【原题】
E. Caisa and Tree
time limit per test
10 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Caisa is now at home and his son has a simple task for him.
Given a rooted tree with n vertices, numbered from 1 to n (vertex 1 is the root). Each vertex of the tree has a value. You should answer q queries. Each query is one of the following:
You are given all the queries, help Caisa to solve the problem.
Input
The first line contains two space-separated integers n, q (1?≤?n,?q?≤?105).
The second line contains n integers a1,?a2,?...,?an (1?≤?ai?≤?2·106), where ai represent the value of node i.
Each of the next n?-?1 lines contains two integers xi and yi (1?≤?xi,?yi?≤?n; xi?≠?yi), denoting the edge of the tree between vertices xi and yi.
Each of the next q lines contains a query in the format that is given above. For each query the following inequalities hold: 1?≤?v?≤?n and 1?≤?w?≤?2·106. Note that: there are no more than 50 queries that changes the value of a vertex.
Output
For each query of the first type output the result of the query.
Sample test(s)
input
4 610 8 4 31 22 33 41 11 21 31 42 1 91 4
output
-112-11
Note
gcd(x,?y) is greatest common divisor of two integers x and y.
【分析】这道题是做现场赛的。本来能A的，但是太紧张了=而且也不会用vector，边表搞的麻烦死了。
开始看到修改操作才50次、时间又松，真是爽！估计每次可以暴力重构这颗树，然后对于每个质因子记录最优值。
首先每次不能sqrt的效率枚举一个数的因子，我们可以预处理出每个数的所有质因子。（其实有更省空间的）
剩下来要解决的问题是：因为我是用dfs的，怎么把某个子树的信息在搜完后再去掉？（以免影响其他子树）HHD表示用vector一点也不虚。其实应该也可以用边表类似的思路，但是麻烦= =
【代码】
#include#include#include #include #define N 100005#define S 2000005#define push push_back#define pop pop_backusing namespace std;vector fac[S],f[S];int data[N],ans[N],end[N],pf[S],deep[N];int C,cnt,n,Q,i,x,y,opt;struct arr{int go,next;}a[N*2];inline void add(int u,int v){a[++cnt].go=v;a[cnt].next=end[u];end[u]=cnt;}inline void init(){ int H=2000000; for (int i=2;ideep[ans[k]]) ans[k]=f[go][temp-1]; f[go].push(k); } for (int i=end[k];i;i=a[i].next) if (a[i].go!=fa) dfs(a[i].go,k); for (int i=0;i