uva10420 - List of Conquests

本文介绍了一个算法问题,旨在帮助角色Leporello统计其主人Don Giovanni在不同国家所追求女性的数量。通过读取输入文件中指定格式的数据,该程序能够快速准确地统计每个国家的女性数量,并按字母顺序输出结果。

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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

开始没看清题目以为可能有重复的人名,直接单个关键字按国家排序即可,不用按人数排序再按国家排序简单很多。

#include <stdio.h>
#include <string.h>
char country[2000][100];
int sum=0;
void sort(int l,int r,int a[])
{int i=l,j=r,x=a[i],ii;
 char ch[100];
 strcpy(ch,country[i]);
 if (l>=r) return;
 while (i<j)
 {while (i<j && (strcmp(country[j],ch)>0)) --j;
  strcpy(country[i],country[j]); a[i]=a[j];
  while (i<j && (strcmp(country[i],ch)<0)) ++i;
  strcpy(country[j],country[i]); a[j]=a[i];
 }
 a[i]=x;  strcpy(country[i],ch);
 sort(l,i-1,a);
 sort(i+1,r,a);
};
void main()
{char s[100],s1[100];
 int t,num[2000],i,j,f;
 scanf("%d\n",&t);
 while (t--)
 {scanf("%s ",&s);
  gets(s1);
  f=1;
  for (i=0;i<sum;i++)
  if (strcmp(country[i],s)==0)
  {++num[i];
   f=0;
  }
  if (f)
  {num[sum]=1;
   strcpy(country[sum],s);
   ++sum;
  }
 }
 sort(0,sum-1,num);
 for (i=0;i<sum;i++)
 printf("%s %d\n",country[i],num[i]);
}

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