2073. Log Files

Time limit: 1.0 second
Memory limit: 64 MB
Nikolay has decided to become the best programmer in the world! Now he regularly takes part in various programming contests, attentively listens to problems analysis and upsolves problems. But the point is that he had participated in such a number of contests that got totally confused, which problems had already been solved and which had not. So Nikolay conceived to make a program that could read contests’ logs and build beautiful summary table of the problems. Nikolay is busy participating in a new contest so he has entrusted this task to you!

Input

The first line contains an integer n (1 ≤ n ≤ 100). It‘s the number of contests‘ descriptions. Then descriptions are given. The first line of description consists of from 1 to 30 symbols — Latin letters, digits and spaces — and gives the name of contest. It‘s given that the name doesn‘t begin and doesn’t end with a space. In the second line of description the date of contest in DD.MM.YY format is given. It‘s also given that the date is correct and YY can be from 00 to 99 that means date from 2000 till 2099. In the third line of description there are numbers p and sseparated by space (1 ≤ p ≤ 13, 0 ≤ s ≤ 100). It‘s amount of problems and Nikolay’s submits in the contest. Then s lines are given. These are submits’ descriptions. Description of each submit consists of the problem‘s letter and the judge verdict separated by space. The letter of the problem is the title Latin letter and all problems are numbered by first p letters of English alphabet. The judge verdict can be one of the following: Accepted, Wrong Answer, Runtime Error, Time Limit Exceeded, Memory Limit Exceeded, Compilation Error.

Output

Print the table, which consists of n+1 lines and 3 columns. Each line (except the first) gives the description of the contest. The first column gives the name of the contest, the second column gives the date of the contest (exactly as it was given in the input), the third column gives the description of the problems. Every description of problems is the line of 13 characters, where thei-th character correlate with the i-th problem. If the problem got verdict Accepted at least one time, this character is ’o’. If the problem was submitted at least once but wasn’t accepted, the character is ’x’. If the problem was just given at the contest but wasn’t submitted, the character is ’.’. Otherwise, the character is ’ ’ (space). Contests in the table must be placed in the same order as in input.
Column with the name of the contest consists of 30 symbols (shorter names must be extended by spaces added to the right to make this length). Columns with the date and description of problems consist of 8 and 13 characters accordingly.
The first line of the table gives the names of columns. The boundaries of the table are formatted by ’|’, ’-’ и ’+’ symbols. To get detailed understanding of the output format you can look at the example.

Sample

input output
2
Codeforces Gamma Round 512
29.02.16
5 4
A Accepted
B Accepted
C Accepted
E Accepted
URKOP
17.10.15
12 11
A Accepted
B Wrong Answer
B Time Limit Exceeded
J Accepted
B Accepted
J Time Limit Exceeded
J Accepted
F Accepted
E Runtime Error
H Accepted
E Runtime Error
+------------------------------+--------+-------------+
|Contest name |Date |ABCDEFGHIJKLM|
+------------------------------+--------+-------------+
|Codeforces Gamma Round 512 |29.02.16|ooo.o |
+------------------------------+--------+-------------+
|URKOP |17.10.15|oo..xo.o.o.. |
+------------------------------+--------+-------------+
Problem Author: Kirill Borozdin (prepared by Kirill Borozdin, Alexey Danilyuk)
Problem Source: Ural Regional School Programming Contest 2015
Difficulty: 158 
 
大模拟
 #include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <ctime>
#include <iomanip>
using namespace std; const int M = ;
char state[M]; inline void solve()
{
char str[M];
for(int i = ; i < M; i++) str[i] = ' '; gets(str);
for(int i = strlen(str); i < M; i++) str[i] = ' ';
printf("|");
for(int i = ; i < ; i++) printf("%c", str[i]); gets(str);
for(int i = strlen(str); i < M; i++) str[i] = ' ';
printf("|");
for(int i = ; i < ; i++) printf("%c", str[i]); int n, m;
scanf("%d%d", &n, &m);
getchar();
for(int i = ; i < M; i++) state[i] = ' ';
for(int i = ; i < n; i++) state[i] = '.';
while(m--)
{
char pro = getchar();
getchar();
gets(str);
int x = pro - 'A';
if(str[] == 'A') state[x] = 'o';
else if(state[x] != 'o') state[x] = 'x';
}
printf("|");
for(int i = ; i < ; i++) printf("%c", state[i]); printf("|\n");
} int main()
{
puts("+------------------------------+--------+-------------+");
puts("|Contest name |Date |ABCDEFGHIJKLM|"); int t;
scanf("%d", &t);
getchar();
while(t--)
{
puts("+------------------------------+--------+-------------+");
solve();
}
puts("+------------------------------+--------+-------------+");
return ;
}

ural 2073. Log Files的更多相关文章

  1. URAL 2073 Log Files (模拟)

    题意:给定 n 场比赛让你把名称,时间,比赛情况按要求输出. 析:很简单么,按照要求输出就好,注意如果曾经AC的题再交错了,结果也是AC的. 代码如下: #pragma comment(linker, ...

  2. How to configure Veritas NetBackup (tm) to write Unified and Legacy log files to a different directory

    Problem DOCUMENTATION: How to configure Veritas NetBackup (tm) to write Unified and Legacy log files ...

  3. How to delete expired archive log files using rman?

    he following commands will helpful to delete the expired archive log files using Oracle Recovery Man ...

  4. Common Linux log files name and usage--reference

    reference:http://www.coolcoder.in/2013/12/common-linux-log-files-name-and-usage.html if you spend lo ...

  5. 14.7.2 Changing the Number or Size of InnoDB Redo Log Files 改变InnoDB Redo Log Files的数量和大小

    14.7.2 Changing the Number or Size of InnoDB Redo Log Files 改变InnoDB Redo Log Files的数量和大小 改变 InnoDB ...

  6. 14.5.2 Changing the Number or Size of InnoDB Redo Log Files 改变InnoDB Redo Log Files的数量

    14.5.2 Changing the Number or Size of InnoDB Redo Log Files 改变InnoDB Redo Log Files的数量 改变InnoDB redo ...

  7. How to Collect Bne Log Files for GL Integrators

    In this Document   Goal   Solution APPLIES TO: Oracle General Ledger - Version 11.0 and laterInforma ...

  8. EBS R12 LOG files 位置

    - Apache, OC4J and OPMN: $LOG_HOME/ora/10.1.3/Apache$LOG_HOME/ora/10.1.3/j2ee$LOG_HOME/ora/10.1.3/op ...

  9. 手动创建binary log files和手动编辑binary log index file会有什么影响

    基本环境:官方社区版MySQL 5.7.19 一.了解Binary Log结构 1.1.High-Level Binary Log Structure and Contents • Binlog包括b ...

随机推荐

  1. windows端口备忘

    FTP 端口号21 SSH 端口号22 Telnet 端口号23

  2. java链式编程设计

    一般情况下,对一个类的实例和操作,是采用这种方法进行的: Channel channel = new Channel(); channel.queueDeclare(QUEUE_NAME, true, ...

  3. oracle dataguard (一)

    一.什么是data guard及data guard的工作原理 Data Guard 是一个集合,由一个primary数据库(生产数据库)及一个或多个standby数据库(最多9个)组成.组成Data ...

  4. iOS - 日期的时间差(某年某月某日的某一天。。。)

    //首先创建格式化对象 NSDateFormatter *dateFormatter = [[NSDateFormatter alloc] init]; [dateFormatter setDateF ...

  5. Jmeter 中使用非GUI启动进行压力测试

    使用非 GUI 模式,即命令行模式运行 JMeter 测试脚本能够大大缩减所需要的系统资源.使用命令jmeter -n -t <testplan filename> -l <list ...

  6. 数据结构和算法 – 4.字符串、 String 类和 StringBuilder 类

    4.1.String类的应用 class String类应用 { static void Main(string[] args) { string astring = "Now is The ...

  7. 【C#】Json数据 排版算法

    我从服务器上取得一串Json数据,然后想表示到画面上.不过服务器上取下的Json数据肯定是经过压缩的,空格和换行都没有.如果直接看,可读性非常差. 由于我这个软件是内部管理用的,使用者既能直接看懂Js ...

  8. GoLang文件增删遍历基本操作

    先学一学GO语言实用的一面. package main import ( "path/filepath" "flag" "os" " ...

  9. POJ3294 Life Forms(后缀数组)

    引用罗穗骞论文中的话: 将n 个字符串连起来,中间用不相同的且没有出现在字符串中的字符隔开,求后缀数组.然后二分答案,用和例3 同样的方法将后缀分成若干组,判断每组的后缀是否出现在不小于k 个的原串中 ...

  10. annotation-config 和 component-scan 的区别

    <context:annotation-config> 和 <context:component-scan>是Spring Core里面的两个基础概念,每个使用者都有必要理解怎 ...