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本来是一道水题，模拟的。但是我当时把一个int变量定义成char型的，结果是死活都调试不出来！一直WA 。最后还是自己检查出来了。
1 #include <stdio.h>
2#include <memory.h>
3#include <stdlib.h>
4#define DEBUG 1
5const int N = 1005 ;
6
7int main()
8   {
9 #if DEBUG
10 freopen("C:\\Documents and Settings\\Administrator\\桌面\\in.txt","r",stdin);
11 freopen("C:\\Documents and Settings\\Administrator\\桌面\\out.txt","w",stdout);
12 #endif
13
14 char a[N], b[N], ans[N], *p, aaa[N] ;
15 int i, j, k, num, geshu ;
16 double nn ;
17 //预先把数据写好
18  for( k=0,i=20; i<=70; i+=10,++k ){
19 for( j=1; j<=3; ++j,++k )
20 b[i+j] = k+'A' ;
21 --k ;
22 }
23 b[74] = 'S' ;
24 for( k=19,i=80; i<=90; i+=10,++k ){
25 for( j=1; j<=3; ++j,++k )
26 b[i+j] = k+'A' ;
27 --k ;
28 }
29 b[94] = 'Z' ;
30 while( scanf("%s", a ) != EOF ){
31 memset(ans,0,sizeof(ans));
32 memset(aaa,0,sizeof(aaa));
33 p = a ;
34 for( i=0; *p; ++i ){
35 num = *p-'0' ;
36 ++p ;
37 num = num*10 + *p-'0' ;
38 ans[i] = b[num] ;
39 ++p ;
40 }
41 ans[i] = '\0' ;
42 geshu = i ;
43 for( j=0; j<=geshu; ++j ){
44 switch ( ans[j] ){
45 case 'Q': ans[j] = 'A' ;break ;
46 case 'W': ans[j] = 'B' ;break ;
47 case 'E': ans[j] = 'C' ;break ;
48 case 'R': ans[j] = 'D' ;break ;
49 case 'T': ans[j] = 'E' ;break ;
50 case 'Y': ans[j] = 'F' ;break ;
51 case 'U': ans[j] = 'G' ;break ;
52 case 'I': ans[j] = 'H' ;break ;
53 case 'O': ans[j] = 'I' ;break ;
54 case 'P': ans[j] = 'J' ;break ;
55 case 'A': ans[j] = 'K' ;break ;
56 case 'S': ans[j] = 'L' ;break ;
57 case 'D': ans[j] = 'M' ;break ;
58 case 'F': ans[j] = 'N' ;break ;
59 case 'G': ans[j] = 'O' ;break ;
60 case 'H': ans[j] = 'P' ;break ;
61 case 'J': ans[j] = 'Q' ;break ;
62 case 'K': ans[j] = 'R' ;break ;
63 case 'L': ans[j] = 'S' ;break ;
64 case 'Z': ans[j] = 'T' ;break ;
65 case 'X': ans[j] = 'U' ;break ;
66 case 'C': ans[j] = 'V' ;break ;
67 case 'V': ans[j] = 'W' ;break ;
68 case 'B': ans[j] = 'X' ;break ;
69 case 'N': ans[j] = 'Y' ;break ;
70 case 'M': ans[j] = 'Z' ;break ;
71 case '\0': ans[j] = '\0' ;break ;
72 }
73 }
74
75 //考虑奇偶的问题
76 nn = geshu/(double)2 ;
77 geshu /= 2 ;
78 if( geshu != nn )
79 ++geshu ;
80 //归并
81 for( i=0,j=0; i<geshu; ++i,j+=2 ){
82 aaa[j] = ans[i] ;
83 aaa[j+1] = ans[i+geshu] ;
84 }
85 aaa[j]='\0';
86 //逆序输出
87 for( i=0; aaa[i]; ++i )
88 ;
89 for( j=i-1; j>=0; --j )
90 printf("%c", aaa[j] ) ;
91 printf("\n") ;
92 memset(a,0,sizeof(a));
93 }
94 return 0 ;
95 }
96
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