HDU 2816 I Love You Too

　　　　　　　　　　　　　　　I Love You Too

Problem Description

This is a true story. A man showed his love to a girl,but the girl didn't replied clearly ,just gave him a Morse Code:
****-/*----/----*/****-/****-/*----/---**/*----/****-/*----/-****/***--/****-/*----/----*/**---/-****/**---/**---/***--/--***/****-/   He was so anxious that he asked for help in the Internet and after one day a girl named "Pianyi angel" found the secret of this code. She translate this code as this five steps:
1.First translate the morse code to a number string:4194418141634192622374
2.Second she cut two number as one group 41 94 41 81 41 63 41 92 62 23 74,according to standard Mobile phone can get this alphabet:GZGTGOGXNCS

3.Third she change this alphabet according to the keyboard:QWERTYUIOPASDFGHJKLZXCVBNM = ABCDEFGHIJKLMNOPQRSTUVWXYZ
So ,we can get OTOEOIOUYVL
4.Fourth, divide this alphabet to two parts: OTOEOI and OUYVL, compose again.we will get OOTUOYEVOLI
5.Finally,reverse this alphabet the answer will appear : I LOVE YOU TOO

I guess you might worship Pianyi angel as me,so let's Orz her.
Now,the task is translate the number strings.

 

 

Input

A number string each line(length <= 1000). I ensure all input are legal.

 

 

Output

An upper alphabet string.

 

 

Sample Input

4194418141634192622374

41944181416341926223

 

 

Sample Output

ILOVEYOUTOO

VOYEUOOTIO

 

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;

char key1[9][5]={" ","ABC","DEF","GHI","JKL","MNO","PQRS","TUV","WXYZ"};
char key2[28]={" QWERTYUIOPASDFGHJKLZXCVBNM"};

int main()
{
    int i,j,k,t;
    char s[1005];
    char temp[1005];
    char ans[1005];
    while(scanf("%s",s)!=EOF)
    {
        int len=strlen(s);
        for(j=0,i=0;i<len;i+=2)
        temp[j++]=key1[s[i]-'0'-1][s[i+1]-'0'-1];
        temp[j]='\0';
        t=0;
        for(i=0;i<j;i++)
        {
            for(k=1;k<27;k++)
            if(temp[i]==key2[k])
            ans[t++]=(char)(k+'A'-1);
        }
        ans[t]='\0';
        for(j=0,i=ceil((double)t/2);i<t;i++)
        temp[j++]=ans[i];
        temp[j]='\0';
        for(i=ceil((double)t/2)-1;i>0;i--)
        ans[i*2]=ans[i];
        for(j=0,i=1;i<t;i+=2)
        ans[i]=temp[j++];
        ans[t]=='\0';
        strrev(ans);
        printf("%s\n",ans);
    }
    return 0;
}