CF Drazil and His Happy Friends
2 seconds
256 megabytes
standard input
standard output
Drazil has many friends. Some of them are happy and some of them are unhappy. Drazil wants to make all his friends become happy. So he invented the following plan.
There are n boys and m girls among his friends. Let's number them from 0 to n - 1 and 0 to m - 1 separately. In i-th day, Drazil invites
-th boy and
-th girl to have dinner together (as Drazil is programmer, i starts from 0). If one of those two people is happy, the other one will also become happy. Otherwise, those two people remain in their states. Once a person becomes happy (or if he/she was happy originally), he stays happy forever.
Drazil wants to know whether he can use this plan to make all his friends become happy at some moment.
The first line contains two integer n and m (1 ≤ n, m ≤ 100).
The second line contains integer b (0 ≤ b ≤ n), denoting the number of happy boys among friends of Drazil, and then follow b distinct integers x1, x2, ..., xb (0 ≤ xi < n), denoting the list of indices of happy boys.
The third line conatins integer g (0 ≤ g ≤ m), denoting the number of happy girls among friends of Drazil, and then follow g distinct integers y1, y2, ... , yg (0 ≤ yj < m), denoting the list of indices of happy girls.
It is guaranteed that there is at least one person that is unhappy among his friends.
If Drazil can make all his friends become happy by this plan, print "Yes". Otherwise, print "No".
2 3
0
1 0
Yes
2 4
1 0
1 2
No
2 3
1 0
1 1
Yes
#include <iostream>
#include <cmath>
#include <cstring>
#include <cstdio>
#include <string>
#include <algorithm>
#include <cctype>
#include <queue>
#include <map>
using namespace std; const int SIZE = ; int main(void)
{
int n,m,box;
int h_n,h_m;
int s_n[SIZE] = {};
int s_m[SIZE] = {}; cin >> n >> m; cin >> h_n;
for(int i = ;i < h_n;i ++)
{
cin >> box;
s_n[box] = ;
}
cin >> h_m;
for(int i = ;i < h_m;i ++)
{
cin >> box;
s_m[box] = ;
} for(int i = ;i < ;i ++)
if(s_n[i % n] || s_m[i % m])
s_n[i % n] = s_m[i % m] = ; int flag_1 = ;
int flag_2 = ;
for(int i = ;i < n;i ++)
if(!s_n[i])
{
flag_1 = ;
break;
}
for(int i = ;i < m;i ++)
if(!s_m[i])
{
flag_2 = ;
break;
}
if(flag_1 && flag_2)
cout << "Yes" << endl;
else
cout << "No" << endl; return ;
}
CF Drazil and His Happy Friends的更多相关文章
- CF Drazil and Factorial (打表)
Drazil and Factorial time limit per test 2 seconds memory limit per test 256 megabytes input standar ...
- CF Drazil and Date (奇偶剪枝)
Drazil and Date time limit per test 1 second memory limit per test 256 megabytes input standard inpu ...
- CF R631 div2 1330 E Drazil Likes Heap
LINK:Drazil Likes Heap 那天打CF的时候 开场A读不懂题 B码了30min才过(当时我怀疑B我写的过于繁琐了. C比B简单多了 随便yy了一个构造发现是对的.D也超级简单 dp了 ...
- 【Cf #292 D】Drazil and Morning Exercise(树的直径,树上差分)
有一个经典的问题存在于这个子问题里,就是求出每个点到其他点的最远距离. 这个问题和树的直径有很大的关系,因为事实上距离每个点最远的点一定是直径的两个端点.所以我们可以很容易地进行$3$遍$Dfs$就可 ...
- Codeforces Round #292 (Div. 1) - B. Drazil and Tiles
B. Drazil and Tiles Drazil created a following problem about putting 1 × 2 tiles into an n × m gri ...
- CF数据结构练习
1. CF 438D The Child and Sequence 大意: n元素序列, m个操作: 1,询问区间和. 2,区间对m取模. 3,单点修改 维护最大值, 取模时暴力对所有>m的数取 ...
- ORA-00494: enqueue [CF] held for too long (more than 900 seconds) by 'inst 1, osid 5166'
凌晨收到同事电话,反馈应用程序访问Oracle数据库时报错,当时现场现象确认: 1. 应用程序访问不了数据库,使用SQL Developer测试发现访问不了数据库.报ORA-12570 TNS:pac ...
- cf之路,1,Codeforces Round #345 (Div. 2)
cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅..... ...
- cf Round 613
A.Peter and Snow Blower(计算几何) 给定一个点和一个多边形,求出这个多边形绕这个点旋转一圈后形成的面积.保证这个点不在多边形内. 画个图能明白 这个图形是一个圆环,那么就是这个 ...
随机推荐
- mysqldump造成Buffer Pool污染的研究
前言: 最近Oracle MySQL在其官方Blog上贴出了 5.6中一些变量默认值的修改.其中innodb_old_blocks_time 的默认值从0替换成了1000(即1s) 关于该参数的作用摘 ...
- No module named BeautifulSoup
遇到 No module named BeautifulSoup 错误,但是的确从官方下载了BeautifulSoup,并安装成功. 后来才发现,有两个BeautifulSoup的版本,一个是2012 ...
- ACM OJ Collection
浙江大学(ZJU):http://acm.zju.edu.cn/ 北京大学(PKU):http://acm.pku.edu.cn/JudgeOnline/ 同济大学(TJU):http://acm.t ...
- POJ 1251 && HDU 1301 Jungle Roads (最小生成树)
Jungle Roads 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/A http://acm.hust.edu.cn/vju ...
- JSF 2.0 hello world example
In this tutorial, we will show you how to develop a JavaServer Faces (JSF) 2.0 hello world example, ...
- VIM技巧(1)
VIM技巧(1) 替换 36s/^\(.* = \)entity.\(.*\)$/\1this.GetShowName("\2",\2); 删除空行 %g/^$/d %g/^\s* ...
- STL学习系列六:List容器
List简介 list是一个双向链表容器,可高效地进行插入删除元素. list不可以随机存取元素,所以不支持at.(pos)函数与[]操作符.it++(ok), it+5(err) #include ...
- C/C++ 不带参数的回调函数 与 带参数的回调函数 函数指针数组 例子
先来不带参数的回调函数例子 #include <iostream> #include <windows.h> void printFunc() { std::cout<& ...
- MVC神韵---你想在哪解脱!(十二)
追加一条电影信息 运行应用程序,在浏览器中输入“http://localhost:xx/Movies/Create”,在表单中输入一条电影信息,然后点击追加按钮,如图所示. 点击追加按钮进行提交,表单 ...
- java对cookie的操作_01
/** * 读取所有cookie * 注意二.从客户端读取Cookie时,包括maxAge在内的其他属性都是不可读的,也不会被提交.浏览器提交Cookie时只会提交name与value属性.maxAg ...