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 }

  

posted @ 2011-08-29 12:49  沐阳  阅读(462)  评论(0编辑  收藏  举报