E - I Love You Too解题报告（张宇）

E - I Love You Too

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 2816

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

 

题意：给出一串长度不大于1000的数字，按每两个数字对应一个字母来解密（如图，即手机按键上的字母，如22代表B），解密后在根据解密对应的编码【3】来得到字母串，然后拆成两半（前部分不短于后部分），再按照【4】重新得到字母串，最后反序输出该字母串就ok啦。（貌似此密码很让人头疼呀。。。。别急，一步一步来，题目并不难）

[cpp] view plaincopy

1. #include<iostream>  
2. using namespace std;  
3. int main()  
4. {  
5.     char a[1010];  
6.     char b[505];  
7.     char c[252];  
8.     char d[252];  
9.     char ans[505];  
10.     char x[8][5]={"KXV","MCN","OPH","QRS","ZYI","JADL","EGW","BUFT"}; //已经按照密码定义了  
11.     while(scanf("%s",a)!=EOF)  
12.     {  
13.         int i,j,k;  
14.         for(i=0,j=0;i<strlen(a);i+=2,j++)  
15.         {     
16.             a[i]-='0';            //把字符转化为数字  
17.             a[i+1]-='0';  
18.             b[j]=x[a[i]-2][a[i+1]-1];   //求出每两个数字对应的一个字母  
19.         }  
20.         //要把字母串分成两个部分,而且前部分不短于后部分  
21.         for(i=0;i<((strlen(a)/2)+1)/2;i++)  
22.             c[i]=b[i];  
23.         for(j=i,k=0;j<strlen(a)/2;j++,k++)  
24.             d[k]=b[j];  
25.         j=0;k=0;  
26.         //再把两个字母串合并  
27.         for(i=0;i<strlen(a)/2;i++)  
28.         {  
29.             if((i+1)%2!=0)  
30.                 ans[i]=c[j++];  
31.             else  
32.                 ans[i]=d[k++];  
33.         }  
34.         for(i=strlen(a)/2-1;i>=0;i--)  
35.             printf("%c",ans[i]);  
36.         printf("\n");  
37.     }  
38.     return 0;  
39. }