DISCO Presents Discovery Channel Code Contest 2020 Qual Task E. Majority of Balls
Not able to solve this problem during the contest (virtual participation).
The first observation is that if we can identify $N-1$ balls of which half is blue and the other half is red, then with these $N - 1$ balls we can identify the color of the rest $N+1$ balls.
It's not hard to note that if there are more blue balls than red balls among balls numbered from $l$ to $l + N - 1$ but less among balls numbered from $l + 1$ to $l + N$, we then know that
- The $l$-th ball and the $l+N$-th ball must be of different colors.
- The $l$-th ball must be blue and the $l+N$-th ball must be red.
- There are equal number of blue and red balls among balls numbered from $l + 1$ to $l+N - 1$.
The problem is whether such $l$ even exists? The answer is ,fortunately, YES.
Here comes the second interesting observation:
Let's assume without loss of generality, there are more blue balls than red balls among the first $N$ balls, then there must be more red balls among the last $N$ balls. So such $l$ as described above must exist, and we can find one using binary search which costs at most $\log_2 N + 1$ queries. When the binary search finishes, we get $l$ and the color of the $l$-th and $l+N$-th balls.
When $l$ is found, for each ball numbered from $1$ to $l - 1$ or from $l + N + 1$ to $2N$, we can know its color with a query. Note that exactly half of these $N - 1$ known balls are blue, so we can use these balls to identify color of balls numbered from $l + 1$ to $l + N -1$ in a similar way.
code
int main() {
int n;
cin >> n;
auto ask = [&](int l) {
cout > res;
return res.front();
};
char ml = ask(1);
int l = 1, r = n + 1;
while (r - l > 1) {
int mid = l + (r - l) / 2;
if (ask(mid) == ml) {
l = mid;
} else {
r = mid;
}
}
vector ans(2 * n + 1);
ans[l] = ml;
ans[l + n] = ml == 'R' ? 'B' : 'R';
auto ask2 = [&](int pos) {
cout << '?';
for (int i = 1; i < n; i++) {
cout << ' ' << l + i;
}
cout << ' ' << pos << endl;
string res;
cin >> res;
return res.front();
};
// [l + 1, l + n)
rng (i, 1, 2 * n + 1) {
if (i < l || i > l + n) {
ans[i] = ask2(i);
}
}
auto ask3 = [&](int pos) {
cout << '?';
rng (i, 1, 2 * n + 1) {
if (i < l || i > l + n) {
cout << ' ' << i;
}
}
cout << ' ' << pos << endl;
string res;
cin >> res;
return res.front();
};
rng (i, l + 1, l + n) {
ans[i] = ask3(i);
}
cout << "! ";
for (int i = 1; i <= 2 * n; i++) {
cout << ans[i];
}
cout << '\n';
return 0;
}
DISCO Presents Discovery Channel Code Contest 2020 Qual Task E. Majority of Balls的更多相关文章
- DISCO Presents Discovery Channel Code Contest 2020 Qual题解
传送门 \(A\) 咕咕 int x,y; int c[4]={0,300000,200000,100000}; int res; int main(){ cin>>x>>y; ...
- IOCCC(The International Obfuscated C Code Contest)
国际 C 语言混乱代码大赛(IOCCC, The International Obfuscated C Code Contest)是一项国际编程赛事,从 1984 年开始,每年举办一次(1997年.1 ...
- 【AtCoder】CODE FESTIVAL 2016 qual A
CODE FESTIVAL 2016 qual A A - CODEFESTIVAL 2016 -- #include <bits/stdc++.h> #define fi first # ...
- 【AtCoder】CODE FESTIVAL 2016 qual B
CODE FESTIVAL 2016 qual B A - Signboard -- #include <bits/stdc++.h> #define fi first #define s ...
- 【AtCoder】CODE FESTIVAL 2016 qual C
CODE FESTIVAL 2016 qual C A - CF -- #include <bits/stdc++.h> #define fi first #define se secon ...
- CODE FESTIVAL 2017 qual B B - Problem Set【水题,stl map】
CODE FESTIVAL 2017 qual B B - Problem Set 确实水题,但当时没想到map,用sort后逐个比较解决的,感觉麻烦些,虽然效率高很多.map确实好写点. 用map: ...
- CODE FESTIVAL 2017 qual B C - 3 Steps【二分图】
CODE FESTIVAL 2017 qual B C - 3 Steps 题意:给定一个n个结点m条边的无向图,若两点间走三步可以到,那么两点间可以直接连一条边,已经有边的不能连,问一共最多能连多少 ...
- M-SOLUTIONS Programming Contest 2020 题解
M-SOLUTIONS Programming Contest 2020 题解 目录 M-SOLUTIONS Programming Contest 2020 题解 A - Kyu in AtCode ...
- Atcoder CODE FESTIVAL 2017 qual B D - 101 to 010 dp
题目链接 题意 对于一个\(01\)串,如果其中存在子串\(101\),则可以将它变成\(010\). 问最多能进行多少次这样的操作. 思路 官方题解 转化 倒过来考虑. 考虑,最终得到的串中的\(' ...
随机推荐
- 2018 Nowcoder Multi-University Training Contest 1
Practice Link J. Different Integers 题意: 给出\(n\)个数,每次询问\((l_i, r_i)\),表示\(a_1, \cdots, a_i, a_j, \cdo ...
- 解决Virtualbox的根分区容量不够用问题
现在Virtualbox新建一块磁盘.容量一定要比原来的大.然后执行克隆命令. 把原来的磁盘内容克隆到新磁盘上.然后重新启动电脑. 运行相关扩容命令即可. #克隆磁盘 cd C:\Program Fi ...
- Po类设计
0.承接MySQL 表设计,同样地,这篇博客中一部分内容是Deolin的个人观点和习惯. 1.一般Po类的域是和DB表字段一一对应的, 而由于每个信息表和关联表都有id.insert_time.upd ...
- c++ 使用类生成随机数
// generate algorithm example #include <iostream> // cout #include <algorithm> // genera ...
- 三十、CentOS 7之systemd
一.系统启动流程 POST --> bootloader --> MBR工作 --> kernel(initramfs/initrd) --> ro rootfs --> ...
- Echarts案例-柱状图
一:先在官网下载 https://www.echartsjs.com/zh/download.html 然后再建立工程,导入这两个包: 写代码: <!DOCTYPE html> <h ...
- Python学习日记(八)—— 模块一(sys、os、hashlib、random、time、RE)
模块,用一砣代码实现了某个功能的代码集合. 类似于函数式编程和面向过程编程,函数式编程则完成一个功能,其他代码用来调用即可,提供了代码的重用性和代码间的耦合.而对于一个复杂的功能来,可能需要多个函数才 ...
- mysql 常见面试题
附录: https://mp.weixin.qq.com/s/pC0_Y7M7BkoUmlRwneZZdA 一.为什么用自增列作为主键 1.如果我们定义了主键(PRIMARY KEY),那么InnoD ...
- pwn学习日记Day22 《程序员的自我修养》读书笔记
知识杂项 软连接 命令: ln -s 原文件 目标文件 特征: 1.相当于windows的快捷方式 2.只是一个符号连接,所以软连接文件大小都很小 3.当运行软连接的时候,会根据连接指向找到真正的文件 ...
- RabbitMQ JAVA客户端调用例子
1.安装erlang 下载地址:http://www.erlang.org/downloads 设置ERLANG环境变量 2.安装RabbitMQ 下载地址: http://www.rabbitmq. ...