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CF 题目集锦 PART 7 #264 div 2 E_html/css_WEB-ITnose

程序员文章站 2022-05-05 11:29:08
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【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:

  • Format of the query is "1 v". Let's write out the sequence of vertices along the path from the root to vertex v: u1,?u2,?...,?uk (u1?=?1; uk?=?v). You need to output such a vertex ui that gcd(value of ui,?value of v)?>?1 and i?
  • Format of the query is "2 v w". You must change the value of vertex v to w.
  • 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;vectorfac[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