B. Drazil and His Happy Friends time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard 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.
Input 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.
Output If Drazil can make all his friends become happy by this plan, print “Yes”. Otherwise, print “No”.
Sample test(s) input 2 3 0 1 0 output Yes input 2 4 1 0 1 2 output No input 2 3 1 0 1 1 output Yes Note 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.
题目描述 有一些男孩和女孩，其中有一些是happy的。主角从男生0号和女生0号开始每次叫一个来家里吃饭 = = 如果其中有一个是happy的那么两人都变happy 否则就都unhappy回家 问最后主角能让所有人都happy么。
解题思路 因为数据很小，暴力模拟也可。 也可以求出男生女生人数的gcd，如果gcd=1那么所有人中有一个Happy的最后都可以让大家happy。如果gcd!=1假设gcd=t那么把t个人绑在一起看成一个人也是可以的，只要看每t个人中的i号，在所有组中是不是至少有一个是happy的。如果i号一个happy的都没有，那么肯定这个i号到最后也没人和他happy = =
代码如下
maxn = 110;int boy[maxn];int girl[maxn];int gcd(int a,int b) {return b==0?a:gcd(b,a%b);}int main(){int b,g;scanf(“%d%d”,&b,&g);int cnt;scanf(“%d”,&cnt);int chk1 = cnt;while(cnt–) {int a;scanf(“%d”,&a);boy[a] = 1;}scanf(“%d”,&cnt);int chk2 = cnt;while(cnt–) {int a;scanf(“%d”,&a);girl[a] = 1;}int t = gcd(b,g);if(t == 1) {if(chk1 || chk2) printf(“Yes\n”);else printf(“No\n”);}else {for(int i = 0 ; i < t ; i ++) {//检查每个t里第i个是否有快乐的bool chck = false;for(int j = i ; j < b ; j += t) {if(boy[j]) chck = true;}for(int j = i ; j < g ; j += t) {if(girl[j]) chck = true;}if(chck == false) {printf(“No\n”);break;}if(i == t-1) printf(“Yes\n”);}}return 0;}
，放弃等于又一次可以选择的机会。
