POJ1298解题报告
来源:互联网 发布:网络教育文凭考公务员 编辑:程序博客网 时间:2024/04/28 11:15
The Hardest Problem Ever
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 17404 Accepted: 9598
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.
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:
Following the final data set will be a single line, "ENDOFINPUT".
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
STARTNS BFW, JAJSYX TK NRUTWYFSHJ FWJ YMJ WJXZQY TK YWNANFQ HFZXJXENDSTARTN BTZQI WFYMJW GJ KNWXY NS F QNYYQJ NGJWNFS ANQQFLJ YMFS XJHTSI NS WTRJENDSTARTIFSLJW PSTBX KZQQ BJQQ YMFY HFJXFW NX RTWJ IFSLJWTZX YMFS MJENDENDOFINPUT
Sample Output
IN WAR, EVENTS OF IMPORTANCE ARE THE RESULT OF TRIVIAL CAUSESI WOULD RATHER BE FIRST IN A LITTLE IBERIAN VILLAGE THAN SECOND IN ROMEDANGER KNOWS FULL WELL THAT CAESAR IS MORE DANGEROUS THAN HE
这道题目的名字蛮好笑,最难的题目-----呵呵,其实是最简单的字符串问题
不过,注意一下scanf("%s",c)和gets(s)的区别,一个是按空格隔开读取,一个则是读取一整行。
#include<stdio.h>#include<limits.h>#include<string.h>#define MAXNUM 100#define LEN 26const char ciper[]={'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'};const char plain[]={'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'};char input1[MAXNUM];int main(){ scanf("%s",input1); int len; int i,j,tag; //判断是否结束输入 while(strcmp(input1,"ENDOFINPUT")!=0) { if(strcmp(input1,"START")==0) { } else if(strcmp(input1,"END")==0) { printf("\n"); } else { len=strlen(input1); for(i=0;i<len;i++) { tag=1; for(j=0;j<LEN&&tag;j++) { if(ciper[j]==input1[i]) { input1[i]=plain[j]; tag=0; } } } printf("%s",input1); } gets(input1); } system("pause"); return 0;}
- POJ1298解题报告
- POJ1298解题思路
- POJ1298
- POJ1298
- POJ1298
- poj1298
- poj1298
- 解题报告
- 解题报告
- 解题报告
- 解题报告
- 解题报告
- 解题报告
- 解题报告
- POJ1298模拟
- Antiprime解题报告
- expr解题报告
- 华容道解题报告
- 模拟实现WPF的依赖属性及绑定通知机制(2)--依赖对象的准备 .
- 工作代码片段-cmake
- 模拟实现WPF的依赖属性及绑定通知机制(3)--依赖对象
- 多线程模拟生产者消费者关系
- 模拟实现WPF的依赖属性及绑定通知机制(4)--模拟实现绑定连动机制 .
- POJ1298解题报告
- 正在背诵软件工程里面的一些重要概念,写下来增加理解!!~~
- Silverlight的依赖属性与附加属性
- java.sql.SQLException: Parameter index out of range (2 > number of parameters, which is 1).
- 查询数据库中的慢差查询
- 函数指针
- 超纯超美的曲线(Peter De Jong Attractor)
- 获取服务器地址
- Web服务中文综述阅读整理