i] == '_') t3[i3++] = s[i];if(i1)fix(t1, k1, i1);if(i2)fix(t2, k2, i2);if(i3)fix(t3, k3, i3);i1 = i2 = i3 = 0;for(i = 0; i < len; i++)if(s[i] >= 'a' && s[i] <= 'i')  s[i] = t1[i1++];else if(s[i] >= 'j' && s[i] <= 'r') s[i] = t2[i2++];else if( (s[i] >= 's' && s[i] <= 'z' ) || s[i] == '_') s[i] = t3[i3++];printf("%s\n", s);}return 0;
}

POJ1107 ZOJ1042 UVALive2291 W's Cipher【密码+模拟】相关推荐

Bailian2819 W的密码【密码+模拟】
2819:W的密码总时间限制: 1000ms 内存限制: 65536kB 描述加密一条信息需要三个整数码, k1, k2 和 k3.字符[a-i] 组成一组, [j-r] 是第二组, 其它所有字符 ...
【Python】忘记登录密码？遍历数字字母组合密码+模拟提交表单
准备忘记了密码?试一试暴力破解吧打开F12,填写用户名.密码后,点击登录,看到请求头信息.提交的表单信息,例如下图: 也可以看到此次登录请求发送到的 url: 也可以看到此次请求返回的结果: 遍历 ...
selenium使用账号密码模拟登陆淘宝，使用账号密码！
1,需要相关的库如下 from selenium import webdriver from selenium.webdriver.common.action_chains import Action ...
selenium使用账号密码模拟登陆淘宝，使用账号密码
1,需要相关的库如下 from selenium import webdriver from selenium.webdriver.common.action_chains import Action ...
w ndows7管理员密码,Windows7默认管理员账户用户名和密码是什么
很多小伙伴更新了系统,下载了深度技术win7系统,但是却不知道这个Windows7默认管理员账户用户名和密码是什么,其实这个没有密码,Windows7默认账户用户名是:Administrator,密码 ...
W ndows10开机密码设置,windows10密码,小编告诉你win10开机密码怎么设置
对于用户来说,最重要的莫过于用户的个人的隐私了,可见和不可见的,都是最要紧的,为了更好的保护自己的个人隐私,给电脑设置开机密码是必不可少的一道关卡了,不过这对于一些新手来说,是很比较艰难的操作,下面, ...
北大考研复试上机——W's Cipher
题目描述 Weird Wally's Wireless Widgets, Inc. manufactures an eclectic assortmentof small, wireless, net ...
Liunx操作系统的引导过程（系统操作引导过程，模拟MBR,GRUB故障,root密码遗忘解决，优化启动过程，运行级别的分类）
文章目录 Liunx操作系统的引导过程引导过程 Blos自检 MBR 引导 grub引导菜单(Boot Loader) 内核启动启动init进程,依据inittab文件设定运行级别系统初始化进程 ...
挖藕！模拟sql抓取qq密码
挖藕!模拟sql查看qq密码模拟抓取到的qq数据通过封装实现数据库的连接 1. 创建配置文件(包含驱动,url 密码,用户名) 2. 实现封装工具类(涉及类加载机制复习) 3. 利用封装工具类实现 ...

POJ1107 ZOJ1042 UVALive2291 W's Cipher【密码+模拟】

POJ1107 ZOJ1042 UVALive2291 W's Cipher【密码+模拟】相关推荐

最新文章

热门文章