B. Drazil and His Happy Friends
来源:互联网 发布:软件评分表 编辑:程序博客网 时间:2024/06/07 04:00
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".
2 301 0
Yes
2 41 01 2
No
2 31 01 1
Yes
By we define the remainder of integer division of i by k.
In first sample case:
- On the 0-th day, Drazil invites 0-th boy and 0-th girl. Because 0-th girl is happy at the beginning, 0-th boy become happy at this day.
- On the 1-st day, Drazil invites 1-st boy and 1-st girl. They are both unhappy, so nothing changes at this day.
- On the 2-nd day, Drazil invites 0-th boy and 2-nd girl. Because 0-th boy is already happy he makes 2-nd girl become happy at this day.
- On the 3-rd day, Drazil invites 1-st boy and 0-th girl. 0-th girl is happy, so she makes 1-st boy happy.
- On the 4-th day, Drazil invites 0-th boy and 1-st girl. 0-th boy is happy, so he makes the 1-st girl happy. So, all friends become happy at this moment.
解题说明:此题是一道模拟题,按顺序进行遍历,然后加以判断即可。
#include<cstdio>#include <cstring>#include<cmath>#include<iostream>#include<algorithm>#include<vector>using namespace std;int main(){ int n,m,b,g,i=0,x,j=0;int B[100]={0},G[100]={0}; scanf("%d%d",&n,&m); scanf("%d",&b); for(i=0;i<b;i++){ scanf("%d",&x); B[x]=1; } scanf("%d",&g); for(i=0;i<g;i++){ scanf("%d",&x); G[x]=1; } for(i=0;i<m*n+m+n;i++){ if(B[i%n]==0&&G[i%m]==1) {B[i%n]=1;}else if(B[i%n]==1&&G[i%m]==0) {G[i%m]=1;} } for(i=0;i<n&&B[i];i++); for(j=0;j<m&&G[j];j++); if(i==n&&j==m){printf("Yes\n");}else {printf("No\n");} return 0;}
- B. Drazil and His Happy Friends
- 515B. Drazil and His Happy Friends
- B. Drazil and His Happy Friends
- B. Drazil and His Happy Friends
- codeforces 515B B. Drazil and His Happy Friends(模拟)
- Codeforce 515 B . Drazil and His Happy Friends 暴力
- CodeForces - 515B Drazil and His Happy Friends
- Codeforce 515 B . Drazil and His Happy Friends
- CodeForces515B Drazil and His Happy Friends (数学)
- Codeforces Round #292 (Div. 2) -- B. Drazil and His Happy Friends
- [数论] Codeforces 516E. Drazil and His Happy Friends
- [数论] Codeforces 819D R #421 D.Mister B and Astronomers & 516E R #292 E. Drazil and His Happy Friends
- codeforces#292B_Drazil and His Happy Friends-暴力水题
- Problem F: ZZY and his little friends
- CSU1323: ZZY and his little friends
- CF 66D Petya and His Friends
- sdut 3256 BIGZHUGOD and His Friends II
- CSU1323: ZZY and his little friends
- ssh能够连接而sftp不能连接的解决方法
- leetcode--Search a 2D Matrix
- 怎样理解最小二乘法原理及其用途
- Java的基本数据类型和Java的变量类型
- POJ 1611 The Suspects
- B. Drazil and His Happy Friends
- 数据预处理之独热编码(One-Hot Encoding)
- Java源码之HashMap
- 【UEditor】百度UE富文本自定义按钮添加文本
- HDU5536 Chip Factory(字典树)
- Codeforces Round 610
- The connection to adb is down, and a severe error has occured.的错误.
- Django+Query
- 离散数学及其应用 前言