UVA10420 List of Conquests

今天除夕哈～祝acmer新年多多AC～～

这次又因为英文没看懂题目，以为名字有多个。这个题目没有考虑重名的情况，应该是个小bug吧。虽然是简单的题，不过也学习了qsort套用结构体排序，可以参考七种qsort排序方法。

题目：

Problem B - List of Conquests

Problem B
List of Conquests
Input: standard input
Output: standard output
Time Limit: 2 seconds

In Act I, Leporello is telling Donna Elvira about his master's long list of conquests:

``This is the list of the beauties my master has loved, a list I've made out myself: take a look, read it with me. In Italy six hundred and forty, in Germany two hundred and thirty-one, a hundred in France, ninety-one in Turkey; but in Spain already a thousand and three! Among them are country girls, waiting-maids, city beauties; there are countesses, baronesses, marchionesses, princesses: women of every rank, of every size, of every age.'' (Madamina, il catalogo è questo)

As Leporello records all the ``beauties'' Don Giovanni ``loved'' in chronological order, it is very troublesome for him to present his master's conquest to others because he needs to count the number of ``beauties'' by their nationality each time. You are to help Leporello to count.

Input

The input consists of at most 2000 lines, but the first. The first line contains a number n, indicating that there will be n more lines. Each following line, with at most 75 characters, contains a country (the first word) and the name of a woman (the rest of the words in the line) Giovanni loved. You may assume that the name of all countries consist of only one word.

Output

The output consists of lines in alphabetical order. Each line starts with the name of a country, followed by the total number of women Giovanni loved in that country, separated by a space.

Sample Input

3

Spain Donna Elvira

England Jane Doe

Spain Donna Anna

Sample Output

England 1

Spain 2

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Problem-setter: Thomas Tang, Queens University, Canada

 

 

“Failure to produce a reasonably good and error free problem set illustrates two things a) Lack of creativity b) Lack of commitment”

解答：

 

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<stdlib.h>
 4 struct list
 5 {
 6     int cnt;
 7     char name[100];
 8 };
 9 int cmp_list(const void *a,const void *b)
10 {
11     return strcmp(((list *)a)->name,((list *)b)->name);
12 }
13 list con[2013];
14 char str[100],tmp[100];
15 int main()
16 {
17     int t,i,j,num=0;
18     scanf("%d",&t);
19     for(i=0;i<t;i++)
20     {
21         scanf("%s",str);
22         gets(tmp);
23         int flag=0;
24         for(j=0;j<num;j++)
25         {
26             if(strcmp(str,con[j].name)==0)
27             {
28                 flag=1;
29                 con[j].cnt++;
30                 break;
31             }
32         }
33         if(flag==0)
34         {
35             sprintf(con[num].name,"%s",str);
36             con[num].cnt++;
37             num++;
38         }
39     }
40     qsort(con,num,sizeof(list),cmp_list);
41     for(i=0;i<num;i++)
42         printf("%s %d\n",con[i].name,con[i].cnt);
43 }