hd2816 I Love You Too

I Love You Too

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1768 Accepted Submission(s): 1061



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
A number string each line(length <= 1000). I ensure all input are legal.

[align=left]Output[/align]
An upper alphabet string.

[align=left]Sample Input[/align]

4194418141634192622374
41944181416341926223

[align=left]Sample Output[/align]

ILOVEYOUTOO
VOYEUOOTIO

嗯，题意不是很好懂，主要是因为是英文的。。。
嗯，大意就是给一字符串，每两个为一组对应手机9键的字母，然后再把键盘上的QWERTYUIOPASDFGHJKLZXCVBNM 对应为ABCDEFGHIJKLMNOPQRSTUVWXYZ找到一字符串，再分成两个字符串s1,s2，长度若是奇数，s1比s2 多1个字符，并且先输出这个字符；然后输出s2的最后一个字符和s1的剩下的最后一个字符，如此交错输出即为结果

#include<cstdio>
#include<cstring>
char st[1010],sr[1010],str[1010],ss1[1010],ss2[1010];
char ori[30]="QWERTYUIOPASDFGHJKLZXCVBNM ";
char aft[30]="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
char vv[10][5]={" "," ","ABC","DEF","GHI","JKL","MNO","PQRS","TUV","WXYZ"};
int main()
{
int len,i,j,flag;
while(~scanf("%s",st))
{
len=strlen(st);
for(i=0,j=0;i<len;i+=2)
{
sr[j++]=vv[st[i]-'0'][st[i+1]-'1'];
}
len=j;
for(i=0;i<len;++i)
{
for(j=0;j<26;++j)
{
if(sr[i]==ori[j])
{
str[i]=aft[j];
}
}
}
len=strlen(str);
flag=len/2;
if(len&1)printf("%c",str[flag]);
for(i=1;i<=flag;++i)printf("%c%c",str[len-i],str[flag-i]);
printf("\n");
memset(sr,'\0',sizeof(sr));
memset(st,'\0',sizeof(st));
memset(str,'\0',sizeof(str));
}
return 0;
}