Can you help them in solving this problem?
InputThere are several test cases, please process till EOF.
Each test case starts with a line containing two integers N(1 <= N <= 10 5) and K(0 <=K < 10 6 + 3). The following line contains n numbers v i(1 <= v i < 10 6 + 3), where vi indicates the integer on vertex i. Then follows N - 1 lines. Each line contains two integers x and y, representing an undirected edge between vertex x and vertex y.OutputFor each test case, print a single line containing two integers a and b (where a < b), representing the two endpoints of the chain. If multiply solutions exist, please print the lexicographically smallest one. In case no solution exists, print “No solution”(without quotes) instead.
For more information, please refer to the Sample Output below.Sample Input
5 60
2 5 2 3 3
1 2
1 3
2 4
2 5
5 2
2 5 2 3 3
1 2
1 3
2 4
2 5
Sample Output
3 4
No solution
Hint
树 点分治
自己想到了点分治，自己想到了逆元，自己想到了更新答案的方法，然后TLE到飞起，去看了别人的题解才发现没有初始化邻接表，真是感人至深。
其实就是个比较普通的点分治，把已经处理的链的结果存到map里（直接用数组就好），在处理新结点的时候匹配map里的值。
需要注意的是在mod意义下找乘积为K的点，需要求逆元。逆元递推式见代码
1 /*by SilverN*/ 2 #include<algorithm> 3 #include<iostream> 4 #include<cstring> 5 #include<cstdio> 6 #include<cmath> 7 #include<vector> 8 #include<map> 9 #define LL long long 10 using namespace std; 11 const int mod=1e6+3; 12 const int mxn=120010; 13 int read(){ 14 int x=0,f=1;char ch=getchar(); 15 while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();} 16 while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();} 17 return x*f; 18 } 19 struct edge{int v,nxt;}e[mxn<<1]; 20 int hd[mxn],mct=0; 21 void add_edge(int u,int v){e[++mct].v=v;e[mct].nxt=hd[u];hd[u]=mct;return;} 22 int n,K; 23 LL w[mxn]; 24 // 25 int mc[mxn],sz[mxn],rt,smm; 26 LL dis[mxn]; 27 bool vis[mxn]; 28 //map<int,int>mp; 29 int mp[mxn*11]; 30 // 31 LL inv[mxn*11]; 32 void init(){ 33 inv[1]=1; 34 for(int i=2;i<mod;i++){ 35 inv[i]=((-mod/i)*inv[mod%i]%mod+mod)%mod; 36 } 37 return; 38 } 39 void DFS_sz(int u,int fa){ 40 sz[u]=1;mc[u]=0; 41 for(int i=hd[u];i;i=e[i].nxt){ 42 if(e[i].v==fa || vis[e[i].v])continue; 43 int v=e[i].v; 44 DFS_sz(v,u); 45 sz[u]+=sz[v]; 46 mc[u]=max(mc[u],sz[v]); 47 } 48 mc[u]=max(mc[u],smm-mc[u]); 49 if(mc[u]<mc[rt])rt=u; 50 return; 51 } 52 int st[mxn],top=0; 53 int ans[2]; 54 inline void upd(int x,int y){ 55 if(x>y)swap(x,y); 56 if(!ans[0] || ans[0]>x || (ans[0]==x && ans[1]>y)){ 57 ans[0]=x;ans[1]=y; 58 } 59 return; 60 } 61 void clc(int u,int fa,int mode){ 62 if(!mode){ 63 LL tmp=(K*inv[dis[u]]%mod+mod)%mod; 64 if(mp[tmp]){ upd(u,mp[tmp]); } 65 } 66 else{ 67 if(!mp[dis[u]]){mp[dis[u]]=u;st[++top]=dis[u];} 68 else mp[dis[u]]=min(mp[dis[u]],u); 69 } 70 for(int i=hd[u];i;i=e[i].nxt){ 71 if(e[i].v==fa || vis[e[i].v])continue; 72 int v=e[i].v; 73 dis[v]=dis[u]*w[v]%mod; 74 clc(v,u,mode); 75 } 76 return; 77 } 78 void solve(int x){ 79 vis[x]=1; 80 mp[w[x]]=x; 81 for(int i=hd[x];i;i=e[i].nxt){ 82 if(vis[e[i].v])continue; 83 int v=e[i].v; 84 dis[v]=w[v]%mod; 85 clc(v,x,0); 86 dis[v]=w[x]*w[v]%mod; 87 clc(v,x,1); 88 } 89 while(top){ 90 mp[st[top--]]=0; 91 } 92 mp[w[x]]=0; 93 for(int i=hd[x];i;i=e[i].nxt){ 94 if(vis[e[i].v])continue;int v=e[i].v; 95 smm=sz[v]; 96 rt=0; 97 DFS_sz(v,0); 98 solve(rt); 99 } 100 return; 101 } 102 int main(){ 103 int i,j,u,v; 104 init(); 105 while(scanf("%d%d",&n,&K)!=EOF){ 106 memset(hd,0,sizeof hd);mct=0; 107 memset(vis,0,sizeof vis); 108 ans[0]=ans[1]=0; 109 for(i=1;i<=n;i++)w[i]=read(); 110 for(i=1;i<n;i++){ 111 u=read();v=read(); 112 add_edge(u,v); 113 add_edge(v,u); 114 } 115 ans[0]=ans[1]=0; 116 smm=n;mc[rt=0]=mod; 117 DFS_sz(1,0); 118 solve(rt); 119 if(!ans[0])printf("No solution "); 120 else printf("%d %d ",ans[0],ans[1]); 121 } 122 return 0; 123 }