Problem Link : https://www.codechef.com/problems/DIVMAC

Solution Idea:

In this problem, a tricky observation is the primary key of the solution. That observation is the input number is in range 1 to 10^6. So We can perform update operation on an index at most 20 times. Because if we divide a number of range 1 to 10^6 with it’s prime divisors we can divide them at most 20 times.

So in the update section of segment tree we can make a check whether the maximum least prime divisor of this sub tree is 1 or not. If it is a 1 then we don’t need to update in the sub tree. Otherwise we will explore this sub tree and update them.

After this observation this problem is a simple RMQ problem.

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cerr<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define LINE_PRINT_CASE  printf("Case %d:\n",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;

//using namespace __gnu_pbds;
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;


/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

#define mx 100005
#define segment_tree int l=n*2,r=l+1,mid=(b+e)/2
bitset<mx/2>vis;

vector<int>prime;

void sieve()
{
    int x=mx/2,y=sqrt(mx)/2;
    for(int i=1; i<y; i++)
    {
        if(vis[i])
        {
            for(int j=i*(i+1)*2; j<x; j+=(2*i+1))
                vis[j]=1;
        }
    }
    prime.pb(2);
    for(int i=3; i<mx; i+=2)
        if(vis[i/2]==0) prime.pb(i);
}

deque<int>dq[mx];

int tree[3*mx];

void init(int n, int b, int e)
{
    if(b==e)
    {
        tree[n]=dq[b].front();
        return;
    }
    segment_tree;
    init(l,b,mid);
    init(r,mid+1,e);
    tree[n]=max(tree[l],tree[r]);
}

void update(int n, int b, int e, int i, int j)
{
    if(b>j || e<i) return;
    if(tree[n]==1) return;
    if(b==e)
    {
        dq[b].pop_front();
        tree[n]=dq[b].front();
        return;
    }
    segment_tree;
    update(l,b,mid,i,j);
    update(r,mid+1,e,i,j);
    tree[n]=max(tree[l],tree[r]);
}

int query(int n, int b, int e, int i, int j)
{
    if(b>j || e<i) return 0;
    if(b>=i && e<=j) return tree[n];
    segment_tree;
    int p=query(l,b,mid,i,j);
    int q=query(r,mid+1,e,i,j);
    return max(p,q);
}

int main()
{

//    freopen("in.txt","r",stdin);
//	  freopen("out.txt","w",stdout);

    sieve();

//    for(int i=0;i<100;i++)
//        D(prime[i]);

    int t;
    sf(t);
    TEST_CASE(t)
    {
        int n,m;
        sff(n,m);
        for(int i=1; i<=n; i++)
        {
            int a;
            sf(a);
            int root=sqrt(a);
            dq[i].clear();
            for(int j=0;prime[j]<=root;j++)
            {
                if(a%prime[j]==0)
                {
                    while(a%prime[j]==0)
                    {
                        dq[i].pb(prime[j]);
                        a/=prime[j];
                    }
                    root=sqrt(a);
                }
            }
            if(a>1)
                dq[i].pb(a);
            dq[i].pb(1);
        }

        init(1,1,n);
        bool check=0;
        while(m--)
        {
            int a,b,c;
            sfff(a,b,c);
            if(a==0)
            {
                update(1,1,n,b,c);
            }
            else
            {
                if(check==0)
                {
                    printf("%d",query(1,1,n,b,c));
                    check=1;
                }
                else
                printf(" %d",query(1,1,n,b,c));
            }
        }
        printf("\n");
    }

    return 0;
}