排序检索B - List of Conquests
2015-08-18 16:03
246 查看
这题题目长不难
每句话第一个单词是国家
输出是统计国家的名字和出现的次数
输出之前要根据国家名字排序
因为要对应国家的名字和次数
用结构体非常方便
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 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. The rst 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 rst 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.
SampleInput
3
Spain Donna Elvira
England Jane Doe
Spain Donna Anna
SampleOutput
England 1
Spain 2
每句话第一个单词是国家
输出是统计国家的名字和出现的次数
输出之前要根据国家名字排序
因为要对应国家的名字和次数
用结构体非常方便
#include <stdio.h> #include <string.h> #include <ctype.h> #define N 100000 struct list{ char name[2000]; int num; }s , ss; int main(){ int n, mark, num = 0; char c[2000], city[2000]; scanf("%d", &n); getchar(); while (n--) { gets(c); for (int i = 0; i < strlen(c); i++) if (c[i] == ' ') { mark = i; break; } for (int i = 0; i < mark; i++) city[i] = c[i]; city[mark] = '\0'; int k; for (k = 0; k < num; k++) if (strcmp(s[k].name, city) == 0) { s[k].num++; break; } if (k == num) { strcpy(s[k].name, city); s[k].num++; num++; } for (int i = 0; i < num; i++) for (int j = 0; j < num; j++) { if (strcmp(s[i].name, s[j].name) < 0) { ss = s[i]; s[i] = s[j]; s[j] = ss; } } } for (int i = 0; i < num; i++) printf("%s %d\n", s[i].name, s[i].num); return 0; }
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 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. The rst 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 rst 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.
SampleInput
3
Spain Donna Elvira
England Jane Doe
Spain Donna Anna
SampleOutput
England 1
Spain 2
相关文章推荐
- iOS开发系列--UITableView全面解析
- hdoj 1509 Windows Message Queue
- iOS开发 -- UIViewContentMode各类型效果
- UIAlertView的基本用法
- 黑马程序员_String类型以及StringBuilder
- POJ 1986 DIstance Query LCA水题
- RequireJS Optimizer 的使用和配置
- Reveal UI 分析工具简单使用
- 【Mockplus教程】安装Mockplus
- 【Mockplus教程】界面闪烁花屏怎么办?
- 【Mockplus教程】MAC 安全提示无法安装怎么办?
- Unable to determine the principal end of an association between the types '***. The principal end of this association must be explicitly configured using either the relationship fluent API or data annotations.
- 【Mockplus教程】为什么 Windows XP 无法安装?
- 【Mockplus视频教程】《10分钟玩转Mockplus》
- IOS UITableView拖动排序功能
- [leetcode] 187.Repeated DNA Sequences
- POJ---2299-Ultra-QuickSort
- ios-用xib和UI table View controller 的团购网站
- GUID全球唯一ID
- android中更新UI的方法