Codechef: Dividing Machine (DIVMAC)

0

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 &lt;bits/stdc++.h&gt;
#include &lt;ext/pb_ds/assoc_container.hpp&gt;
#include &lt;ext/pb_ds/tree_policy.hpp&gt;

#define pii              pair &lt;int,int&gt;
#define pll              pair &lt;long long,long long&gt;
#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&lt;&lt;#x &quot; = &quot;&lt;&lt;(x)&lt;&lt;endl
#define VI               vector &lt;int&gt;
#define DBG              pf(&quot;Hi\n&quot;)
#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(&quot;%d&quot;,&amp;a)
#define sfl(a)           scanf(&quot;%lld&quot;,&amp;a)
#define sff(a,b)         scanf(&quot;%d %d&quot;,&amp;a,&amp;b)
#define sffl(a,b)        scanf(&quot;%lld %lld&quot;,&amp;a,&amp;b)
#define sfff(a,b,c)      scanf(&quot;%d %d %d&quot;,&amp;a,&amp;b,&amp;c)
#define sfffl(a,b,c)     scanf(&quot;%lld %lld %lld&quot;,&amp;a,&amp;b,&amp;c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i&lt;n;i++)
#define loop1(i,n)       for(int i=1;i&lt;=n;i++)
#define REP(i,a,b)       for(int i=a;i&lt;b;i++)
#define RREP(i,a,b)      for(int i=a;i&gt;=b;i--)
#define TEST_CASE(t)     for(int z=1;z&lt;=t;z++)
#define PRINT_CASE       printf(&quot;Case %d: &quot;,z)
#define LINE_PRINT_CASE  printf(&quot;Case %d:\n&quot;,z)
#define CASE_PRINT       cout&lt;&lt;&quot;Case &quot;&lt;&lt;z&lt;&lt;&quot;: &quot;
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1&lt;&lt;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&lt;int, null_type, less&lt;int&gt;, rb_tree_tag, tree_order_statistics_node_update&gt; 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
/*------------------------------------------------*/

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

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

vector&lt;int&gt;prime;

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

deque&lt;int&gt;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&gt;j || e&lt;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&gt;j || e&lt;i) return 0;
if(b&gt;=i &amp;&amp; e&lt;=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(&quot;in.txt&quot;,&quot;r&quot;,stdin);
//	  freopen(&quot;out.txt&quot;,&quot;w&quot;,stdout);

sieve();

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

int t;
sf(t);
TEST_CASE(t)
{
int n,m;
sff(n,m);
for(int i=1; i&lt;=n; i++)
{
int a;
sf(a);
int root=sqrt(a);
dq[i].clear();
for(int j=0;prime[j]&lt;=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&gt;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(&quot;%d&quot;,query(1,1,n,b,c));
check=1;
}
else
printf(&quot; %d&quot;,query(1,1,n,b,c));
}
}
printf(&quot;\n&quot;);
}

return 0;
}

```