No Pain No Game HDU - 4630 （线段树离线）

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 a 1, a 2, ..., a n.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 1, a 2, ..., a n. 
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

/*
@Author: Top_Spirit
@Language: C++
*/
#include <bits/stdc++.h>
using namespace std ;
typedef long long ll ;
const int Maxn = 5e4 + 10 ;
const double eps = 1e-6 ;

#define lson l, mid, ri << 1
#define rson mid + 1, r, ri << 1 | 1

struct message{
    int l, r ;
    int id ;
}me[Maxn << 1];

struct interval{
    int l, r ;
    int Max ;
}in[Maxn << 2];

int n, m, a[Maxn] ;
vector < int > ve[Maxn] ;
int path[Maxn], ans[Maxn] ;

void init(){
    for (int i = 1; i <= Maxn; i++){
        for (int j = i; j <= Maxn; j += i){
            ve[j].push_back(i) ;
        }
    }
}

void build (int l, int r, int ri){
    in[ri].l = l ;
    in[ri].r = r ;
    in[ri].Max = 0 ;
    if (l == r) return ;
    int mid = (l + r) >> 1 ;
    build(lson) ;
    build(rson) ;
}

void Update(int pos, int val, int ri){
    in[ri].Max = max(val, in[ri].Max) ;
    if (in[ri].l == in[ri].r) return ;
    int mid = (in[ri].l + in[ri].r) >> 1 ;
    if (mid >= pos) Update(pos, val, ri << 1) ;
    else Update(pos, val, ri << 1 | 1) ;
}

int query(int l, int r, int ri){
    if (in[ri].l == l && in[ri].r == r) return in[ri].Max ;
    int mid = (in[ri].l + in[ri].r) >> 1 ;
    if (mid >= r) query(l, r, ri << 1) ;
    else if (mid + 1 <= l) query(l, r, ri << 1 | 1) ;
    else return max(query(lson), query(rson)) ;
}

int main (){
    int T ;
    init() ;
    cin >> T ;
    while (T--){
        cin >> n ;
        for (int i = 1; i <= n; i++) cin >> a[i] ;
        cin >> m;
        for (int i = 1; i <= m; i++){
            cin >> me[i].l >> me[i].r ;
            me[i].id = i ;
        }
        sort(me + 1, me + m + 1, [](message a, message b){ return a.r < b.r ;}) ;
        memset(path, -1, sizeof(path)) ;
        build (1, n, 1) ;
        for (int i = 1, j = 1; j <= m && i <= n; i++){
            for (int k = 0; k < ve[a[i]].size(); k++){
                int tmp = ve[a[i]][k] ;
                if (path[tmp] != -1){
                    Update(path[tmp], tmp, 1) ;
                }
                path[tmp] = i ;
            }
            while (j <= m && me[j].r == i) {
                ans[me[j].id] = query(me[j].l, me[j].r, 1) ;
                j++ ;
            }
        }
        for (int i = 1; i <= m; i++){
            cout << ans[i] << endl ;
        }
    }
    return 0 ;
}