2013-2014 ACM-ICPC, NEERC, Southern Subregional Contest Problem D. Grumpy Cat 交互题
Problem D. Grumpy Cat
题目连接:
http://www.codeforces.com/gym/100253
Description
This problem is a little bit unusual. Here you are to implement an interaction with a testing system. That
means that you can make queries and get responses in the online mode. Please be sure to use the stream
ushing operation after each query's output in order not to leave part of your output in some buer. For
example, in Ñ++ you've got to use the fflush(stdout) function, in Java call System.out.flush(),
and in Pascal flush(output).
Fall is coming and it's time to elect a new governor in Cattown. After looking at the results of the past
elections the citizens decided to nominate a new candidate Grumpy Cat. They said, all the previous
governors used to spend the Cattown's budget on their own needs and did nothing helpful for the town.
Grumpy Cat, they said, can't steal more money than he needs to feed himself.
The administration of Cattown refused to register Grumpy as a new candidate and it caused a lot of
discontent. People organized the biggest demonstration in the history of Cattown.
Your friend Eugene is a correspondent of a local newspaper. He has a work assignment to talk to
demonstrants to understand their main demands. After talking with several people Eugene realized that
each demonstrant either wants Grumpy Cat to be the new governor or doesn't want Grumpy Cat to be a
registered candidate or just wants to take part in the demonstration and doesn't have any requirements
at all. Let's call these 3 types of people grumpy-lovers, grumpy-haters and grumpy-neutral respectively. It
is known that there is at least one grumpy-lover and at least one grumpy-hater among the demonstrants.
Eugene decided to nd a type of each demonstrant. He can choose a group of demonstrants and ask them
for a number of Grumpy Cat supporters among the group. The people of Cattown don't like journalists,
reporters and correspondents. Each time Eugene asks a group of people, they proceed as follows:
- They talk to each other to understand who is who there. For sure, all grumpy-lovers are counted as
Grumpy Cat supporters and grumpy-haters are not. - If there are more grumpy-lovers than grumpy-haters in this group, all grumpy-neutrals express
support of Grumpy Cat at the time of this survey. - If there are more grumpy-haters than grumpy-lovers in this group, all grumpy-neutral people do not
support Grumpy Cat at the time of this survey. - If there are equal numbers of grumpy-haters and grumpy-lovers in this group, each grumpy-neutral
demonstrant at the time of this survey decides to support or not to support independently on his
own account. - After all grumpy-neutral people decide their position regarding Grumpy Cat, somebody tells the
correspondent the number of supporters.
A fact that a grumpy-neutral person has supported or hasn't supported Grumpy Cat doesn't aect its
decision in the future. Eugene can think that the surveys are completely independent.
Eugene doesn't have much time to do too many surveys. He can do at most ⌊
2πn
3
⌋ surveys, where π is
the ratio of a circle's circumference to its diameter and ⌊x⌋ is x rounded down. It seems too dicult for
him! Help him and write a program to interact with demonstrants to nd the type of each demonstrant.
Eugene knows that there is at least one grumpy-lover and there is at least one grumpy-hater among the
demonstrants.
Input
To read answers to the queries your program should use standard input.
The input starts with a line containing a positive integer t the number of testcases in the test.
Each testcase starts with a line containing a single integer n (2 ≤ n ≤ 100) the number of demonstrants.
The following lines will contain one integer each the number of Grumpy Cat supporters according to
the preceding survey.
The total number of demonstrants in all testcases in the test doesn't exceed 1000
Output
The program should use the standard output to print queries. Each query describes a single survey. It
should contain exactly two lines: the rst line should contain g (1 ≤ g ≤ n) the number of demonstrants
in an interviewed group, the second line should contain g distinct positive integer numbers t1, t2, . . . , tg
(1 ≤ ti ≤ n) the numbers of the demonstrants in the group. The demonstrants are numbered from 1
to n.
After your program found the types of all the demonstrants it should print exactly two lines: the rst line
should contain the only integer -1, the second line should contain exactly n integer numbers f1, f2, . . . , fn
(1 ≤ fi ≤ 3), where fi = 1 if the i-th demonstrant is a grumpy-lover, fi = 2 if the i-th demonstrant is a
grumpy-hater and fi = 3 in case of the i-th demonstrant is grumpy-neutral.
After the output of each line your program should execute the flush operation. Use single space to
separate integers in a line. Each line should end with end-of-line.
The program should write queries for the succeeding testcase after printing two lines described in the
second paragraph for the previous testcase. The program should terminate normally after the last testcase.
Sample Input
2
3
2
1
0
5
0
3
Sample Output
3
1 2 3
3
1 2 3
1
2
-1
1 2 3
3
2 4 3
3
5 3 1
-1
1 2 3 2 3
Hint
题意
有三种人,一种是兹瓷的人,一种是不兹瓷的人,和一种摇摆不定的人。
你每次可以询问一个集合,问这个集合的人里面有多少个兹瓷的人。
摇摆不定的人会哪边人多,兹瓷哪边。如果一样多,就随便回答。
然后你最多问2n次,让你确定所有人的身份。
题解:
先每个人都问一遍,这样你可以把所有人分成两类:兹瓷和摇摆的集合A,不支持和摇摆的集合B。
然后对于每一个集合A的元素都和集合B的所有人问一遍,如果有一个兹瓷的,说明那个人是兹瓷,否则就是摇摆不定的。
反之同理。
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1005;
int ans[maxn];
void solve()
{
memset(ans,0,sizeof(ans));
int n;
scanf("%d",&n);
vector<int> A;A.clear();
vector<int> B;B.clear();
for(int i=1;i<=n;i++)
{
printf("1\n%d\n",i);
fflush(stdout);
int x;scanf("%d",&x);
if(x==1)A.push_back(i);
else B.push_back(i);
}
for(int i=0;i<A.size();i++)
{
printf("%d\n",1+B.size());
fflush(stdout);
printf("%d",A[i]);
fflush(stdout);
for(int j=0;j<B.size();j++)
printf(" %d",B[j]),fflush(stdout);
printf("\n");
fflush(stdout);
int x;scanf("%d",&x);
if(x==0)ans[A[i]]=3;
else ans[A[i]]=1;
}
for(int i=0;i<B.size();i++)
{
printf("%d\n",1+A.size());
fflush(stdout);
printf("%d",B[i]);
fflush(stdout);
for(int j=0;j<A.size();j++)
printf(" %d",A[j]),fflush(stdout);
printf("\n");
fflush(stdout);
int x;scanf("%d",&x);
if(x==A.size()+1)ans[B[i]]=3;
else ans[B[i]]=2;
}
printf("-1\n");fflush(stdout);
for(int i=1;i<=n;i++)
{
if(i==1)printf("%d",ans[i]);
else printf(" %d",ans[i]);
fflush(stdout);
}
printf("\n");
fflush(stdout);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)solve();
return 0;
}
2013-2014 ACM-ICPC, NEERC, Southern Subregional Contest Problem D. Grumpy Cat 交互题的更多相关文章
- 2010-2011 ACM-ICPC, NEERC, Moscow Subregional Contest Problem K. KMC Attacks 交互题 暴力
Problem K. KMC Attacks 题目连接: http://codeforces.com/gym/100714 Description Warrant VI is a remote pla ...
- 2018-2019 ICPC, NEERC, Southern Subregional Contest
目录 2018-2019 ICPC, NEERC, Southern Subregional Contest (Codeforces 1070) A.Find a Number(BFS) C.Clou ...
- Codeforces 2018-2019 ICPC, NEERC, Southern Subregional Contest
2018-2019 ICPC, NEERC, Southern Subregional Contest 闲谈: 被操哥和男神带飞的一场ACM,第一把做了这么多题,荣幸成为7题队,虽然比赛的时候频频出锅 ...
- 2018-2019 ICPC, NEERC, Southern Subregional Contest (Online Mirror) Solution
从这里开始 题目列表 瞎扯 Problem A Find a Number Problem B Berkomnadzor Problem C Cloud Computing Problem D Gar ...
- Codeforces1070 2018-2019 ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred)总结
第一次打ACM比赛,和yyf两个人一起搞事情 感觉被两个学长队暴打的好惨啊 然后我一直做傻子题,yyf一直在切神仙题 然后放一波题解(部分) A. Find a Number LINK 题目大意 给你 ...
- codeforce1070 2018-2019 ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred) 题解
秉承ACM团队合作的思想懒,这篇blog只有部分题解,剩余的请前往星感大神Star_Feel的blog食用(表示男神汉克斯更懒不屑于写我们分别代写了下...) C. Cloud Computing 扫 ...
- 2018-2019 ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred)
A. Find a Number 找到一个树,可以被d整除,且数字和为s 记忆化搜索 static class S{ int mod,s; String str; public S(int mod, ...
- 2018.10.20 2018-2019 ICPC,NEERC,Southern Subregional Contest(Online Mirror, ACM-ICPC Rules)
i207M的“怕不是一个小时就要弃疗的flag”并没有生效,这次居然写到了最后,好评=.= 然而可能是退役前和i207M的最后一场比赛了TAT 不过打得真的好爽啊QAQ 最终结果: 看见那几个罚时没, ...
- 2018-2019 ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred) Solution
A. Find a Number Solved By 2017212212083 题意:$找一个最小的n使得n % d == 0 并且 n 的每一位数字加起来之和为s$ 思路: 定义一个二元组$< ...
随机推荐
- bzoj千题计划226:bzoj2763: [JLOI2011]飞行路线
http://www.lydsy.com/JudgeOnline/problem.php?id=2763 这也算分层图最短路? dp[i][j]到城市i,还剩k次免费次数的最短路 #include&l ...
- 蓝桥杯 带分数 DFS应用
问题描述 100 可以表示为带分数的形式:100 = 3 + 69258 / 714. 还可以表示为:100 = 82 + 3546 / 197. 注意特征:带分数中,数字1~9分别出现且只出现一次( ...
- JSBinding+Bridge.NET:Inspector拖变量支持
之前的文档说了,JSB的设计是不允许gameObject上挂逻辑脚本的.原因很简单,在Js工程中根本就不存在C#形式的逻辑脚本,如果在Cs工程中挂上了,到了Js工程这边,直接Missing. 实际在使 ...
- 20155314 2016-2017-2 《Java程序设计》第8周学习总结
20155314 2016-2017-2 <Java程序设计>第8周学习总结 教材学习内容总结 了解NIO 会使用Channel.Buffer与NIO2 会使用日志API.国际化 会使用正 ...
- MySQL忘记密码了怎么办?
接手一个项目时,如果上一位负责人没有把项目文档.账号密码整理好是一件很头疼的事情.. 例如,当你想打开MySQL数据库的时候 输入: mysql -u root -p 一回车想输入密码,发现密码错误! ...
- 转载一篇介绍CUDA
鉴于自己的毕设需要使用GPU CUDA这项技术,想找一本入门的教材,选择了Jason Sanders等所著的书<CUDA By Example an Introduction to Genera ...
- linux的lemon安装示范
\(First\): 1. 准备好 lemon原文件 2. 解压压缩包(名字必须是lemon) 第二步:找到readme.md这个文件 第三步:了解一下安装指南 第四步:打开终端 注意:源代码目录就是 ...
- google浏览器测试时清理缓存、强制不用缓存刷新快捷键(常用、效率)
Ctrl+Shift+Del 清除Google浏览器缓存的快捷键 Ctrl+Shift+R 重新加载当前网页而不使用缓存内容
- Python输出9*9 乘法表
for i in range(1,10): for j in range(1,i+1): print(str(j) + str("*") + str(i)+"=" ...
- python2.7中MySQLdb的安装与使用详解
Python2.7中MySQLdb的使用 import MySQLdb #1.建立连接 connect = MySQLdb.connect( '127.0.0.1', #数据库地址 'root', # ...