hdu2093

这一题其实不难，要注意输出和 ‘-’ 的次数以及排序，未用qsort的时候一直超时，晕死了！！！

 1 #include <stdio.h>
 2 #include <stdlib.h>
 3 #include <string.h>
 4 
 5 struct infor{
 6     char name[11];
 7     int count,time;
 8 }people[1000];
 9 
10 int m;
11 
12 int cmp( const void *a,const void *b ){
13     struct infor *c =( struct infor * )a;
14     struct infor *d =( struct infor * )b;
15     if( c ->count != d ->count )return d ->count - c ->count;
16     else if( c->time != d -> time )return c -> time - d -> time;
17     else return c -> name - d -> name;
18 }
19 
20 int main()
21 {
22     int deal( char p[] );
23     int n, i, j, k;
24     char p[100];
25 
26     memset( people, 0, sizeof( people ) );
27 
28     scanf( "%d%d", &n, &m );
29 
30     k = 0;
31     while(scanf("%s",people[k].name)!=EOF){
32         for( j = 0; j < n; j ++ ){
33             scanf( "%s", p );
34             if( deal( p ) > 0 ){
35                 people[k].count ++;
36                 people[k].time += deal( p );
37             }
38         }
39         k ++;
40     }
41 
42     qsort( people, k, sizeof( people[0] ), cmp );
43 
44     for( i = 0; i < k; i ++ ){
45         printf( "%-10s %2d %4d\n", people[i].name, people[i].count, people[i].time);
46     }
47     return 0;
48 }
49 int deal( char p[] ){
50     int i, j, lenth, wrong, time, t;
51 
52     if( p[0] == '-' )
53         return 0;
54 
55     for( i = 0; p[i] != '(' && p[i] != '\0'; i ++ );
56 
57     lenth = strlen( p );
58     wrong = time = 0;
59     for( j = i - 1, t = 1; j >= 0; j -- ){
60         time += ( p[j] - '0' ) * t;
61         t *= 10;
62     }
63     if( p[lenth-1] == ')' ){
64         for( i = lenth - 2, t = 1; p[i] != '('; i -- ){
65             wrong += ( p[i] - '0' ) * t;
66             t *= 10;
67         }
68         time += wrong * m;
69     }
70 
71     return time;
72 }

 

posted @ 2013-03-01 19:28  hacker_hzh  阅读(180)  评论(0)    收藏  举报