zrq495
www.zrq495.com

开始用字符串TLE,后来改成整数AC。先排序,再计数。

代码:

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<stdlib.h>
 4 
 5 typedef struct
 6 {
 7     int num;
 8     int count;
 9 }node;
10 
11 int cmp(const void *a, const void *b)
12 {
13     node *pa=(node *)a;
14     node *pb=(node *)b;
15     return pa->num-pb->num;
16 }
17 
18 int main()
19 {
20     int T, n, i, j, flag=0;
21     char s[50];
22     node tel[100010];
23     scanf("%d", &T);
24     while(T--)
25     {
26         scanf("%d%*c", &n);
27         for (i=0; i<n; i++)
28         {
29             tel[i].num=0;
30             tel[i].count=1;
31             gets(s);
32             for (j=0; j<strlen(s); j++)
33             {
34                 if (s[j] != '-')        //可以这样 int b[]={2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,7,7,7,8,8,8,9,9,9,9};
35                 {                  //
36                     if (s[j]>='0' && s[j]<='9')  //下面的就可一简写很多了。
37                         tel[i].num=tel[i].num*10+s[j]-'0';
38                     if (s[j]=='A' || s[j]=='B' || s[j]=='C')
39                         tel[i].num=tel[i].num*10+2;
40                     if (s[j]=='D' || s[j]=='E' || s[j]=='F')
41                         tel[i].num=tel[i].num*10+3;
42                     if (s[j]=='G' || s[j]=='H' || s[j]=='I')
43                         tel[i].num=tel[i].num*10+4;
44                     if (s[j]=='J' || s[j]=='K' || s[j]=='L')
45                         tel[i].num=tel[i].num*10+5;
46                     if (s[j]=='M' || s[j]=='N' || s[j]=='O')
47                         tel[i].num=tel[i].num*10+6;
48                     if (s[j]=='P' || s[j]=='R' || s[j]=='S')
49                         tel[i].num=tel[i].num*10+7;
50                     if (s[j]=='T' || s[j]=='U' || s[j]=='V')
51                         tel[i].num=tel[i].num*10+8;
52                     if (s[j]=='W' || s[j]=='X' || s[j]=='Y')
53                         tel[i].num=tel[i].num*10+9;
54                 }
55             }
56         }
57         qsort(tel, n, sizeof(node), cmp);    //快排,如果不用可能会TLE。
58         flag=0;
59         for (i=0; i<n-1;)
60         {
61             for (j=i+1; j<n; j++)
62             {
63                 if (tel[i].num==tel[j].num)
64                 {
65                     tel[i].count++;
66                 }
67                 else
68                     break;
69             }
70             if (tel[i].count>1)
71             {
72                 flag=1;
73                 printf("%03d-%04d %d\n", tel[i].num/10000,tel[i].num%10000,tel[i].count);
74             }
75             i=j;
76         }
77         if (!flag)
78             printf("No duplicates.\n");
79         if (T)
80             printf("\n");
81     }
82     return 0;
83 }
posted on 2012-07-18 11:37  zrq495  阅读(243)  评论(0编辑  收藏  举报