思路：

点分治

先预处理好1e6 + 3以内到逆元

然后用map 映射以分治点为起点的链的值a 成他的下标 u 

然后暴力跑出以分治点儿子为起点的链的值b，然后在map里查找inv[b]*k

代码：

#include
     
      using namespace std;#define fi first#define se second#define pi acos(-1.0)#define LL long long//#define mp make_pair#define pb push_back#define ls rt<<1, l, m#define rs rt<<1|1, m+1, r#define ULL unsigned LL#define pll pair
      
       #define pii pair
       
        #define mem(a, b) memset(a, b, sizeof(a))#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);//headconst int MOD = 1e6 + 3;const int INF = 0x7f7f7f7f;const int N = 1e5 + 5;int inv[MOD + 5], mp[MOD + 5], head[N], mxsz[N], sz[N], v[N], cnt = 0, rt = 0, n, k, ans1, ans2;int deep[N], dis[N], id[N], top = 0;bool vis[N];struct edge {    int to, nxt;}edge[N*2];void add_edge(int u, int v) {    edge[cnt].to = v;    edge[cnt].nxt = head[u];    head[u] = cnt++;}void init() {    inv[1] = 1;    for (int i = 2; i < MOD; i++) inv[i] = (MOD - MOD/i) * 1LL * inv[MOD%i] % MOD;}void update(int x, int y) {    int t = (1LL * inv[x] * k) % MOD;    int now = mp[t];    if(!now) return ;    if(now > y) swap(now, y);    if(now < ans1 || now == ans1 && y < ans2) ans1 = now, ans2 = y;}void get_rt(int o, int u) {    sz[u] = 1, mxsz[u] = 0;    for (int i = head[u]; ~i; i = edge[i].nxt) {        if(edge[i].to != o && !vis[edge[i].to]) {            get_rt(u, edge[i].to);            sz[u] += sz[edge[i].to];            mxsz[u] = max(mxsz[u], sz[edge[i].to]);        }    }    mxsz[u] = max(mxsz[u], n - sz[u]);    if(mxsz[u] < mxsz[rt]) rt = u;}void get_d(int o, int u) {    deep[++top] = dis[u];    id[top] = u;    for (int i = head[u]; ~i; i = edge[i].nxt) {        if(!vis[edge[i].to] && edge[i].to != o) {            dis[edge[i].to] = (1LL * dis[u] * v[edge[i].to])%MOD;            get_d(u, edge[i].to);        }    }}void solve(int u) {    vis[u] = true;    mp[v[u]] = u;    for (int i = head[u]; ~i; i = edge[i].nxt) {        if(!vis[edge[i].to]) {            top = 0, dis[edge[i].to] = v[edge[i].to];            get_d(u, edge[i].to);            for (int j = 1; j <= top; j++) update(deep[j], id[j]);            top = 0, dis[edge[i].to] = (1LL * v[u] * v[edge[i].to])%MOD;            get_d(u, edge[i].to);            for (int j = 1; j <= top; j++) {                int t = deep[j];                if(!mp[t] || id[j] < mp[t]) mp[t] = id[j];            }        }    }    mp[v[u]] = 0;    for (int i = head[u]; ~i; i = edge[i].nxt) {        if(!vis[edge[i].to]) {            top = 0, dis[edge[i].to] = (1LL * v[u] * v[edge[i].to])%MOD;            get_d(u, edge[i].to);            for (int j = 1; j <= top; j++) mp[deep[j]] = 0;        }    }    for (int i = head[u]; ~i; i = edge[i].nxt) {        if(!vis[edge[i].to]) {            mxsz[0] = n =  sz[edge[i].to];            get_rt(rt = 0, edge[i].to);            solve(rt);        }    }}int main() {    init();    int u, V;    while(~scanf("%d%d", &n, &k)) {        mem(head, -1);        mem(vis, false);        mem(mp, 0);        cnt = 0;        ans1 = ans2 = INF;        for (int i = 1; i <= n; i++) scanf("%d", &v[i]);        for (int i = 1; i < n; i++) scanf("%d%d", &u, &V), add_edge(u, V), add_edge(V, u);        mxsz[0] = n;        get_rt(rt = 0, 1);        solve(rt);        if(ans1 == INF) printf("No solution\n");        else printf("%d %d\n", ans1, ans2);    }    return 0;}