HDOJ I Love You Too （第一周）

I Love You Too

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 11   Accepted Submission(s) : 5

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

  

 

读懂了就好做了，一次A
ac代码：

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
#include<string.h>
using namespace std;
char s1[1050][6]={{"ABC"},{"DEF"},{"GHI"},{"JKL"},{"MNO"},{"PQRS"},{"TUV"},{"WXYZ"}};//step1
char s2[4][1050]={{"QWERTYUIOPASDFGHJKLZXCVBNM"},{"ABCDEFGHIJKLMNOPQRSTUVWXYZ"}};//step2
int main()
{
    char num[1050],s[1050];
    char last[1050];
    int i,j;
    while(scanf("%s",s)!=EOF)
    {
        int len=strlen(s);
        int c=0;//step1
        for(i=0;i<len;)
        {
            num[c++]=s1[s[i]-'0'-2][s[i+1]-'0'-1];
            i+=2;
        }
        for(j=0;j<c;j++)
        {
            for(i=0;i<26;i++)
            {
                if(s2[0][i]==num[j])
                {
                    num[j]=s2[1][i];
                    break;
                }
            }
        }
        int k;
        if(c%2==1)
        k=c/2+1;
        else
        k=c/2;
        int l=0;
        for(i=0,j=k;i<k;i++,j++)
        {
            last[l++]=num[i];
            if(j<c)
            last[l++]=num[j];
        }
        i=l-1;
        while(i>=0)
        {
            printf("%c",last[i]);
            i--;
        }
        printf("\n");
    }
    return 0;
}