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HDU-1048-The Hardest Problem Ever(C++ && 偶尔一水......)

时间:2015-06-14 12:32:31      阅读:132      评论:0      收藏:0      [点我收藏+]

标签:acm   c++   hdu   字符串处理   水题   

The Hardest Problem Ever

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19343    Accepted Submission(s): 9061


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
 

Source
 

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题目的意思很明确:就是按照题目给定的要求解密字符串,解密的规则如下:

Cipher text(原密码)
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(破译密码)
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 

不难发现大于或者等于‘E‘的字符都是向前推进5个字符,而小于等于‘E‘的字符也是向前推进5个字符,不过超出的部分从‘Z‘往前面推!

另外注意一下输入格式:(包括多组输入数据)\

1.ENDINPUT表示程序的结束!

2.每组输入数据以START开始,END结束.

最后上代码:

#include <stdio.h>
#include <string.h>
int main(void)
{
    char str[200];
    int i;
    while (gets(str) != NULL && strcmp(str, "ENDOFINPUT"))
    {
        if (!strcmp(str, "START") || !strcmp(str, "END"))
            continue;
        for(i=0;i<strlen(str);i++)
        {
            if(str[i]>='A')
            {
                if(str[i]<='E')
                {
                    str[i]='V'+str[i]-'A';
                    printf("%c",str[i]);
                }
                else
                {
                    printf("%c",str[i]-5);
                }
            }
            else
            {
                printf("%c",str[i]);
            }
        }
        printf("\n");
    }
    return 0;
}



 

HDU-1048-The Hardest Problem Ever(C++ && 偶尔一水......)

标签:acm   c++   hdu   字符串处理   水题   

原文地址:http://blog.csdn.net/qq_16542775/article/details/46489955

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