Codeforces Round #430 (Div. 2) 842C Ilya And The Tree(暴力)

这题我个人觉得是有问题的.
(修改,这题没问题,set也不是暴力,而是去重作用.因为真正的gcd,不会太多.)
想了很久,感觉想不出下至n上至nlogn的算法,非常难受,然后上网看题解,发现居然是一个暴力,遍历所有情况,居然不会超时,n^2logn的也行,真是无言以对.既然这样,这题也失去了意义.
(好在我没打这场比赛啊,A题都过不了,b题估计也要wa个好几次,C就跟别提了,少说要掉100分,T_T)

/*  xzppp  */#include <iostream>#include <vector>#include <cstdio>#include <string.h>#include <algorithm>#include <queue>#include <map>#include <string>#include <cmath>#include <set>using namespace std;#define FFF freopen("in.txt","r",stdin);freopen("out.txt","w",stdout);#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define MP make_pair#define PB push_backtypedef long long  LL;typedef unsigned long long ULL;typedef pair<int,int > pii;const int MAXN = 2*1e5+17;const int MAXM = 20;const int MAXV = 1001+17;const int INF = 0x7fffffff;const int MOD = 1e9+7;int gcd(int a,int b){    return b==0?a:gcd(b,a%b);}int a[MAXN];vector<int > G[MAXN];set<int > s[MAXN];int vis[MAXN];void dfs(int cur,int pre,int all){    if(vis[cur]) return;    vis[cur] = 1;    for (set<int>::iterator i = s[pre].begin(); i != s[pre].end(); ++i)        s[cur].insert(gcd(a[cur],*i));    s[cur].insert(all);    all = gcd(all,a[cur]);    for (int i = 0; i < G[cur].size(); ++i)        dfs(G[cur][i],cur,all);}int main(){    #ifndef ONLINE_JUDGE     FFF    #endif    int n;    cin>>n;    for (int i = 0; i < n; ++i)    {        scanf("%d",a+i);    }    for (int i = 0; i < n-1; ++i)    {        int u,v;        scanf("%d%d",&u,&v);        u--;v--;        G[u].PB(v);        G[v].PB(u);    }    dfs(0,0,0);    s[0].insert(a[0]);    for (int i = 0; i < n; ++i)    {        printf("%d\n",*s[i].rbegin());    }    return 0;}