UVa 10420 List of Conquests
题意就是有N个pl妹子,然后每行第一个单词是妹子的国籍,后面是妹子的名字。
你的任务就是统计相同国籍妹子的个数,然后按字母表顺序输出。
我首先把所有的国籍都读入,然后用qsort()按字母表顺序排序。
List of Conquests
Input: standard input
Output: standard output
Time Limit: 2 seconds
In Act I, Leporello is telling Donna Elviraabout his master's long list of conquests:
``This is the list of the beauties my master has loved, a list I've madeout myself: take a look, read it with me. In Italy six hundred and forty, inGermany 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 hismaster's conquest to others because he needs to count the number of``beauties'' by their nationality each time. You are to help Leporello tocount.
Input
The input consists of at most 2000 lines, but the first. The first linecontains a number n,indicating that there will be n more lines. Each following line, withat most 75 characters, contains a country (thefirst 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 onlyone word.
Output
The output consists of lines in alphabetical order. Eachline starts with the name of a country, followed by the total number of womenGiovanni 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
Problem-setter: Thomas Tang,Queens University, Canada
AC代码:
//#define LOCAL
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std; int main(void)
{
#ifdef LOCAL
freopen("10420in.txt", "r", stdin);
#endif int cmp(const void *a, const void *b);
int N, i, j;
char country[][], c[];
scanf("%d", &N);
for(i = ; i < N; ++i)
{
scanf("%s", country[i]);
gets(c);
}
qsort(country, N, sizeof(country[]), cmp); i = ;
while(i < N)
{
j = i;
while(strcmp(country[j], country[i]) == && j < N)
{
++j;
}
cout << country[i] << " " << j - i << endl;
i = j;
}
return ;
}
int cmp(const void *a, const void *b)
{
return strcmp((char *)a, (char *)b);
}
代码君
UVa 10420 List of Conquests的更多相关文章
- [算法练习] UVA 10420 - List of Conquests?
UVA Online Judge 题目10420 - List of Conquests 问题描述: 题目很简单,给出一个出席宴会的人员列表,包括国籍和姓名(姓名完全没用).统计每个国家有多少人参加, ...
- UVA题目分类
题目 Volume 0. Getting Started 开始10055 - Hashmat the Brave Warrior 10071 - Back to High School Physics ...
- Volume 1. Sorting/Searching(uva)
340 - Master-Mind Hints /*读了老半天才把题读懂,读懂了题输出格式没注意,结果re了两次. 题意:先给一串数字S,然后每次给出对应相同数目的的一串数字Si,然后优先统计Si和S ...
- 刘汝佳 算法竞赛-入门经典 第二部分 算法篇 第五章 3(Sorting/Searching)
第一题:340 - Master-Mind Hints UVA:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Item ...
- uva 1354 Mobile Computing ——yhx
aaarticlea/png;base64,iVBORw0KGgoAAAANSUhEUgAABGcAAANuCAYAAAC7f2QuAAAgAElEQVR4nOy9XUhjWbo3vu72RRgkF5
- UVA 10564 Paths through the Hourglass[DP 打印]
UVA - 10564 Paths through the Hourglass 题意: 要求从第一层走到最下面一层,只能往左下或右下走 问有多少条路径之和刚好等于S? 如果有的话,输出字典序最小的路径 ...
- UVA 11404 Palindromic Subsequence[DP LCS 打印]
UVA - 11404 Palindromic Subsequence 题意:一个字符串,删去0个或多个字符,输出字典序最小且最长的回文字符串 不要求路径区间DP都可以做 然而要字典序最小 倒过来求L ...
- UVA&&POJ离散概率与数学期望入门练习[4]
POJ3869 Headshot 题意:给出左轮手枪的子弹序列,打了一枪没子弹,要使下一枪也没子弹概率最大应该rotate还是shoot 条件概率,|00|/(|00|+|01|)和|0|/n谁大的问 ...
- UVA计数方法练习[3]
UVA - 11538 Chess Queen 题意:n*m放置两个互相攻击的后的方案数 分开讨论行 列 两条对角线 一个求和式 可以化简后计算 // // main.cpp // uva11538 ...
随机推荐
- 一个奇怪的网络故障 默认网关为0.0.0.0(Windows)
用IPCONFIG命令看到的情况是这样: Windows IP 配置 以太网适配器 本地连接 : 连接特定的 DNS 后缀 . . . . . . . : 本地链接 IPv6 地址. . . . . ...
- HDOJ 1709 The Balance(母函数)
The Balance Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total ...
- SQL分页查询总结{转}
开发过程中经常遇到分页的需求,今天在此总结一下吧.简单说来方法有两种,一种在源上控制,一种在端上控制.源上控制把分页逻辑放在SQL层:端上控制一次性获取所有数据,把分页逻辑放在UI上(如GridVie ...
- ssh 远程 centos 乱码
今天,帮我们同学处理一下中文显示乱码的问题.这个是个国内Linux用户烦恼的问题,由于大部分的Linux发行版都是以英语为主体的,而且英文在通用性和稳定性上都比中文要好一些,各种奇怪的BUG也要少一点 ...
- POC
大概就是原型验证的意思 验证概念 编辑 概念验证(Proof of concept,简称POC)是对某些想法的一个不完整的实现,以证明其可行性,示范其原理,其目的是为了验证一些概念或理论.在计算机安全 ...
- Ubuntu下开启ssh服务
网上有很多介绍在Ubuntu下开启SSH服务的文章,但大多数介绍的方法测试后都不太理想,均不能实现远程登录到Ubuntu上,最后分析原因是都没有真正开启ssh-server服务.最终成功的方法如下: ...
- jquery ajax post 传递数组 ,多checkbox 取值
jquery ajax post 传递数组 ,多checkbox 取值 http://w8700569.iteye.com/blog/1954396 使用$.each(function(){});可以 ...
- 评论 “App死亡潮:400万应用僵尸超八成,周期仅10月”
点这里 原文: App死亡潮:400万应用僵尸超八成,周期仅10月 时间 2015-04-05 22:48:19 和讯科技相似文章 (16)原文 http://tech.hexun.com/201 ...
- poj 3317 Stake Your Claim 极大极小搜索
思路:为了方便,当c1>c2时将0变为1,1变为0. 空格最多有10个,每个空格有3个状态,如果不状态压缩,会TLE的.所以最多有3^10种情况 代码如下: #include<iostre ...
- tomcat下context.xml中JNDI数据源配置
jndi(Java Naming and Directory Interface,Java命名和目录接口)是一组在Java应用中访问命名和目录服务的API.命名服务将名称和对象联系起来,使得我们可以用 ...