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【HDOJ】3242 List Operations

时间:2015-02-04 01:59:56      阅读:148      评论:0      收藏:0      [点我收藏+]

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注意以下数据:
[] ++ []
[] -- [1,2]
[1,2] ++ []
[] ++ [1]
[2] ++ []
[2,3] -- []
结果是
[]
[]
[1,2]
[1]
[2]
[2,3]

 1 /* 3242 */
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 
 6 #define MAXN 205
 7 
 8 typedef struct {
 9     char s[81];
10     bool f;
11 } word_t;
12 
13 word_t words[2][MAXN];
14 int n[2], len[2];
15 char src[2][MAXN];
16 char op[5];
17 
18 int main() {
19     int i, j, k;
20     
21     #ifndef ONLINE_JUDGE
22         freopen("data.in", "r", stdin);
23     #endif
24     
25     while (scanf("%s", src[0])!=EOF && (src[0][0]!=.)) {
26         scanf("%s %s", op, src[1]);
27         n[0] = n[1] = 0;
28         for (i=0; i<2; ++i) {
29             k = 0;
30             for (j=1; src[i][j]; ++j) {
31                 if (src[i][j]==, || src[i][j]==]) {
32                     if (src[i][j-1] != [) {
33                         words[i][n[i]].s[k] = \0;
34                         words[i][n[i]].f = true;
35                         n[i]++;
36                         k = 0;
37                     }
38                 } else {
39                     words[i][n[i]].s[k++] = src[i][j];
40                 }
41             }
42             len[i] = j;
43         }
44         
45         if (op[0] == +) {
46             src[0][len[0]-1] = \0;
47             printf("%s", src[0]);
48             if (n[1] == 0) {
49                 printf("]\n");
50             } else {
51                 if (n[0] == 0)
52                     printf("%s\n", src[1]+1);
53                 else
54                     printf(",%s\n", src[1]+1);
55             }
56         } else {
57             for (j=0; j<n[1]; ++j) {
58                 for (i=0; i<n[0]; ++i)  {
59                     if (words[0][i].f && strcmp(words[0][i].s, words[1][j].s)==0) {
60                         words[0][i].f = false;
61                         break;
62                     }
63                 }
64             }
65             printf("[");
66             i = 0;
67             while (i < n[0]) {
68                 if (words[0][i].f)
69                     break;
70                 ++i;
71             }
72             if (i < n[0]) {
73                 printf("%s", words[0][i].s);
74                 ++i;
75                 while (i < n[0]) {
76                     if (words[0][i].f)
77                         printf(",%s", words[0][i].s);
78                     ++i;
79                 }
80             }
81             printf("]\n");
82         }
83     }
84     
85     return 0;
86 }

 

【HDOJ】3242 List Operations

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原文地址:http://www.cnblogs.com/bombe1013/p/4271425.html

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