这里的算法非常简单,就是“模拟”,注意编写每个传令官的算法时先分开测试,放在一起就会混淆。

POJ - 2041Unreliable Message

Time Limit: 1000MS Memory Limit: 30000KB 64bit IO Format: %I64d & %I64u

Description

The King of a little Kingdom on a little island in the Pacific Ocean frequently has childish ideas. One day he said, "You shall make use of a message relaying game when you inform me of something." In response to the King's statement, six servants were selected as messengers whose names were Mr. J, Miss C, Mr. E, Mr. A, Dr. P, and Mr. M. They had to relay a message to the next messenger until the message got to the King. Messages addressed to the King consist of digits ('0'-'9') and alphabet characters ('a'-'z', 'A'-'Z'). Capital and small letters are distinguished in messages. For example, "ke3E9Aa" is a message. Contrary to King's expectations, he always received wrong messages, because each messenger changed messages a bit before passing them to the next messenger. Since it irritated the King, he told you who are the Minister of the Science and Technology Agency of the Kingdom, "We don't want such a wrong message any more. You shall develop software to correct it!" In response to the King's new statement, you analyzed the messengers' mistakes with all technologies in the Kingdom, and acquired the following features of mistakes of each messenger. A surprising point was that each messenger made the same mistake whenever relaying a message. The following facts were observed. Mr. J rotates all characters of the message to the left by one. For example, he transforms "aB23d" to "B23da". Miss C rotates all characters of the message to the right by one. For example, she transforms "aB23d" to "daB23". Mr. E swaps the left half of the message with the right half. If the message has an odd number of characters, the middle one does not move. For example, he transforms "e3ac" to "ace3", and "aB23d" to "3d2aB". Mr. A reverses the message. For example, he transforms "aB23d" to "d32Ba". Dr. P increments by one all the digits in the message. If a digit is '9', it becomes '0'. The alphabet characters do not change. For example, he transforms "aB23d" to "aB34d", and "e9ac" to "e0ac". Mr. M decrements by one all the digits in the message. If a digit is '0', it becomes '9'. The alphabet characters do not change. For example, he transforms "aB23d" to "aB12d", and "e0ac" to "e9ac". The software you must develop is to infer the original message from the final message, given the order of the messengers. For example, if the order of the messengers is A->J->M->P and the message given to the King is "aB23d", what is the original message? According to the features of the messengers' mistakes, the sequence leading to the final message is "32Bad"->"daB23"->"aB23d"->"aB12d"->"aB23d": As a result, the original message should be "32Bad".

Input

The input format is as follows. n The order of messengers The message given to the King ... The order of messengers The message given to the King The first line of the input contains a positive integer n, which denotes the number of data sets. Each data set is a pair of the order of messengers and the message given to the King. The number of messengers relaying a message is between 1 and 6 inclusive. The same person may not appear more than once in the order of messengers. The length of a message is between 1 and 25 inclusive.

Output

The inferred messages are printed each on a separate line.

Sample Input

5

AJMP

aB23d

E

86AE

AM

6

JPEM

WaEaETC302Q

CP

rTurnAGundam1isdefferentf

Sample Output

32Bad

AE86

7

EC302QTWaEa

TurnAGundam0isdefferent

Source Japan 2003,Aizu

 #include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h> int main()
{
int n = ;
scanf("%d", &n);
char str[], order[];
for(int i = ; i < n; i++) {
scanf("%s", str);
scanf("%s", order);
int r = strlen(str);
int len = strlen(order);
char ch;
int t;
for(int i = r-; i >= ; i--) {
switch(str[i]){
printf("str -- %c\n", str[i]);
case 'J':
ch = order[len-];
for(int j = len-; j > ; j--) {
order[j] = order[j-];
}
order[] = ch;
break;
case 'C':
ch = order[];
for(int j = ; j < len-; j++) {
order[j] = order[j+];
}
order[len-] = ch;
break;
case 'E':
t = len/;
for(int i = ; i < t; i++) {
ch = order[i];
if(len%) {
order[i] = order[i+t+];
order[i+t+] = ch;
}
else {
order[i] = order[i+t];
order[i+t] = ch;
}
}
break;
case 'A':
t = len/;
for(int i = ; i < t; i++) {
ch = order[i];
order[i] = order[len--i];
order[len--i] = ch;
}
break;
case 'P':
for(int i = ; i < len; i++) {
if(order[i] >= '' && order[i] <= '') {
if(order[i] == '') order[i] = '';
else order[i]--;
}
}
break;
case 'M':
for(int i = ; i < len; i++) {
if(order[i] >= '' && order[i] <= '') {
if(order[i] == '') order[i] = '';
else order[i]++;
}
}
break;
}
}
printf("%s\n", order);
}
return ;
}

POJ - 2041Unreliable Message的更多相关文章

  1. OpenJudge / Poj 2141 Message Decowding

    1.链接地址: http://poj.org/problem?id=2141 http://bailian.openjudge.cn/practice/2141/ 2.题目: Message Deco ...

  2. Eclipse 4.2 failed to start after TEE is installed

    ---------------  VM Arguments---------------  jvm_args: -Dosgi.requiredJavaVersion=1.6 -Dhelp.lucene ...

  3. POJ 2774 Long Long Message 后缀数组

    Long Long Message   Description The little cat is majoring in physics in the capital of Byterland. A ...

  4. poj 2774 Long Long Message 后缀数组基础题

    Time Limit: 4000MS   Memory Limit: 131072K Total Submissions: 24756   Accepted: 10130 Case Time Limi ...

  5. 后缀数组(模板题) - 求最长公共子串 - poj 2774 Long Long Message

    Language: Default Long Long Message Time Limit: 4000MS   Memory Limit: 131072K Total Submissions: 21 ...

  6. POJ 2774 Long Long Message (Hash + 二分)

    Long Long Message Time Limit: 4000MS   Memory Limit: 131072K Total Submissions: 34473   Accepted: 13 ...

  7. POJ 2774 Long Long Message [ 最长公共子串 后缀数组]

    题目:http://poj.org/problem?id=2774 Long Long Message Time Limit: 4000MS   Memory Limit: 131072K Total ...

  8. POJ 题目2774 Long Long Message(后缀数组,求最长公共子串长度)

    Long Long Message Time Limit: 4000MS   Memory Limit: 131072K Total Submissions: 23696   Accepted: 97 ...

  9. Long Long Message POJ - 2774 后缀数组

    The little cat is majoring in physics in the capital of Byterland. A piece of sad news comes to him ...

随机推荐

  1. Objdump-查看汇编指令

    作用 Objdump可以用来看汇编指令 查看汇编指令 测试文件 编译指令 gcc -g -o objtest 1.8.c objdump -S objtest |more /main 查看结果

  2. [LeetCode]题解(python):039-Combination Sum

    题目来源 https://leetcode.com/problems/combination-sum/ Given a set of candidate numbers (C) and a targe ...

  3. LightOj1285 - Drawing Simple Polygon(连接多边形各点)

    题目链接:http://lightoj.com/volume_showproblem.php?problem=1285 题意:给你一些点,然后把它们用一条线把它们连起来,构成一个多边形,不能有相交,必 ...

  4. 查询数据表,去除符合某些条件的记录,没有自动增长列(not exists)

    select distinct ccode,isnull(cexch_name,''),N'',N'',N'2014.03',0,1,1,1,12 from RP_bankrecp where not ...

  5. MVC4怎么设置@Html.TextBoxFor这样的输入框的css样式

    在传统webForm中,输入框的这样的: <input id="userName" name="userName" type="text&quo ...

  6. iostart命令

    Linux系统中的 iostat命令可以对系统的磁盘IO和CPU使用情况进行监控.iostat属于sysstat软件包,可以用yum -y install sysstat 直接安装. 1.基本使用:i ...

  7. 与Perl兼容的正则表达式函数

    <?php $str = "abc_ABC_123abc"; $patt = '/^abc(.*?)abc$/'; //有一个模式单元就会多一个存储 $arr = null; ...

  8. robotframework 测试结果写入数据库

    即将更新...........

  9. c# 工具

    虚拟机 http://sandcastle.codeplex.com/ 读取excel等的一个工具 http://npoi.codeplex.com/releases/view/115353 host ...

  10. Java I/O解读与使用实例

    林炳文Evankaka原创作品.转载请注明出处http://blog.csdn.net/evankaka 摘要:本文主要讲解了Java I/O解读与使用实例. 一.I/O基本概念 I/O全称是Inpu ...