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：

 1 #include<cstdio>

 2 #include<cstring>

 3 #include<cmath>

 4 using namespace std;

 5 

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

 7 char key2[28]={" QWERTYUIOPASDFGHJKLZXCVBNM"};

 8 

 9 int main()

10 {

11     int i,j,k,t;

12     char s[1005];

13     char temp[1005];

14     char ans[1005];

15     while(scanf("%s",s)!=EOF)

16     {

17         int len=strlen(s);

18         for(j=0,i=0;i<len;i+=2)

19         temp[j++]=key1[s[i]-'0'-1][s[i+1]-'0'-1];

20         temp[j]='\0';

21         t=0;

22         for(i=0;i<j;i++)

23         {

24             for(k=1;k<27;k++)

25             if(temp[i]==key2[k])

26             ans[t++]=(char)(k+'A'-1);

27         }

28         ans[t]='\0';

29         for(j=0,i=ceil((double)t/2);i<t;i++)

30         temp[j++]=ans[i];

31         temp[j]='\0';

32         for(i=ceil((double)t/2)-1;i>0;i--)

33         ans[i*2]=ans[i];

34         for(j=0,i=1;i<t;i+=2)

35         ans[i]=temp[j++];

36         ans[t]=='\0';

37         strrev(ans);

38         printf("%s\n",ans);

39     }

40     return 0;

41 }