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

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

419441814163419262237441944181416341926223

 

Sample Output

ILOVEYOUTOOVOYEUOOTIO

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;}