joj1064

1064: Caeser Comes Back

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Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	1540	821	Standard

Assume you are a spy who is fulfilling a very important task. You have just stole a very important file from enemy's computer. Now you have to send the file to your headquarter by a special equipment. But in order to make safety, you should encode the file before sending it. Because the soldiers of enemy may come at any time, the encode process shouldn't take too much time. Under such situation, the only encode algorithm you could do is Caeser Encode Algorithm.

The Caeser algorithm can be expressed as:

In the encode process, a character is replaced by the 3rd one after it. Here you can assume that the file consists of only 'a' to 'z' and 'A' to 'Z'. So the last 3 characters are replaced by the first 3 characters. For example, consider this string abcdABCDwxyzWXYZ. After the encode process, it becomes defgDEFGzabcZABC. While in the decode process, this algorithm applies reversely.

Input Specification

The input consists of several lines. Each line consists of two strings A and B(at most 80 characters), separated by a space. If A equals to ENCODE, you should encode B while if A equals to DECODE, you should decode B. If A equals to END, that means the end of input, which you shouldn't process. No other values will A equal to except ENCODE, DECODE and END.

Output Specification

For each A=ENCODE or DECODE, you should encode or decode B and output the result on a single line each.

Sample Input

ENCODE abcd
ENCODE WXYZ
DECODE zAbC
ENCODE aA
END

Sample Output

defg
ZABC
wXyZ
dD

Problem Source: 1st JOJ Cup Online VContest Warmup Problem

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This problem is used for contest: 75 

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Problem Set with Online Judge System Version 3.12
Jilin University

Developed by skywind, SIYEE

 

 

 

#include<cstdio>
char string1[7];
char string2[88];
void encode(char *p)
{
    for(int i=0;p[i]!='\0';i++)
    {
        if(p[i]=='x'){p[i]='a';continue;}
        if(p[i]=='y'){p[i]='b';continue;}
        if(p[i]=='z'){p[i]='c';continue;}
        if(p[i]=='X'){p[i]='A';continue;}
        if(p[i]=='Y'){p[i]='B';continue;}
        if(p[i]=='Z'){p[i]='C';continue;}
        p[i]=p[i]+3;
    }
}
void decode(char *p)
{
    for(int i=0;p[i]!='\0';i++)
    {
        if(p[i]=='a'){p[i]='x';continue;}
        if(p[i]=='b'){p[i]='y';continue;}
        if(p[i]=='c'){p[i]='z';continue;}
        if(p[i]=='A'){p[i]='X';continue;}
        if(p[i]=='B'){p[i]='Y';continue;}
        if(p[i]=='C'){p[i]='Z';continue;}
        p[i]=p[i]-3;
    }
}
bool equal(const char*a,const char*b)
{
    int i=0;
    do
    {
        if(a[i]!=b[i])return false;
        i++;
    }while(a[i]!='\0'&&b[i]!='\0');
    return true;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    while(1)
    {
        scanf("%s",string1);
        if(equal(string1,"END"))break;
        getchar();
        gets(string2);
        if(equal(string1,"ENCODE"))encode(string2);
        else decode(string2);
        //printf("%s\n",string1);
       // decode(string2);
        printf("%s\n",string2);
    }
    return 0;
}