A - The Hardest Problem Ever（字符串模拟题）

最新推荐文章于 2024-07-25 16:30:00 发布

ACMer_hades

最新推荐文章于 2024-07-25 16:30:00 发布

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本文介绍了一种基于凯撒密码的解密方法，并通过一个具体的C语言程序实例展示了如何将加密的消息还原为原始明文。该程序能够处理字母向左移动五位的解密过程，特别针对A到E字母的特殊转换。

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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.

题目看上去很烦，但是大概意思就是密码翻译，输入一个字母，然后输出时就相当于解密，而秘钥就是把这个字母向左移动5个单位（-5)，但是要注意A~E的字母，它们移动的时候是（+21）；

最后，注意输入输出格式；

#include<stdio.h>
#include<string.h>
#include<ctype.h>
int main(){
	int i,j,k;
	char s[500],a[250],c;
	while(scanf("%s",a)!=EOF && strlen(a)<6){
		//要注意加上getchar()；因为读入了字符串了； 
		getchar();
		//因为一开始读入的是Start这个字符串，所以要读入其他密文的话，采用一个一个字符读入，然后如果是换行符的话，那么就跳出循环break; 
		if(a[0]=='S'){
			k=0;
			while(c=getchar()){
				if('A'<=c&&c<='E') s[k++]=c+21;
				else if(c>'E'&&c<='Z') s[k++]=c-5;
				else if(c=='\n') break;
				else 	s[k++]=c;
			}
		}
		else {
			s[k]='\0';		//别忘了这一步； 
			puts(s);
		}
		if(strcmp(a,"ENDOFINPUT")==0) break;
	}
}