No Pain No Game（树状数组）

Problem Description

Life is a game,and you lose it,so you suicide.

But you can not kill yourself before you solve this problem:

Given you a sequence of number a1, a2, ..., an.They are also a permutation of 1...n.

You need to answer some queries,each with the following format:

If we chose two number a,b (shouldn't be the same) from interval [l, r],what is the maximum gcd(a, b)? If there's no way to choose two distinct number(l=r) then the answer is zero.

Input

First line contains a number T(T <= 5),denote the number of test cases. Then follow T test cases. For each test cases,the first line contains a number n(1 <= n <= 50000). The second line contains n number a[sub]1[/sub], a[sub]2[/sub], ..., a[sub]n[/sub]. The
third line contains a number Q(1 <= Q <= 50000) denoting the number of queries. Then Q lines follows,each lines contains two integer l, r(1 <= l <= r <= n),denote a query.

Output

For each test cases,for each query print the answer in one line.

Sample Input

1

10

8 2 4 9 5 7 10 6 1 3

5

2 10

2 4

6 9

1 4

7 10

Sample Output

5

2

2

4

3

题意：

给你个数字数字序列，然后m次查询，求区间[l,r]最大的两个数字的公因子。

思路：先预处理一下，求出每一个数字的因子，维护 c[j] 代表 [j,i] 出现 2 次以上的最大数。

#if 1
#include<bits/stdc++.h>
using namespace std;
const int MAXX=5e4+5;
int c[MAXX],a[MAXX],ans[MAXX],pre[MAXX];
vector<int> factor[MAXX];
void pre_do()
{
for(int i=1; i<MAXX; i++)
{
for(int j=i; j<MAXX; j+=i)
{
factor[j].push_back(i);
}
}
}
struct node
{
int l,r,id;
bool operator <(const node &a)const
{
return r<a.r;
}
}nodes[MAXX];
int lowbit(int x)
{
return x&(-x);
}

void add(int x,int v)
{
for(; x>0; x-=lowbit(x))
{
c[x]=max(c[x],v);
}
}
int sum(int x)
{
int M=0;
for(; x<MAXX; x+=lowbit(x))
{
M=max(c[x],M);
}
return M;
}

int main()
{

ios::sync_with_stdio(false);
pre_do();
int t;
cin>>t;
while(t--)
{
memset(pre,0,sizeof(pre));
memset(c,0,sizeof(c));
int n,m;
cin>>n;
for(int i=1; i<=n; i++)
{
cin>>a[i];
}
cin>>m;
for(int i=0; i<m; i++)
{
cin>>nodes[i].l>>nodes[i].r;
nodes[i].id=i;
}
sort(nodes,nodes+m);
int j=0;
for(int i=1; i<=n&&j<m; i++)
{
int x=a[i];
for(int k=0; k<(int)factor[x].size(); k++)
{
int val=factor[x][k];
if(pre[val])
add(pre[val],val);  //出现两次
pre[val]=i;
}

while(j<m&& nodes[j].r==i)
{
ans[nodes[j].id]=sum(nodes[j].l);
j++;
}

}
for(int i=0; i<m; i++)
{
cout<<ans[i]<<endl;
}
}

}
#endif