A. Vasya and Football
2 seconds
256 megabytes
standard input
standard output
Vasya has started watching football games. He has learned that for some fouls the players receive yellow cards, and for some fouls they receive red cards. A player who receives the second yellow card automatically receives a red card.
Vasya is watching a recorded football match now and makes notes of all the fouls that he would give a card for. Help Vasya determine all the moments in time when players would be given red cards if Vasya were the judge. For each player, Vasya wants to know
only thefirst moment of time when he would receive a red card from Vasya.
The first line contains the name of the team playing at home. The second line contains the name of the team playing away. Both lines are not empty. The lengths of both lines do not exceed 20. Each line contains only of large English letters. The names of the
teams are distinct.
Next follows number n (1 ≤ n ≤ 90)
— the number of fouls.
Each of the following n lines contains information about a foul in the following form:
- first goes number t (1 ≤ t ≤ 90)
— the minute when the foul occurs; - then goes letter "h" or letter "a"
— if the letter is "h", then the card was given to a home team player, otherwise the card was given to an away team player; - then goes the player's number m (1 ≤ m ≤ 99);
- then goes letter "y" or letter "r"
— if the letter is "y", that means that the yellow card was given, otherwise the red card was given.
The players from different teams can have the same number. The players within one team have distinct numbers. The fouls go chronologically, no two fouls happened at the same minute.
For each event when a player received his first red card in a chronological order print a string containing the following information:
- The name of the team to which the player belongs;
- the player's number in his team;
- the minute when he received the card.
If no player received a card, then you do not need to print anything.
It is possible case that the program will not print anything to the output (if there were no red cards).
MC
CSKA
9
28 a 3 y
62 h 25 y
66 h 42 y
70 h 25 y
77 a 4 y
79 a 25 y
82 h 42 r
89 h 16 y
90 a 13 r
MC 25 70
MC 42 82
CSKA 13 90
#include <iostream>
#include <cstring>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
using namespace std; int b[100], c1[110], c2[110], c3[110], c4[110]; struct node {
int t;
char c;
int m;
char d;
}a[100]; int cmp(node a, node b) {
return a.t < b.t;
} int main() {
string str1, str2;
cin >> str1 >> str2;
int n;
cin >> n;
memset(a, 0, sizeof(a)); for (int i = 0; i < n; i++)
cin >> a[i].t >> a[i].c >> a[i].m >> a[i].d;
sort(a, a + n, cmp); memset(b, 0, sizeof(b));
memset(c1, 0, sizeof(c1));
memset(c2, 0, sizeof(c2));
memset(c3, 0, sizeof(c2));
memset(c4, 0, sizeof(c2));
int num = 0;
//int flag1 = 0, flag2 = 0;
for (int i = 0; i < n; i++) {
if (a[i].d == 'r' && a[i].c == 'h' && c3[a[i].m] == 0) {
b[num++] = i;
c3[a[i].m] ++; }
else if (a[i].d == 'r' && a[i].c == 'a' && c4[a[i].m] == 0) {
b[num++] = i;
c4[a[i].m] ++; } else if (a[i].d == 'y' && c1[a[i].m] == 1 && a[i].c == 'h' && c3[a[i].m] == 0) {
b[num++] = i;
c3[a[i].m] ++;
}
else if (a[i].d == 'y' && c2[a[i].m] == 1 && a[i].c == 'a' && c4[a[i].m] == 0) {
b[num++] = i;
c4[a[i].m] ++;
}
else if (a[i].d == 'y' && a[i].c == 'h')
c1[a[i].m] ++;
else if (a[i].d == 'y' && a[i].c == 'a')
c2[a[i].m] ++;
}
for (int i = 0; i < num; i++) {
if (a[b[i]].c == 'h')
cout << str1 << " " << a[b[i]].m << " " << a[b[i]].t << endl;
else
cout << str2 << " " << a[b[i]].m << " " << a[b[i]].t << endl;
}
return 0;
}
A. Vasya and Football的更多相关文章
- cf493A Vasya and Football
A. Vasya and Football time limit per test 2 seconds memory limit per test 256 megabytes input standa ...
- Codeforces Round #281 (Div. 2) A. Vasya and Football 模拟
A. Vasya and Football 题目连接: http://codeforces.com/contest/493/problem/A Description Vasya has starte ...
- Codeforces Round #281 (Div. 2) A. Vasya and Football 暴力水题
A. Vasya and Football time limit per test 2 seconds memory limit per test 256 megabytes input standa ...
- Codeforces Round #281 (Div. 2) A. Vasya and Football 暴力
A. Vasya and Football Vasya has started watching football games. He has learned that for some foul ...
- codeforces 493A. Vasya and Football 解题报告
题目链接:http://codeforces.com/contest/493/problem/A 题目意思:给出两个字符串,分别代表 home 和 away.然后有 t 个player,每个playe ...
- Codeforces Round #281 (Div. 2) A. Vasya and Football(模拟)
简单题,却犯了两个错误导致WA了多次. 第一是程序容错性不好,没有考虑到输入数据中可能给实际已经罚下场的人再来牌,这种情况在system测试数据里是有的... 二是chronologically这个词 ...
- CF493A Vasya and Football 题解
Content 有两个球队在踢足球,现在给出一些足球运动员被黄牌或红牌警告的时间,求每个队员第一次被红牌警告的时间. 注意:根据足球比赛规则,两张黄牌自动换成一张红牌. 数据范围:比赛时间 \(90\ ...
- Codeforces Round #281 (Div. 2)
题目链接:http://codeforces.com/contest/493 A. Vasya and Football Vasya has started watching football gam ...
- Codeforces Round #281 (Div. 2) A 模拟
A. Vasya and Football time limit per test 2 seconds memory limit per test 256 megabytes input standa ...
随机推荐
- MySQL sql语句获取当前日期|时间|时间戳
1.1 获得当前日期+时间(date + time)函数:now() mysql> select now();+———————+| now() |+———————+| 2013-04-08 20 ...
- Java SE 8 流库(三)
1.7. Optional类型 容器对象,可能包含或不包含非空值.如果存在一个值,isPresent()将返回true,get()将返回值.还提供了依赖于包含值是否存在的附加方法,如orElse()( ...
- 回顾2017系列篇(一):最佳的11篇UI/UX设计文章
2017已经接近尾声,在这一年中,设计领域发生了诸多变化.也是时候对2017年做一个总结,本文主要是从2017设计文章入手,列出了个人认为2017设计行业里最重要的UI/UX文章的前11名,供大家参考 ...
- 利用VSTS跟Kubernetes进行CI/CD
准备VSTS管理环境 首先我们需要到www.visualstudio.com下申请好的VSTS账号,然后在账号下创建一个用Git作为代码管理的项目 创建好项目后我们就可以利用git clone将代码库 ...
- 跟我一起,利用bitcms内容管理系统从0到1学习小程序开发:一、IIS下SSL环境搭建
缘起 1.从事互联网十来年了,一直想把自己的从事开发过程遇到的问题给写出来,分享给大家.可是可是这只是个种想法,想想之后就放下了,写出来的类文章是少之又少.古人说无志之人常立志,有志之人立长志.今天, ...
- Linux中ls对文件进行按大小排序和按时间排序,设置ls时间格式
1 按文件大小排序 使用 ll -S | grep '^[^d]' // 格式化文件大小形式 ll -Sh | grep '^[^d]' 2 按文件修改时间排序显示 使用 ll -rt 3 设置ls ...
- 《深入理解java虚拟机》 - 需要一本书来融汇贯通你的经验(下)
上一章讲到了类的加载机制,主要有传统派的 双亲委派模型 和 现代主义激进派的 osgi 类加载器.接下来继续. 第8章 虚拟机字节码执行引擎 局部变量表,用于存储方法参数和方法内部定义的局部变量. 操 ...
- webpack之loader实践
初识前端模板概念的开发者,通常都使用过underscore的template方法,非常简单好用,支持赋值,条件判断,循环等,基本可以满足我们的需求. 在使用Webpack搭建开发环境的时候,如果要使用 ...
- JSP EL隐含对象
JSP 内置对象 JSP EL隐含对象 描述 page pageScope page 作用域 request requestScope request 作用域 session sessionScope ...
- Python的可变类型与不可变类型
Python基础知识,自己写一写比较不容易忘 Python的每个对象都分为可变和不可变,主要的核心类型中,数字.字符串.元组是不可变的,列表.字典是可变的. 对不可变类型的变量重新赋值,实际上是重新创 ...