HDU-2816 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): 865    Accepted Submission(s): 531

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

 

　　该题算是一个模拟题了，题目不难，这里用了map来直接映射。

　　

 1 #include <cstring>
 2 #include <cstdlib>
 3 #include <cstdio>
 4 #include <map>
 5 using namespace std;
 6 
 7 char s[1005];
 8 
 9 int rec[505], word[505];
10 
11 int main()
12 {
13     map< int, char >mp;
14     mp[21] = 'K', mp[22] = 'X', mp[23] = 'V';
15     mp[31] = 'M', mp[32] = 'C', mp[33] = 'N';
16     mp[41] = 'O', mp[42] = 'P', mp[43] = 'H';
17     mp[51] = 'Q', mp[52] = 'R', mp[53] = 'S';
18     mp[61] = 'Z', mp[62] = 'Y', mp[63] = 'I';
19     mp[71] = 'J', mp[72] = 'A', mp[73] = 'D', mp[74] = 'L';
20     mp[81] = 'E', mp[82] = 'G', mp[83] ='W';
21     mp[91] = 'B', mp[92] = 'U', mp[93] = 'F', mp[94] = 'T';
22     while( scanf( "%s", s ) != EOF )
23     {
24         int len = strlen( s ), cnt = 0;
25         memset( rec, 0, sizeof( rec ) );
26         for( int i = 0; i < len; i += 2, ++cnt )
27         {
28             rec[cnt] += s[i] - '0';
29             rec[cnt] = rec[cnt] * 10 + s[i+1] - '0';
30         }
31         int lim = ( cnt - 1 ) >> 1;
32         for( int i = lim, j = cnt - 1, k = 0; i >= 0; --i, --j, k += 2 )
33         {
34             if( cnt & 1 )
35             {
36                 word[k] = rec[i];
37                 if( j > lim )
38                     word[k+1] = rec[j];
39             }
40             else
41             {
42                 word[k] = rec[j], word[k+1] = rec[i];
43             }
44         }
45         for( int i = 0; i < cnt; ++i )
46         {
47             printf( "%c", mp[word[i]] );
48         }
49         puts( "" );
50     }
51     return 0;
52 }