刘汝佳算法竞赛-入门经典第二部分算法篇第五章 1（String）

第一题：401 - Palindromes

UVA ： http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=96&page=show_problem&problem=342

解题思路：此题很水，只要把 mirrored string 类的对应关系搞对，基本就可以了！但是细节要注意，首先只有一个元素的时候需要单独判断，一个字符是回文串，是不是 mirrored string 则需要判断，另外最后输出中间隔了一行！

解题代码：

 1 // File Name: 刘汝佳-基础/part2/Palindromes/401.cpp
 2 // Author: sheng
 3 // Created Time: 2013年07月20日 星期六 23时37分45秒
 4 
 5 #include <iostream>
 6 #include <stdio.h>
 7 #include <string.h>
 8 using namespace std;
 9 
10 const int max_n = 100000;
11 const int max_f = 100;
12 
13 int main ()
14 {
15     bool mir[max_f][max_f];
16     char ch[max_n];
17     memset (mir, false, sizeof (mir));
18     mir['A']['A'] = true;
19     mir['E']['3'] = mir['3']['E'] = true;
20     mir['H']['H'] = true;
21     mir['I']['I'] = 1;
22     mir['J']['L'] = mir['L']['J'] = 1;
23     mir['M']['M'] = 1;
24     mir['O']['O'] = 1;
25     mir['S']['2'] = mir['2']['S'] = 1;
26     for (int i = 'T'; i <= 'Y'; i ++)
27         mir[i][i] = 1;
28     mir['Z']['5'] = mir['5']['Z'] = 1;
29     mir['1']['1'] = 1;
30     mir['8']['8'] = 1;
31     while (~scanf ("%s", ch))
32     {
33         bool ms = true, ps = true;
34         int len = strlen (ch);
35 //        cout << len << endl;
36         if (len == 1 && (!mir[ch[0]][ch[0]]))
37             ms = false;
38         for (int i = 0; i < len/2; i ++)
39         {
40             if ( ch[i] != ch[len-1-i] )
41                 ps = false;
42             if (!mir[ch[i]][ch[len-1-i]])
43                 ms = false;
44             if ((!ms) && (!ps))
45                 break;
46         }
47         if (ms && ps)
48             cout << ch << " -- is a mirrored palindrome.\n\n";
49         else if (ms)
50             cout << ch << " -- is a mirrored string.\n\n";
51         else if (ps)
52             cout << ch << " -- is a regular palindrome.\n\n";
53         else  cout << ch << " -- is not a palindrome.\n\n";
54     }
55     return 0;
56 }

View Code

第二题：10010 - Where's Waldorf?

UVA：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=951

解题思路：因为数据范围很小所以可以暴力搜索，总共八个方向，只要有一个方向可以组成当前单词就标记此开始点并跳出，寻找下一个单词！

水题一道，却花了一下午的时间，英语差是一方面，最不该就是使用了goto，害得程序陷入死循环，而我却在没有bug的地方找死循环的原因找了一下午，哎！还需使劲练习啊！

解题代码：

  1 // File Name    :刘汝佳-基础/part2/10010 - Where's Waldorf?
  2 // Author        :Freetion
  3 // Created Time    :2013年07月22日 星期一 14时32分18秒
  4 
  5 #include <string.h>
  6 #include <stdio.h>
  7 
  8 char map[52][52];
  9 bool tag[52][52];
 10 int T;
 11 int n, m, k, q;
 12 char ser[23][100];
 13 int len;
 14 
 15 struct CUN
 16 {
 17     int x, y;
 18 }cun[23];
 19 
 20 const int alt_x[8] = {-1, -1, 1, 1, 0, 0, -1, 1};
 21 const int alt_y[8] = {-1, 1, -1, 1, -1, 1, 0, 0};
 22 
 23 int serch(int x, int y, int l)
 24 {
 25     for (int i = 0; i < 8; i ++)//八个方向
 26     {
 27         int xx = x + alt_x[i];
 28         int yy = y + alt_y[i];
 29         l = 1;
 30         while (l < len && xx > 0 && xx <= n && yy > 0 && yy <= m)//向一个方向找
 31         {
 32             int temp_x = xx + alt_x[i];
 33             int temp_y = yy + alt_y[i];
 34             if (ser[q][l] == map[xx][yy])
 35             {
 36                 l ++;
 37                 xx = temp_x;
 38                 yy = temp_y;
 39             }
 40             else break;
 41         }
 42         if (l == len)
 43             return 1;//找到返回1
 44     }
 45     return 0;//没有返回0
 46 }
 47 
 48 void init()
 49 {
 50     for (int i = 1; i <= n; i ++)
 51     {
 52         for (int j = 1; j <= m; j ++)
 53         {
 54             char ch;
 55             scanf ("%c", &ch);
 56             if (ch <= 'Z')
 57                 ch = ch - 'A' + 'a';
 58             map[i][j] = ch;
 59         }
 60         map[i][m+1] = '\0';
 61         getchar();
 62     }
 63 /*
 64     for (int i = 1; i <= n; i++)
 65     {
 66         for (int j = 1; j <= m; j ++)
 67             printf("%c", map[i][j]);
 68         printf("\n");
 69     }
 70 */
 71     scanf ("%d", &k);
 72     getchar();
 73     int len;
 74     char ch[100];
 75     for (int i = 1; i <= k; i ++)
 76     {
 77         scanf ("%s", ch);
 78         len = strlen(ch);
 79         for (int j = 0; j < len; j ++)
 80         {
 81             if (ch[j] <= 'Z')
 82                 ch[j] = ch[j] - 'A' + 'a';
 83         }
 84         strcpy(ser[i], ch);
 85         ser[i][len] = '\0';
 86     }
 87 /*
 88     for (int i = 1; i <= k; i ++)
 89         printf("len = %d %s\n", strlen(ser[i]), ser[i]);
 90 */
 91 }
 92 
 93 int main ()
 94 {
 95     scanf ("%d", &T);
 96     while (T--)
 97     {
 98         scanf ("%d%d", &n, &m);
 99         getchar();
100         init();
101         for (q = 1; q <= k; q ++)//就是在这里使用了goto导致q一直为1
102         {
103             int tag = 0;
104             len = strlen(ser[q]);
105             char ch = ser[q][0];
106             for (int i = 1; i <= n; i ++)
107             {
108                 for (int j = 1; j <= m; j ++)
109                 {
110                     if (ch == map[i][j])
111                         if (serch(i, j, 1))
112                         {
113                             cun[q] = (CUN){i, j};
114                             tag = 1;
115                         }
116                     if (tag)
117                         break;
118                 }
119                 if (tag)
120                     break;
121             }
122         }
123         for (int i = 1; i <= k; i ++)
124             printf ("%d %d\n", cun[i].x, cun[i].y);
125         if (T)
126             printf ("\n");
127     }
128     return 0;
129 }

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第三题：10361 - Automatic Poetry

UVA：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1302

解题思路：水题一道！

解题代码：

 1 // File Name    :刘汝佳-基础/part2/10361 - Automatic Poetry
 2 // Author        :Freetion
 3 // Created Time    :2013年07月22日 星期一 20时10分53秒
 4 
 5 #include <stdio.h>
 6 #include <string>
 7 #include <string.h>
 8 #include <iostream>
 9 //#define LOCAL
10 using namespace std;
11 
12 const int max_l = 102;
13 char str1[max_l], str2[max_l], str3[max_l], str4[max_l], str5[max_l], str6[max_l];
14 
15 int main()
16 {
17 #ifdef LOCAL
18     freopen("data.in", "r", stdin);
19     freopen("data.out", "w", stdout);
20 #endif
21     int n, num1, num2, num3, num4;
22     while (~scanf("%d", &n))
23     {
24         getchar();
25         for (int i = 0; i < n; i ++)
26         {
27             gets(str1);
28             gets(str2);
29             int len1 = strlen(str1);
30             int tag = 0;
31             num1 = num2 = num3 = num4 = 0;
32             for (int j = 0; j < len1; j ++)
33             {
34                 if (str1[j] != '<' && str1[j] != '>')
35                     printf ("%c", str1[j]);
36                 else if (str1[j] == '<')
37                     tag ++;
38                 else if (str1[j] == '>')
39                     tag ++;
40                 if (tag == 1)
41                     str3[num1++] = str1[j];
42                 else if (tag == 2)
43                     str5[num3++] = str1[j];
44                 else if (tag == 3)
45                     str4[num2++] = str1[j];
46                 else if (tag == 4)
47                     str6[num4++] = str1[j];
48 
49             }
50             printf("\n");
51             int len2 = strlen(str2);
52             for (int j = 0; j < len2-3; j ++)
53                 printf ("%c", str2[j]);
54             for (int j = 1; j < num2; j ++)
55                 printf("%c", str4[j]);
56             for (int j = 1; j < num3; j ++)
57                 printf ("%c", str5[j]);
58             for (int j = 1; j < num1; j ++)
59                 printf ("%c", str3[j]);
60             for (int j = 1; j < num4; j ++)
61                 printf ("%c", str6[j]);
62             printf ("\n");
63 
64         }
65 
66     }
67     return 0;
68 }

View Code

第四题：537 - Artificial Intelligence?

UVA：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=478

解题思路：按关键字将数据提取出来即可，水题！还有，输出是每个案列后都会有一个空行，这里wa三次！ UVA为什么没有PE呢？

解题代码：好吧，可能太长，有点凌乱！

  1 // File Name    :刘汝佳-基础/part2/537 - Artificial Intelligence?
  2 // Author        :Freetion
  3 // Created Time    :2013年07月22日 星期一 22时25分10秒
  4 
  5 #include <stdio.h>
  6 #include <string.h>
  7 
  8 const int max_l = 1000;
  9 char phy[max_l];
 10 
 11 int main()
 12 {
 13     int t, T;
 14     scanf ("%d", &t);
 15     T = t;
 16     getchar();
 17     double U, I, P;
 18     int tag_u = 0, tag_i = 0, tag_p = 0;
 19     while(t--)
 20     {
 21         gets(phy);
 22         int len = strlen(phy);
 23         U = I = P = 0;
 24         for (int i = 0; i < len; i ++)
 25         {
 26             if (phy[i] == 'U' && phy[++i] == '=')
 27             {
 28                 tag_u = 1;
 29                 int tag = 1;
 30                 int tm;
 31                 i ++;
 32                 while(phy[i] == '.' || (phy[i] >= '0' && phy[i] <= '9'))
 33                 {
 34                     if (phy[i] == '.')
 35                     {
 36                         tm = 10;
 37                         tag = 0;
 38                         i ++;
 39                     }
 40                     if (tag)
 41                         U = U*10 + phy[i] - '0';
 42                     else
 43                     {
 44                         U = U + (phy[i] - '0')*1.0/tm;
 45                         tm *= 10;
 46                     }
 47                     i ++;
 48                 }
 49                 if (phy[i] == 'm')
 50                     U = U/1000;
 51                 else if (phy[i] == 'k')
 52                     U = U*1000;
 53                 else if (phy[i] == 'M')
 54                     U = U*1000*1000;
 55             }
 56             if (phy[i] == 'I' && phy[++i] == '=')
 57             {
 58                 tag_i = 1;
 59                 int tag = 1;
 60                 int tm;
 61                 i ++;
 62                 while(phy[i] == '.' || (phy[i] >= '0' && phy[i] <= '9'))
 63                 {
 64                     if (phy[i] == '.')
 65                     {
 66                         tm = 10;
 67                         tag = 0;
 68                         i ++;
 69                     }
 70                     if (tag)
 71                         I = I*10 + phy[i] - '0';
 72                     else
 73                     {
 74                         I = I + (phy[i] - '0')*1.0/tm;
 75                         tm *= 10;
 76                     }
 77                     i ++;
 78                 }
 79                 if (phy[i] == 'm')
 80                     I = I/1000;
 81                 else if (phy[i] == 'k')
 82                     I = I*1000;
 83                 else if (phy[i] == 'M')
 84                     I = I*1000*1000;
 85             }
 86             if (phy[i] == 'P' && phy[++i] == '=')
 87             {
 88                 tag_p = 1;
 89                 int tag = 1;
 90                 int tm;
 91                 i ++;
 92                 while(phy[i] == '.' || (phy[i] >= '0' && phy[i] <= '9'))
 93                 {
 94                     if (phy[i] == '.')
 95                     {
 96                         tm = 10;
 97                         tag = 0;
 98                         i ++;
 99                     }
100                     if (tag)
101                         P = P*10 + phy[i] - '0';
102                     else
103                     {
104                         P = P + (phy[i] - '0')*1.0/tm;
105                         tm *= 10;
106                     }
107                     i ++;
108                 }
109                 if (phy[i] == 'm')
110                     P = P/1000;
111                 else if (phy[i] == 'k')
112                     P = P*1000;
113                 else if (phy[i] == 'M')
114                     P = P*1000*1000;
115             }
116             if (tag_p && tag_u)
117                 break;
118             if (tag_u && tag_i)
119                 break;
120             if (tag_p && tag_i)
121                 break;
122 
123         }
124         printf ("Problem #%d\n", T - t);
125         if (tag_u && tag_i)
126         {
127             tag_u = tag_i = 0;
128             printf ("P=%.2fW\n", U*I);
129         }
130         else if (tag_u && tag_p)
131         {
132             tag_u = tag_p = 0;
133             printf("I=%.2fA\n", P/U);
134         }
135         else
136         {
137             printf("U=%.2fV\n", P/I);
138             tag_i = tag_p = 0;
139         }
140         //if (t)
141             printf ("\n");
142     }
143     return 0;
144 }

View Code

第五题：409 - Excuses, Excuses!

UVA：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=96&page=show_problem&problem=350

解题思路：暴力搜素即可，但须注意判断条件。

解题代码：

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <map>
 4 #include <algorithm>
 5 #include <string.h>
 6 using namespace std;
 7 //#define LOCAL
 8 
 9 struct SENT
10 {
11     int sign, num;
12     bool operator < (const SENT s) const
13     {
14         if (num == s.num)
15             return sign < s.sign;
16         return num > s.num;
17     }
18 }sen[23];
19 
20 int main()
21 {
22 #ifdef LOCAL
23     freopen("data.in", "r", stdin);
24     freopen("data.out", "w", stdout);
25 #endif
26     char key[23][23], sent[23][1000];
27     int n, m, T = 0;
28     while (~scanf("%d%d", &n, &m))
29     {
30         getchar();
31         for (int i = 0; i < n; i++)
32         {
33             scanf ("%s", key[i]);
34             getchar();
35             int len = strlen(key[i]);
36             for (int j = 0; j < len; j ++)
37             {
38                 if (key[i][j] <= 'Z')
39                     key[i][j] = key[i][j] - 'A' + 'a';
40             }
41         }
42         bool tag[23];
43         for (int i = 0; i < m; i ++)
44         {
45             memset(tag, true, sizeof(tag));
46             sen[i].num = 0;
47             sen[i].sign = i;
48             gets(sent[i]);
49         //    puts(sent[i]);
50             int len = strlen(sent[i]);
51             for (int j = 0; j < len; j ++)
52             {
53                 if (j == 0 || ((sent[i][j-1] > 'Z' || sent[i][j-1] < 'A') && (sent[i][j-1] < 'a' || sent[i][j-1] > 'z')))
54                 {//不能仅用空格作为判断条件 
55                     for (int k = 0; k < n; k ++)
56                     {
57                         if ((key[k][0] == sent[i][j] || key[k][0] == sent[i][j]-'A'+'a') && tag[k])
58                         {    
59                             int len_k;
60                             int l;
61                             len_k = strlen(key[k]);
62                             for (l = 1; l < len_k; l ++)
63                             {
64                                 if (j+l < len && key[k][l] != sent[i][j+l] && key[k][l] != sent[i][j+l]-'A'+'a')
65                                     break;
66                             }
67                             if (l == len_k)
68                             {
69                                 tag[k] = false;
70                                 sen[i].num ++;
71                                 j = j+l;
72                             }
73                         }
74                     }
75                 }
76             }
77         }
78         sort(sen, sen+m);
79         printf("Excuse Set #%d\n", ++T);
80         for (int i = 0; i < m; i ++)
81             if (sen[i].num == sen[0].num)
82                 puts(sent[sen[i].sign]);
83         puts("");            
84     }
85     return 0;
86 }

View Code

第六题：10878 - Decode the tape

UVA：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=96&page=show_problem&problem=1819

解题思路：找出输入字符与ASCII码的关系就可输出，这里以output打印出 A 作介绍：对应的字符串：

\|		o				.			o	\|
0	1	2	3	4	5	6	7	8	9	10

这里列出打印 A 字符的依据，并标上列数， A 字符的ASCII码值为 65 而 65 = 2^{（10 -} ^4）+ 2^{(10 - 10)}所以打印 A字符，其余字符也是如此计算！

解题代码：

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <math.h>
 4 using namespace std;
 5 //#define LOCAL
 6 int main()
 7 {
 8 #ifdef LOCAL
 9     freopen("data.in", "r", stdin);
10     freopen("data.out", "w", stdout);
11 #endif
12     char ch;
13     int tag = 0;
14     while (1)
15     {
16         int s = 0;
17         for (int i = 0; i < 11; i++)
18         {
19             scanf ("%c", &ch);
20             if (ch == '_')
21                 tag ++;
22             if (ch == 'o')
23             {
24                 if (i < 6)
25                     s += pow(2, 8-i);
26                 else s += pow(2, 9 - i);
27             }
28         }
29         getchar();
30         if (s)
31             printf("%c", s);
32         if (tag == 22)
33             break;
34     }
35 }

View Code

第七题：10815 - Andy's First Dictionary

uva：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1756

解题思路：输入字符串，找出所有单词，找过过程中，判断是否保存过，保存过就抛弃，没有则保存，并标记已保存，然后排序输出就可。

解题代码：

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <math.h>
 4 #include <algorithm>
 5 #include <string.h>
 6 #include <string>
 7 #include <map>
 8 using namespace std;
 9 
10 #define LOCAL
11 
12 #ifdef WINDOWS
13     #define LL __int64
14     #define LLD "%I64d"
15 #else
16     #define LL long long
17     #define LLD "%lld"
18 #endif
19  
20 int main()
21 {
22 
23 #ifdef LOCAL
24     freopen("data.in", "r", stdin);
25     freopen("data.out", "w", stdout);
26 #endif
27     
28     map <string, int> word;
29     string str[10000], tm1_str, tm2_str;
30     int cun = 0;
31     while (cin >> tm1_str)
32     {
33         int len = tm1_str.length();
34         
35         for (int i = 0; i < len; i ++)
36         {
37             tm2_str.clear();
38             while((tm1_str[i] <= 'Z' && tm1_str[i] >= 'A') || (tm1_str[i] <= 'z' && tm1_str[i] >= 'a'))
39             {
40                 if (tm1_str[i] <= 'Z')
41                     tm2_str += (tm1_str[i] - 'A' + 'a');
42                 else tm2_str += tm1_str[i];
43                 i ++;
44                 if (i >= len)
45                     break;
46             }
47             if (!tm2_str.empty())
48             {
49                 if (word[tm2_str] == 0)
50                 {
51                     str[cun++] = tm2_str;
52                     word[tm2_str] = 1;
53                 }
54             }
55         }
56         
57     }
58     sort(str, str+cun);
59     for (int i = 0; i < cun; i ++)
60         cout << str[i] << endl;
61     return 0;
62 }

View Code

第八题：644 - Immediate Decodability

UVA：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=585

解题思路：先按字典序或字符串长短排个序，然后就挨个找就可以了，字典序也可以是因为比如0000，01后者肯定不是前者的前缀，至于00， 000这一类就需要进行判断！

解题代码：

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <math.h>
 4 #include <algorithm>
 5 #include <string.h>
 6 #include <string>
 7 #include <map>
 8 using namespace std;
 9 
10 //#define LOCAL  //Please annotate this line when you submit
11 
12 #ifdef WINDOWS
13     #define LL __int64
14     #define LLD "%I64d"
15 #else
16     #define LL long long
17     #define LLD "%lld"
18 #endif
19 
20 string str, str1[1000];
21 map <string, bool> IM;
22 
23 int solve(int cun)
24 {
25     int tag = 1;
26     sort(str1, str1+cun);
27     for (int i = 0; i < cun && tag; i ++)
28     {
29         string temp;
30         temp.clear();
31         int len = str1[i].length();
32         for (int j = 0; j < len && tag; j ++)
33         {
34             temp += str1[i][j];
35             if (IM[temp])
36             {
37                 tag = 0;
38                 break;
39             }
40         }
41         IM[str1[i]] = 1;
42         str1[i].clear();
43     }
44     return tag;
45 }
46 
47 int main()
48 {
49 
50 #ifdef LOCAL
51     freopen("data.in", "r", stdin);
52     freopen("data.out", "w", stdout);
53 #endif
54 
55     int T = 0;
56     bool tag = 1;
57     int cun = 0;
58     while (cin>>str)
59     {
60         if (str[0] == '9')
61         {
62             tag = solve(cun);
63             if (tag)
64                 printf("Set %d is immediately decodable\n", ++T);
65             else printf("Set %d is not immediately decodable\n", ++T);
66             tag = 1;
67             IM.clear();
68             cun = 0;
69         }
70         else
71             str1[cun ++] = str;
72     }
73     return 0;
74 }

View Code

第九题：10115 - Automatic Editing

UVA：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1056

解题思路：字符串匹配与替换，水题

解题代码：

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <math.h>
 4 #include <algorithm>
 5 #include <string.h>
 6 #include <string>
 7 #include <map>
 8 using namespace std;
 9 
10 //#define LOCAL  //Please annotate this line when you submit
11 
12 #ifdef WINDOWS
13     #define LL __int64
14     #define LLD "%I64d"
15 #else
16     #define LL long long
17     #define LLD "%lld"
18 #endif
19 
20 const int max_n = 1000;
21 
22 int n;
23 char rep[max_n][max_n], by[max_n][max_n];
24 char str[2][max_n];
25 
26 int init()
27 {
28     if(~scanf ("%d", &n) && n)
29     {
30         getchar();
31         for (int i = 0; i < n; i++)
32         {
33             gets(rep[i]);
34             gets(by[i]);
35         }
36         gets(str[0]);
37     //    puts(str[0]);
38         return 1;
39     }
40     return 0;
41 }
42 
43 int solve()
44 {
45     int pos = 0;
46     for (int i = 0; i < n; i ++)
47     {
48         int len = strlen(str[pos]);
49         for(int j = 0; j < len; j ++)
50         {
51             if(str[pos][j] == rep[i][0])
52             {
53                 int k;
54                 int tm_len = strlen(rep[i]);
55                 for (k = 1; k < tm_len && j+k < len; k ++)
56                 {
57                     if (str[pos][j+k] != rep[i][k])
58                         break;
59                 }
60                 if (tm_len == k)
61                 {
62                     int cun = 0;
63                     for (int r = 0; r < len; r ++)
64                     {
65                         if (r != j)
66                             str[pos^1][cun++] = str[pos][r];
67                         else
68                         {
69                             int by_len = strlen(by[i]);
70                             for(int l = 0; l < by_len; l ++)
71                                 str[pos^1][cun++] = by[i][l];
72                             r = j + tm_len - 1;
73                         }
74                     }
75                     str[pos^1][cun] = '\0';
76                     j = 0;
77                     len = strlen(str[pos^1]);
78                     pos ^= 1;
79                 }
80             }
81         }
82     }
83     return pos;    
84 }
85 
86 int main()
87 {
88     
89 #ifdef LOCAL
90     freopen("data.in", "r", stdin);
91     freopen("data.out", "w", stdout);
92 #endif
93     while(init())
94     {
95         int pos = solve();
96         puts(str[pos]);
97     }
98     return 0;
99 }

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posted on 2013-07-21 01:46 圣手摘星阅读(291) 评论(0) 收藏举报

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