zoukankan      html  css  js  c++  java
  • hdu 1048

    The Hardest Problem Ever

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 11459    Accepted Submission(s): 5165


    Problem Description
    Julius Caesar lived in a time of danger and intrigue. The hardest situation Caesar ever faced was keeping himself alive. In order for him to survive, he decided to create one of the first ciphers. This cipher was so incredibly sound, that no one could figure it out without knowing how it worked. 
    You are a sub captain of Caesar's army. It is your job to decipher the messages sent by Caesar and provide to your general. The code is simple. For each letter in a plaintext message, you shift it five places to the right to create the secure message (i.e., if the letter is 'A', the cipher text would be 'F'). Since you are creating plain text out of Caesar's messages, you will do the opposite: 

    Cipher text
    A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

    Plain text
    V W X Y Z A B C D E F G H I J K L M N O P Q R S T U 

    Only letters are shifted in this cipher. Any non-alphabetical character should remain the same, and all alphabetical characters will be upper case.
     
    Input
    Input to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the following description, and there will be no blank lines separating data sets. All characters will be uppercase. 

    A single data set has 3 components: 

    Start line - A single line, "START" 

    Cipher message - A single line containing from one to two hundred characters, inclusive, comprising a single message from Caesar 

    End line - A single line, "END" 

    Following the final data set will be a single line, "ENDOFINPUT".
     
    Output
    For each data set, there will be exactly one line of output. This is the original message by Caesar.
     
    Sample Input
    START NS BFW, JAJSYX TK NRUTWYFSHJ FWJ YMJ WJXZQY TK YWNANFQ HFZXJX END START N BTZQI WFYMJW GJ KNWXY NS F QNYYQJ NGJWNFS ANQQFLJ YMFS XJHTSI NS WTRJ END START IFSLJW PSTBX KZQQ BJQQ YMFY HFJXFW NX RTWJ IFSLJWTZX YMFS MJ END ENDOFINPUT
     
    Sample Output
    IN WAR, EVENTS OF IMPORTANCE ARE THE RESULT OF TRIVIAL CAUSES I WOULD RATHER BE FIRST IN A LITTLE IBERIAN VILLAGE THAN SECOND IN ROME DANGER KNOWS FULL WELL THAT CAESAR IS MORE DANGEROUS THAN HE
     1 #include <iostream>
     2 #include <cctype>
     3 #include <cstdlib>
     4 #include <cstdio>
     5 #include <cstring>
     6 
     7 using namespace std;
     8 
     9 int main(void)
    10 {
    11     char a[200+10], b[200+10];
    12 #ifndef ONLINE_JUDGE
    13     freopen("1048.in", "r", stdin);
    14 #endif
    15     while (~scanf("%s", a))
    16     {
    17         getchar();
    18         if (!strcmp(a, "START"))
    19         {
    20             gets(b);
    21             for (int i = 0; b[i] != '\0'; ++i)
    22             {
    23                 if (isalpha(b[i]) && isupper(b[i]))
    24                 {
    25                     if (b[i] < 'F')
    26                         printf("%c", b[i] + 21);
    27                     else printf("%c", b[i] - 5);
    28                 }
    29                 else
    30                     printf("%c", b[i]);
    31             }
    32             printf("\n");
    33         }
    34         else if (!strcmp("END", a))
    35             continue;
    36         else break;
    37     }
    38 
    39     return 0;
    40 }

    这是入门题

  • 相关阅读:
    SVN中trunk、branches、tag的使用
    svn建立分支和svn代码合并的操作方法
    SVN分支的合并和同步
    iOS开发--即时通讯
    iOS 开发--开源图片处理圆角
    ios开发--网易滚动导航栏
    ios开发--高德地图SDK使用简介
    大型网站架构演变和知识体系
    Nginx配置文件nginx.conf中文详解
    nginx+apache+php+mysql服务器集群搭建
  • 原文地址:https://www.cnblogs.com/liuxueyang/p/2870785.html
Copyright © 2011-2022 走看看