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
- linux课程总结
linux课程总结 --20125111 李冰清 转眼间,为期十六周的linux课程已进入尾声,回想起这十六周的课程,不断浮现在脑海里的是娄老师的笑容以及这十六周以来的点点滴滴. 第一次听到娄老师说将 ...
- ubuntu中mysql中文乱码及用python3.x调用
首先声明解决方法也是网上找来的,知识自己记下来以防以后用到. ubuntu版本是14.04使用apt-get命令安装mysql sudo apt-get install mysql-server ...
- Kafka 温故(四):Kafka的安装
Step 1: 下载Kafka > tar -xzf kafka_2.9.2-0.8.1.1.tgz> cd kafka_2.9.2-0.8.1.1 Step 2: 启动服务Kafka用到 ...
- 将网页设置为允许 XMLHttpRequest 跨域访问
在非IE下,使用XMLHttpRequest 不能跨域访问, 除非要访问的网页设置为允许跨域访问. 将网页设置为允许跨域访问的方法如下: Java Response.AddHeader("A ...
- es6笔记(6) Iterator 和 for...of循环
概要 js中的数组.对象,加上ES6中增加的Map.Set四种数据集合. Iterator提供了一种机制,为各种不同的数据结构提供统一的访问机制.任何数据结构只要部署Iterator接口,就可以完成遍 ...
- shell 判断文件出现次数
判断 file 文件中 第一个变量 出现次数 awk '{print $1}' file |sort |uniq -c|sort -k1r
- asp.net mvc 根据路由生成正确的url
假设存在这样一段路由配置: routes.MapRoute( name: "ProductList1_01", url: "pl/{bigSortId}_{smallSo ...
- 2017/05/22 java 基础 随笔
多态:一种事物多种形态 前提:1.子父类继承关系 2.方法复写.重写 3.父类引用指向子类对象 成员变量: package com.huawei; public class Demo1 { publi ...
- 【PE结构】恶意代码数字签名验证
说明 恶意代码数字签名验证功能,WinverityTrust.CryptQueryObject 代码实现 WinVerifyTrust //------------------------------ ...
- MySQL 5.7在线设置复制过滤【转】
转自 MySQL 5.7在线设置复制过滤 - yayun - 博客园 https://www.cnblogs.com/gomysql/p/4991197.html 5.7也GA了,有许多新的特性,其中 ...