Gym 100851E Easy Problemset (模拟题)
Problem E. Easy Problemset
Input file: easy.in
Output file: easy.out
Perhaps one of the hardest problems of any ACM ICPC contest is to create a problemset with a reasonable number of easy problems. On Not Easy European Regional Contest this problem is solved as follows. There are n jury members (judges). They are numbered from 1 to n. Judge number i had prepared pi easy problems before the jury meeting. Each of these problems has a hardness between 0 and 49 (the higher the harder). Each judge also knows a very large (say infinite) number of hard problems (their hardness is 50). Judges need to select k problems to be used on the contest during this meeting.
They start to propose problems in the ascending order of judges numbers. The first judge takes the first problem from his list of remaining easy problems (or a hard problem, if he has already proposed all his easy problems) and proposes it. The proposed problem is selected for the contest if its hardness is greater than or equal to the total hardness of the problems selected so far, otherwise it is considered too easy. Then the second judge does the same etc.; after the n-th judge, the first one proposes his next problem, and so on. This procedure is stopped immediately when k problems are selected. If all judges have proposed all their easy problems, but they still have selected less than k problems, then they take some hard problems to complete the problemset regardless of the total hardness. Your task is to calculate the total hardness of the problemset created by the judges.
Input The first line of the input file contains the number of judges n (2 ≤ n ≤ 10) and the number of problems k (8 ≤ k ≤ 14). The i-th of the following n lines contains the description of the problems prepared by the i-th judge. It starts with pi (1 ≤ pi ≤ 10) followed by pi non negative integers between 0 and 49 — the hardnesses of the problems prepared by the i-th judge in the order they will be proposed.
Output Output the only integer — the total hardness of the selected problems.
Sample input and output
3 8
5 0 3 12 1 10
4 1 1 23 20
4 1 5 17 49
94
3 10
2 1 3
1 1
2 2 5
354
In the first example, three problems with hardnesses of 0, 1, and 1 are selected first. Then the first judge proposes the problem with hardness 3 and it is selected, too. The problem proposed by the second judge with hardness 1 is not selected, because it is too easy. Then the problems proposed by the third, the first, and the second judges are selected (their hardnesses are 5, 12 and 23). The following three proposed problems with hardness of 17, 1 and 20 are not selected, and the problemset is completed with a problem proposed by the third judge with hardness of 49. So the total hardness of the problemset is 94.
In the second example, three problems with hardnesses of 1, 1, and 2 are selected first. The second problem of the first judge (hardness 3) is too easy. The second judge is out of his easy problems, so he proposes a problem with hardness 50 and it is selected. The third judge’s problem with hardness 5 is not selected. The judges decide to take 6 more hard problems to complete the problemset, which gives the total hardness of 54 + 6 · 50 = 354.
题意:n行,每行第一个数是pi,代表这一行有pi个数,一共取k个数,竖着取,要求是题目中的黑体字,所选取的数的总和不能超过要选的数,这个要选的数才能被选。
题解:注意数据范围是0到49,没有的时添50,暴力跑,如果竖着选的时候发现没有数就选50,模拟下就行了。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
using namespace std;
int main()
{
freopen("easy.in","r",stdin);
freopen("easy.out","w",stdout);
int data[],a[][],cnt;
int n,k;
while(cin>>n>>k)
{
memset(a,-,sizeof(a));
memset(data,-,sizeof(data));
cnt=;
int maxlen=-;
for(int i=;i<n;i++)
{
cin>>a[i][];
if(a[i][]>maxlen)
maxlen=a[i][];
for(int j=;j<=a[i][];j++)
cin>>a[i][j];
}
for(int j=;j<=maxlen;j++)
{
for(int i=;i<n;i++)
data[cnt++]=a[i][j];
}
// for(int i=0;i<cnt;i++)
// cout<<data[i]<<" ";
int num=;
int sum=;
for(int i=;i<cnt;i++)
{
if(num==k)
break;
if(sum>=)
break;
if(data[i]==-&&sum<=)
break;
if(sum<=data[i])
{
sum+=data[i];
num++;
}
}
if(num<k)
sum=sum+(k-num)*;
cout<<sum<<endl;
}
return ;
}
Gym 100851E Easy Problemset (模拟题)的更多相关文章
- Gym 100851E Easy Problemset (水题,模拟)
题意:给定 n 个裁判,然后每个都一些题目,现在要从每一个按顺序去选出 k 个题,并且这 k 个要按不递减顺序,如果没有,就用50补充. 析:就按他说的来,直接模拟就好. 代码如下: #pragma ...
- Gym 100801E Easy Arithmetic (思维题)
题目:传送门.(需要下载PDF) 题意:给定一个长度不超过1000的字符串表达式,向该表达式中加入'+'或'-',使得表达式的值最大,输出该表达式. 题解:比如300-456就改成300-4+56,遇 ...
- Gym 100646 Problem C: LCR 模拟题
Problem C: LCR 题目连接: http://codeforces.com/gym/100646/attachments Description LCR is a simple game f ...
- Codeforces Gym 100269B Ballot Analyzing Device 模拟题
Ballot Analyzing Device 题目连接: http://codeforces.com/gym/100269/attachments Description Election comm ...
- poj1472[模拟题]
Instant Complexity Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 2017 Accepted: 698 ...
- 2010-2011 ACM-ICPC, NEERC, Moscow Subregional Contest Problem I. Interest Targeting 模拟题
Problem I. Interest Targeting 题目连接: http://codeforces.com/gym/100714 Description A unique display ad ...
- poj 1008:Maya Calendar(模拟题,玛雅日历转换)
Maya Calendar Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 64795 Accepted: 19978 D ...
- poj 1888 Crossword Answers 模拟题
Crossword Answers Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 869 Accepted: 405 D ...
- CodeForces - 427B (模拟题)
Prison Transfer Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u Sub ...
随机推荐
- BZOJ-1705 Longge的问题 一维GCD SUM 乱搞+质因数分解+...
2705: [SDOI2012]Longge的问题 Time Limit: 3 Sec Memory Limit: 128 MB Submit: 1871 Solved: 1172 [Submit][ ...
- jsp学习(五)
在进行jsp与jdbc连接时,出现这样一个错误,提示如下: java.net.ConnectException: Connection refused: connect 后来发现是由于mysql数据库 ...
- IOS基础之 (十二) 类的扩展
对OC类的扩展总结如下,共有4个: 1.子类 subClass 作用:可以使用类的继承来增添父类的变量和方法. 写法:在.h文件中 @interface Student : Person 2.分类 C ...
- android加载大图片到内存
1)演示效果: 1)代码演示: 布局代码: 权限配置:
- Android 获取本地图片
MainActivity.java public class RegisterActivity extends AppCompatActivity { private ImageView iv; @O ...
- 对称加密和分组加密中的四种模式(ECB、CBC、CFB、OFB)
一. AES对称加密: AES加密 分组 二. 分组密码的填充 分组密码的填充 e.g.: PKCS#5填充方式 三. 流密码: 四. 分组密码加密中的四种模式: 3.1 ECB模式 优点: 1. ...
- 新浪微博客户端(11)-自定义checkBox
在最后一个欢迎界面上添加一个CheckBox. // 2.添加4个UIImageView ; i < NEW_FEATURE_NUMS; i++) { UIImageView *imageVie ...
- Log4Net日志记录两种方式
简介 log4net库是Apache log4j框架在Microsoft .NET平台的实现,是一个帮助程序员将日志信息输出到各种目标(控制台.文件.数据库等)的工具. log4net是Ap ...
- IMAP(Internet Mail Access Protocol,Internet邮件访问协议)以前称作交互邮件访问协议(Interactive Mail Access Protocol)。
IMAP(Internet Mail Access Protocol,Internet邮件访问协议)以前称作交互邮件访问协议(Interactive Mail Access Protocol).IMA ...
- C++ 中宏的使用 --来自:http://blog.csdn.net/hgl868/article/details/7058906
宏在代码中的使用实例: g_RunLog2("Middleware client for Linux, build:%s %s", __DATE__, __TIME__); 下面详 ...