# SPOJ: GSS2 – Can you answer these queries II

0

Solution Idea:

```
#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

int ara[mx+5];
int last[2*mx+5];
ll ans[mx+5];

struct data
{
int l,r,id;
};

vector&lt;data&gt;queries;

struct data1
{
ll sum,lazy,best_lazy,ans;
data1()
{
sum=lazy=best_lazy=ans=0;
}
};

data1 tree[3*mx];

void push_down(data1 &amp;cur, data1 &amp;lft, data1 &amp;rgt)
{
//    if(cur.lazy || cur.best_lazy)
//    {
lft.best_lazy=max(lft.best_lazy,lft.lazy+cur.best_lazy);
lft.lazy+=cur.lazy;
lft.ans=max(lft.ans,lft.sum+cur.best_lazy);
lft.sum=lft.sum+cur.lazy;
rgt.best_lazy=max(rgt.best_lazy,rgt.lazy+cur.best_lazy);
rgt.lazy+=cur.lazy;
rgt.ans=max(rgt.ans,rgt.sum+cur.best_lazy);
rgt.sum+=cur.lazy;
cur.lazy=cur.best_lazy=0;
//    }
}

void push_up(data1 &amp;cur, data1 &amp;lft, data1 &amp;rgt)
{
cur.ans=max(lft.ans,rgt.ans);
cur.sum=max(lft.sum,rgt.sum);
}

void update(int n, int b, int e, int i, int j, ll val)
{
if(b&gt;j || e&lt;i) return;
if(b&gt;=i &amp;&amp; e&lt;=j)
{
tree[n].best_lazy=max(tree[n].best_lazy,tree[n].lazy+val);
tree[n].lazy+=val;
tree[n].ans=max(tree[n].ans,tree[n].sum+val);
tree[n].sum+=val;
return;
}
segment_tree;
push_down(tree[n],tree[l],tree[r]);
update(l,b,mid,i,j,val);
update(r,mid+1,e,i,j,val);
push_up(tree[n],tree[l],tree[r]);
}

data1 query(int n, int b, int e, int i, int j)
{
if(b&gt;j || e&lt;i) return data1();
if(b&gt;=i &amp;&amp; e&lt;=j)
{
return tree[n];
}
segment_tree;
push_down(tree[n],tree[l],tree[r]);
data1 p=query(l,b,mid,i,j);
data1 q=query(r,mid+1,e,i,j);
data1 ret;
push_up(ret,p,q);
return ret;
}

int main()
{

//    freopen(&quot;in.txt&quot;,&quot;r&quot;,stdin);
//	  freopen(&quot;out.txt&quot;,&quot;w&quot;,stdout);
int n;
sf(n);
for(int i=1; i&lt;=n; i++) sf(ara[i]);
int m;
sf(m);
data a;
for(int i=0; i&lt;m; i++)
{
sff(a.l,a.r);
a.id=i;
queries.pb(a);
}

sort(all(queries),[](data a,data b)
{
return a.r&lt;b.r;
});

int idx=0;
for(int i=1; i&lt;=n &amp;&amp; idx&lt;SZ(queries); i++)
{
//        if(last[mx+ara[i]]!=0)
//        {
//            update(1,1,n,last[mx+ara[i]],0);
//        }
int xx=last[mx+ara[i]]+1;
update(1,1,n,xx,i,ara[i]);
last[mx+ara[i]]=i;
while(idx&lt;SZ(queries) &amp;&amp; queries[idx].r==i)
{
ans[queries[idx].id]=query(1,1,n,queries[idx].l,queries[idx].r).ans;
idx++;
}
}

for(int i=0; i&lt;m; i++)
printf(&quot;%lld\n&quot;,ans[i]);

return 0;
}

```

# Light OJ: 1267 – Points in Rectangle (II)

0

Solution Idea:

This problem is similar to Codeforces gym 101484: B. Nicoleta’s Cleaning. We need a line sweep technique and some data structure to solve this problem. We need to convert each rectangle to opening and closing with respect to its x coordinates. Then we need to sort all points and open and close together. then for each opening element we need to count how many points are before opening point let is is X and for each closing, we need to count how many points are before and on the closing point in the range, lower y coordinate to upper y coordinate let it is Y. Then the number of points inside the rectangle is Y-X. For each given point we need to update it when we find it through scan line in x coordinate of it.

```#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
/*------------------------------------------------*/

//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));}
/*------------------------------------------------*/

struct data
{
int type,x,st,ed,id;
};

vector<int>x,y;
vector<data>v;

bool cmp(data a, data b)
{
if(a.x<b.x) return true;
if(a.x==b.x)
return a.type<b.type;
return false;
}

#define mx 300005

int tree[mx];

void update(int idx, int val)
{
for(; idx<mx && idx; idx+=(idx&-idx))
tree[idx]+=val;
}

int query(int idx)
{
int ret=0;
for(; idx; idx-=(idx&-idx))
ret+=tree[idx];
return ret;
}

int ans[mx];

int main()
{
//    freopen("in.txt","r",stdin);
//	  freopen("out.txt","w",stdout);
int t;
sf(t);
TEST_CASE(t)
{
int n,m;
sff(n,m);
for(int i=0; i<n; i++)
{
int a,b;
sff(a,b);
data temp;
temp.type=2;
temp.x=a;
temp.ed=temp.st=b;
v.pb(temp);
x.pb(a);
y.pb(b);
}

for(int i=0; i<m; i++)
{
int a,b,c,d;
sff(a,b);
sff(c,d);
x.pb(a);
x.pb(c);
y.pb(b);
y.pb(d);
data temp;
temp.type=3;
temp.x=c;
temp.st=b;
temp.ed=d;
temp.id=i;
v.pb(temp);
data temp1;
temp1.type=1;
temp1.x=a;
temp1.st=b;
temp1.ed=d;
temp1.id=i;
v.pb(temp1);
}

//    sort(all(x));
sort(all(y));
sort(all(v),cmp);

for(int i=0; i<SZ(v); i++)
{
data temp=v[i];
if(temp.type==1)
{
int a=lower_bound(all(y),temp.st)-y.begin()+1;
int b=lower_bound(all(y),temp.ed)-y.begin()+1;
int xx=query(b);
int yy=query(a-1);
ans[temp.id]=xx-yy;
}
else if(temp.type==3)
{
int a=lower_bound(all(y),temp.st)-y.begin()+1;
int b=lower_bound(all(y),temp.ed)-y.begin()+1;
int xx=query(b);
int yy=query(a-1);
ans[temp.id]=(xx-yy)-ans[temp.id];
}
else
{
int a=lower_bound(all(y),temp.st)-y.begin()+1;
update(a,1);
//            int yy=query(a-1); // Never do this while range update on BIT -_-
//            ans[temp.id]=xx;
}
}

LINE_PRINT_CASE;
for(int i=0; i<m; i++)
printf("%d\n",ans[i]);
ms(tree,0);
x.clear();
y.clear();
v.clear();
}
return 0;
}```

# Codeforces gym 101484: B. Nicoleta’s Cleaning

1

Solution Idea:

To solve this problem we need line sweep technique and some data structure. At first convert each rectangle in two element with respect to it x coordinate which both x coordinate have same y coordinates. Now call first x point opening and 2nd x point closing of a rectangle. Now take a array of custom data type. and store every opening and closing of rectangles and the query points and sort the the array in the basis of x coordinate of each element and in case of a tie take closing element first then query point and then opening point.

Now run a loop on the x axis and update every range by 1 if we find a opening and by -1 if we find a closing and if we find a query point just query the value of that point.

Handle the boundary condition carefully.

```
#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));}
/*------------------------------------------------*/

struct data
{
int type,x,st,ed,id;
};

vector&lt;int&gt;x,y;
vector&lt;data&gt;v;

bool cmp(data a, data b)
{
if(a.x&lt;b.x) return true;
if(a.x==b.x)
return a.type&lt;b.type;
return false;
}

#define mx 300005

int tree[mx];

void update(int idx, int val)
{
for(; idx&lt;mx &amp;&amp; idx; idx+=(idx&amp;-idx))
tree[idx]+=val;
}

int query(int idx)
{
int ret=0;
for(; idx; idx-=(idx&amp;-idx))
ret+=tree[idx];
return ret;
}

int ans[mx];

int main()
{
//    freopen(&quot;in.txt&quot;,&quot;r&quot;,stdin);
//	  freopen(&quot;out.txt&quot;,&quot;w&quot;,stdout);

int n,m;
sff(n,m);
for(int i=0; i&lt;n; i++)
{
int a,b;
sff(a,b);
data temp;
temp.type=2;
temp.x=a;
temp.ed=temp.st=b;
temp.id=i;
v.pb(temp);
x.pb(a);
y.pb(b);
}

for(int i=0; i&lt;m; i++)
{
int a,b,c,d;
sff(a,b);
sff(c,d);
x.pb(a);
x.pb(c);
y.pb(b);
y.pb(d);
data temp;
temp.type=1;
temp.x=c;
temp.st=b;
temp.ed=d;
v.pb(temp);
data temp1;
temp1.type=3;
temp1.x=a;
temp1.st=b;
temp1.ed=d;
v.pb(temp1);
}

//    sort(all(x));
sort(all(y));
sort(all(v),cmp);

for(int i=0; i&lt;SZ(v); i++)
{
data temp=v[i];
if(temp.type==1)
{
int a=lower_bound(all(y),temp.st)-y.begin()+1;
int b=lower_bound(all(y),temp.ed)-y.begin()+1;
update(a+1,-1);
update(b,+1);
}
else if(temp.type==3)
{
int a=lower_bound(all(y),temp.st)-y.begin()+1;
int b=lower_bound(all(y),temp.ed)-y.begin()+1;
update(a+1,+1);
update(b,-1);
}
else
{
int a=lower_bound(all(y),temp.st)-y.begin()+1;
int xx=query(a);
//            int yy=query(a-1); // Never do this while range update on BIT -_-
ans[temp.id]=xx;
}
}

for(int i=0; i&lt;n; i++)
printf(&quot;%d\n&quot;,ans[i]);

return 0;
}

```

# Hackerrank: Solve the Queries!

0

Solution Idea:

The main observation in this problem is the given numbers are <=100 and there are 25 prime number less than 100. As we need to answer whether a range multiplication is multiple of another range multiplication or not.

We know when we multiply two number their prime power is added. For example 12*18 = (2^2 * 3^1) * (2^1 * 3^2)
= (2^(2+1) * 3^(1+2))
= (2^3 * 3^3)
= 216

So we can store prime factors of every given number in each segment tree leaf. And while merging two leaf node we need to add each prime power between two two leafs. So a node in the segment tree hold the information about the nubmer in the subtree of this node has how many ith prime divisor.

Now for each query we can get the power factor of each range from the segment tree and check whether they divides or not and produce the required answer from this prime factors.

```
#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));}
/*------------------------------------------------*/

int prime[30]= {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97};
int mp[100];

#define mx 50005
#define segment_tree int l=n*2,r=l+1,mid=(b+e)/2
struct data
{
int lazy;
int ara[25];
};

data ret;

int numbers[mx];
data tree[3*mx];

void func(int n, data &amp;temp, int mul)
{
ms(temp.ara,0);
for(int i=0; i&lt;25 &amp;&amp; prime[i]*prime[i]&lt;=n; i++)
{
if(n%prime[i]==0)
{
while(n%prime[i]==0)
{
n/=prime[i];
temp.ara[i]++;
}
temp.ara[i]*=mul;
}
}

if(n&gt;1)
temp.ara[mp[n]]+=mul;
}

void push_up(data &amp;a, data &amp;b, data &amp;c)
{
for(int i=0; i&lt;25; i++)
a.ara[i]=b.ara[i]+c.ara[i];
}

void push_down(int n, int b, int e)
{
if(tree[n].lazy)
{
segment_tree;
func(tree[n].lazy,tree[l],mid-b+1);
func(tree[n].lazy,tree[r],e-mid);
tree[l].lazy=tree[n].lazy;
tree[r].lazy=tree[n].lazy;
tree[n].lazy=0;
}
}

void init(int n, int b, int e)
{
if(b==e)
{
tree[n].lazy=0;
func(numbers[b],tree[n],1);
return;
}
segment_tree;
init(l,b,mid);
init(r,mid+1,e);
push_up(tree[n],tree[l],tree[r]);
}

void update(int n, int b, int e, int i, int j, int val)
{
if(b&gt;j || e&lt;i) return;
if(b&gt;=i &amp;&amp; e&lt;=j)
{
func(val,tree[n],e-b+1);
tree[n].lazy=val;
return;
}
segment_tree;
push_down(n,b,e);
update(l,b,mid,i,j,val);
update(r,mid+1,e,i,j,val);
push_up(tree[n],tree[l],tree[r]);
}

data query(int n, int b, int e, int i, int j)
{
if(b&gt;j || e&lt;i) return ret;
if(b&gt;=i &amp;&amp; e&lt;=j)
{
return tree[n];
}
push_down(n,b,e);
segment_tree;
data p=query(l,b,mid,i,j);
data q=query(r,mid+1,e,i,j);
data xx;
push_up(xx,p,q);
return xx;
}

ll bigmod(ll n, ll pow, ll mod)
{
ll rett=1;
while(pow)
{
if(pow%2)
rett=(rett*n)%mod;
n=(n*n)%mod;
pow/=2;
}
return rett;
}

int main()
{

//    freopen(&quot;in.txt&quot;,&quot;r&quot;,stdin);
//	  freopen(&quot;out.txt&quot;,&quot;w&quot;,stdout);
for(int i=0; i&lt;25; i++) mp[prime[i]]=i;

int n;
sf(n);
loop1(i,n) sf(numbers[i]);

init(1,1,n);

int q;
sf(q);
while(q--)
{
int a;
sf(a);
if(a==1)
{
int l,r,x;
sfff(l,r,x);
update(1,1,n,l,r,x);
}
else
{
int l, r, p, q, m;
sff(l,r);
sff(p,q);
sf(m);

data x=query(1,1,n,l,r);
data y=query(1,1,n,p,q);
bool valid=1;
for(int i=0; i&lt;25 &amp;&amp; valid; i++)
{
x.ara[i]-=y.ara[i];
if(x.ara[i]&lt;0)
valid=0;
}

if(valid==0)
{
printf(&quot;-1\n&quot;);
continue;
}

ll ans=1;

for(int i=0;i&lt;25;i++)
{
if(x.ara[i])
{
ans*=bigmod(prime[i],x.ara[i],m);
ans%=m;
}
}

printf(&quot;%lld\n&quot;,ans);

}
}

return 0;
}

```

# Hackerrank: Lexicographically Smaller or Equal Strings

0

Solution Idea:

This problem is an example of basic merge sort tree problem. Merge Sort tree is a segment tree in which every node contains a vector. In this problem we can store string in every leaf node of segment tree as the size of the string is <=15. And while merging two leaf node we will merge them in sorted order. Now as the vectors of every node in the segment tree is sorted we can apply binary search on segment tree node to get how many string is lexicographically smaller than the given string.

So after building merge sort tree we can easily answer every query using binary search.

If you want to learn some wonderful application of segment tree or merge sort tree you can see this article.

```
#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

vector&lt;string&gt;tree[3*mx],v;

void init(int n, int b, int e)
{
if(b==e)
{
tree[n].pb(v[b]);
return;
}
int l=n*2,r=l+1,mid=(b+e)/2;
init(l,b,mid);
init(r,mid+1,e);
merge(all(tree[l]),all(tree[r]),back_inserter(tree[n]));
//    cout&lt;&lt;n&lt;&lt;&quot; &quot;&lt;&lt;b&lt;&lt;&quot; &quot;&lt;&lt;e&lt;&lt;endl;
//    for(int i=0;i&lt;SZ(tree[n]);i++)
//        cout&lt;&lt;tree[n][i]&lt;&lt;endl;
//    cout&lt;&lt;&quot;-----------------------------&quot;&lt;&lt;endl;
}

int query(int n, int b, int e, int i, int j, string &amp;str)
{
if(b&gt;j || e&lt;i) return 0;
if(b&gt;=i &amp;&amp; e&lt;=j)
{
//        int x=upper_bound(all(tree[n]),str)-tree[n].begin();
//        int y=lower_bound(all(tree[n]),str)-tree[n].begin();
//        int z=x-y;
//        return z;
return upper_bound(all(tree[n]),str)-tree[n].begin();
}
int l=n*2,r=l+1,mid=(b+e)/2;
int p=query(l,b,mid,i,j,str);
int q=query(r,mid+1,e,i,j,str);
return p+q;
}

int main()
{
//    freopen(&quot;in.txt&quot;,&quot;r&quot;,stdin);
//	  freopen(&quot;out.txt&quot;,&quot;w&quot;,stdout);

int n;
sf(n);

v.pb(&quot;&quot;);
char sss[30];

for(int i=0;i&lt;n;i++)
{
scanf(&quot; %s&quot;,sss);
v.pb(string(sss));
}

init(1,1,n);

int q;
sf(q);
while(q--)
{
int a,b;
sff(a,b);
scanf(&quot; %s&quot;,sss);

string str=string(sss);
//        cout&lt;&lt;str&lt;&lt;endl;
int ans=query(1,1,n,a,b,str);
printf(&quot;%d\n&quot;,ans);
}

return 0;
}

```

# Toph: Another Update-Query Problem

0

Solution Idea:

The main idea behind the solution of this problem is simplifying the equation. For a query operation we need the answer of this equation-
1A_l + (1+d)A_l+1 + (1+2d)A_l+2 + (1+3d)A_l+3 + … + (1+(r-l)d)A_r
Now let the query range l r is 1 3. and the given array is A=[a1,a2,a3,a4,…]

So our quation will be-
1*a1 + (1+d)*a2 + (1+2d)*a3 + (1+3d)*a4
= a1 + a2 + d*a2 + a3 + 2d*a3 + a4 + 3d*a4
= (a1+a2+a3+a4) + d*(a2+ 2*a3 + 3*a4)

Now We can see that first part of this equation is only sum query which can be done using a segment tree and for 2nd part we can use a trick. We can pree calculate the sequence like this-
1*a1 + 2*a2 + 3*a3 + 4*a4 + 5*a5 + 6*a6 + 7*a7 + ……
When we need the 2nd part of the equation we can use this equation. We can get the 2nd part of the equation from this equation by this way-
(2*a2 + 3*a3 + 4*a4)-(a2+a3+a4) = (a2+ 2*a3 + 3*a4).

Now the above equation is also only a sum equation. We can apply any range sum query or range sum update on this equation and the the first sum equation.

So if we store this two equation in two segment tree then we can perform range update or query on the segment tree and get any range sum query from the segment tree and using the equation we can answer each query of the problem correctly.

```
#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
struct data
{
ll sum, ssum, prop;
data()
{
sum=0,ssum=0,prop=0;
}
};

data tree[3*mx];

ll ara[mx];

ll interval_sum(ll b, ll e)
{
ll x=(e*(e+1))/2;
ll y=(b*(b-1))/2;
return (x-y+MOD)%MOD;
}

void push_down(int n, int b, int e)
{
if(tree[n].prop)
{
segment_tree;
tree[l].sum+=((mid-b+1)*tree[n].prop)%MOD;
tree[r].sum+=((e-mid)*tree[n].prop)%MOD;
tree[l].ssum+=(interval_sum(b,mid)*tree[n].prop)%MOD;
tree[r].ssum+=(interval_sum(mid+1,e)*tree[n].prop)%MOD;
tree[l].prop+=tree[n].prop%MOD;
tree[r].prop+=tree[n].prop%MOD;
tree[l].sum%=MOD;
tree[l].ssum%=MOD;
tree[l].prop%=MOD;
tree[r].sum%=MOD;
tree[r].ssum%=MOD;
tree[r].prop%=MOD;
tree[n].prop=0;
}
}

void init(int n, int b, int e)
{
if(b==e)
{
tree[n].sum=ara[b]%MOD;
tree[n].ssum=((ll)b*ara[b])%MOD;
tree[n].prop=0;
return;
}
segment_tree;
init(l,b,mid);
init(r,mid+1,e);
tree[n].sum=(tree[l].sum+tree[r].sum)%MOD;
tree[n].ssum=(tree[l].ssum+tree[r].ssum)%MOD;
tree[n].prop=0;
}

void update(int n, int b, int e, int i, int j, ll val)
{
if(b&gt;j || e&lt;i) return;
if(b&gt;=i &amp;&amp; e&lt;=j)
{
tree[n].sum+=((ll)(e-b+1)*val)%MOD;
tree[n].ssum+=(interval_sum(b,e)*val)%MOD;
tree[n].prop+=val;
tree[n].sum%=MOD;
tree[n].ssum%=MOD;
tree[n].prop%=MOD;
return;
}
push_down(n,b,e);
segment_tree;
update(l,b,mid,i,j,val);
update(r,mid+1,e,i,j,val);
tree[n].ssum=(tree[l].ssum+tree[r].ssum)%MOD;
tree[n].sum=(tree[l].sum+tree[r].sum)%MOD;
}

data query(int n, int b, int e, int i, int j)
{
if(b&gt;j || e&lt;i) return data();
if(b&gt;=i &amp;&amp; e&lt;=j)
return tree[n];
push_down(n,b,e);
segment_tree;
data p=query(l,b,mid,i,j);
data q=query(r,mid+1,e,i,j);
data ret;
ret.ssum=(p.ssum+q.ssum)%MOD;
ret.sum=(p.sum+q.sum)%MOD;
return ret;
}

int main()
{
//    freopen(&quot;in.txt&quot;,&quot;r&quot;,stdin);
//	  freopen(&quot;out.txt&quot;,&quot;w&quot;,stdout);

int t;
sf(t);
TEST_CASE(t)
{
int n,q;
sff(n,q);
loop1(i,n) sfl(ara[i]);
init(1,1,n);
LINE_PRINT_CASE;
while(q--)
{
int a,b,c,d;
sfff(a,b,c);
sf(d);
if(a==1)
{
update(1,1,n,b,c,d);
}
else
{
data ret=query(1,1,n,b,c);
ll sum=ret.sum;
ret=query(1,1,n,b+1,c);
ll ssum=(ret.ssum-(((ll)(b)*ret.sum)%MOD)+MOD)%MOD;
ssum*=(ll)d%MOD;
ssum%=MOD;
ll ans=(sum+ssum)%MOD;
printf(&quot;%lld\n&quot;,ans);
}
}

}

return 0;
}

```

# SPOJ: SEGSQRSS – Sum of Squares with Segment Tree

0

Solution Idea:

This problem is a nice example of segment tree with lazy propagation. All you need to solve this kind of problem is to do some experiment with the formula and convert them into a suitable form. From which we can drive the segment tree solution.

In this problem given an integer array A=[a1,a2,a3,a4…]. You need to perform 3 kinds of operation on it.

type 0: Set all the element in the interval l to r to x.
type 1: Increase all the element in the interval l to r by x.
type 2: Print the value of square sum of the interval l to r. For example let l=1 and r=3. then you need to print a1^2 + a2^2 + a3^2.

At first think we have a pre-calculated square sum array. Then if we perform type 1 operation which is increase it by x then what scenario will happen?

Our sum now become (a1+x)^2 + (a2+x)^2 + (a3+x)^2. Let expand this equation-

(a1+x)^2 + (a2+x)^2 + (a3+x)^2
= a1^2+ 2*a1*x + x^2 + a2^2+ 2*a2*x + x^2 + a3^2+ 2*a3*x + x^2
= (a1^2 + a2^2 + a3^2) + 3*x^2 + 2*x*(a1 + a2 + a3)

for a range l to r the generalized equation is –

(a_l^2 + a_l+1^2 +….+ a_r^2) + (r-l+1)*x^2 + 2*x*(a_l + a_l+1 +….+ a_r)

We can store the information about sum, square_sum, type_0 lazy and type_1 lazy of an interval in each segment tree node. And perform type 1 operation then we can calculate then sum value for a node over a range by multiplication and square sum value by above equation.

For every type 0 operation we can simply put the value of sum and square sum by using some basic calculation and arithmetic multiplication operation.

And for every type 2 operation we just need to get the sum of square sum value over a range l to r.

So we can perform each of 3 operation using a segment tree with lazy propagation.

```
#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
struct data
{
ll sum,sqrsum,lazy,upd;
};

data tree[3*mx];

ll ara[mx];

void push_down(int n, int b, int e)
{
segment_tree;
if(tree[n].upd)
{
tree[l].lazy=0;
tree[r].lazy=0;
tree[l].sum=(mid-b+1)*tree[n].upd;
tree[l].sqrsum=(mid-b+1)*sqr(tree[n].upd);
tree[r].sum=(e-mid)*tree[n].upd;
tree[r].sqrsum=(e-mid)*sqr(tree[n].upd);
tree[l].upd=tree[n].upd;
tree[r].upd=tree[n].upd;
tree[n].upd=0;
}
if(tree[n].lazy)
{
//        tree[l].lazy+=tree[n].lazy;
//        tree[l].status=1;
//        tree[r].lazy+=tree[n].lazy;
//        tree[r].status=1;
//        tree[n].status=0;
tree[l].sqrsum+=(tree[l].sum*(2*tree[n].lazy))+(mid-b+1)*sqr(tree[n].lazy);
tree[r].sqrsum+=(tree[r].sum*(2*tree[n].lazy))+(e-mid)*sqr(tree[n].lazy);
tree[l].sum+=(mid-b+1)*tree[n].lazy;
tree[r].sum+=(e-mid)*tree[n].lazy;
tree[l].lazy+=tree[n].lazy;
tree[r].lazy+=tree[n].lazy;
tree[n].lazy=0;
}
}

void init(int n, int b, int e)
{
if(b==e)
{
tree[n].sum=ara[b];
tree[n].sqrsum=sqr(ara[b]);
tree[n].lazy=0;
tree[n].upd=0;
return;
}
segment_tree;
init(l,b,mid);
init(r,mid+1,e);
tree[n].sum=tree[l].sum+tree[r].sum;
tree[n].sqrsum=tree[l].sqrsum+tree[r].sqrsum;
tree[n].lazy=0;
tree[n].upd=0;
}

void update(int n, int b, int e, int i, int j, ll val, int type)
{
if(b&gt;j || e&lt;i) return;
if(b&gt;=i &amp;&amp; e&lt;=j)
{
if(type==0)
{
tree[n].sum=(e-b+1)*val;
tree[n].sqrsum=(e-b+1)*sqr(val);
tree[n].lazy=0;
tree[n].upd=val;
}
else
{
tree[n].sqrsum+=(tree[n].sum*2*val)+((e-b+1)*sqr(val));
tree[n].sum+=(e-b+1)*val;
tree[n].lazy+=val;
}
return;
}

push_down(n,b,e);

segment_tree;

update(l,b,mid,i,j,val,type);
update(r,mid+1,e,i,j,val,type);
tree[n].sum=tree[l].sum+tree[r].sum;
tree[n].sqrsum=tree[l].sqrsum+tree[r].sqrsum;
}

ll 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].sqrsum;
push_down(n,b,e);
segment_tree;
ll p=query(l,b,mid,i,j);
ll q=query(r,mid+1,e,i,j);
return p+q;
}

int main()
{
//    freopen(&quot;in.txt&quot;,&quot;r&quot;,stdin);
//	  freopen(&quot;out.txt&quot;,&quot;w&quot;,stdout);

int t;
sf(t);
TEST_CASE(t)
{
int n,q;
sff(n,q);
loop1(i,n) sfl(ara[i]);
init(1,1,n);
LINE_PRINT_CASE;
while(q--)
{
int a,b,c,d;
sfff(a,b,c);
if(a==2)
{
ll ans=query(1,1,n,b,c);
printf(&quot;%lld\n&quot;,ans);
}
else
{
sf(d);
if(a==1)
update(1,1,n,b,c,d,1);
else
update(1,1,n,b,c,d,0);
}
}
}

return 0;
}

```

# SPOJ: BTCODE_G – Coloring Trees

0

Solution Idea:

The solution idea of this problem is not so hard. This one can be solve with Heavy Light Decomposition. But as the time limit is very tight we need a lot of optimization. To solve this problem we need to convert the tree to an array using Euler tour or in-order traversal on the tree. Now there are 2 types of operation. Let consider each tree node as a special node which contain a 30 size array for storing it’s color information.

Now let a paint operation is 1 a b. Now add +1 to array index [b] of discovery_time(a)node and -1 to array index [b] of finishing_time[a) node.

For each query operation 2 a b we need to check for each color. Now at first determine the LCA of node a and b in the original given graph. Now for the i’th color we need to count the summation of array[i] from root node to discovery_time(a) node. Let this summation is x. Then we need count the summation of array[i] from root node to discovery_time(b) node. Let this summation is y. Here we add root to LCA node’s value twice but we don’t need this. So we need to subtract this from the sum. So we need count the summation of array[i] from root node to discovery_time(LCA) node. Let this is z. Now if we subtract 2*z then we will delete the LCA node’s value so wee need to add this value too.

So Total number of i’th color node in the path from a to b is

=x+y-(2*z)+ array[i] value of LCA node.

We can do this for each of 30 color and When we get a color which fill every node in the range a to b then we will print YES otherwise NO.

I have implemented this solution using segment tree but got TLE several time. Then I implement the same idea using BIT and got AC. I think the main idea behind the TLE in segment tree is wee need to copy the 30 size array element several time in segment tree. Which will take some time. As the time limit is very tight, the segment tree solution is inefficient.

```
#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 200005

vector&lt;int&gt;g[mx];
pii times[mx];
int dfsnum;
int parent[mx],level[mx],node_cnt[2*mx];

void dfs(int u, int p)
{
times[u].ff=++dfsnum;
parent[u]=p;
for(int i=0; i&lt;SZ(g[u]); i++)
{
int v=g[u][i];
if(v==p) continue;
level[v]=level[u]+1;
dfs(v,u);
}
times[u].ss=++dfsnum;
}

int dp[mx][21];

int func_lca(int idx, int p)
{
if(p==0)
{
return dp[idx][p]=parent[idx];
}
int &amp;ret=dp[idx][p];
if(ret!=-1) return ret;
int u=func_lca(idx,p-1);
ret=func_lca(u,p-1);
return ret;
}

int lca_query(int p, int q)
{
if(level[q]&gt;level[p]) swap(p,q);
for(int i=20;i&gt;=0;i--)
{
int a=func_lca(p,i);
if(level[a]&gt;=level[q])
p=a;
}

if(p==q) return p;

for(int i=20;i&gt;=0;i--)
{
int a=func_lca(p,i);
int b=func_lca(q,i);
if(a!=b)
{
p=a;
q=b;
}
}

return parent[p];
}

int tree[31][2*mx];

void update(int id, int idx, int val)
{
for(;idx&lt;=dfsnum &amp;&amp; idx;idx+=idx&amp;-idx)
tree[id][idx]+=val;
}

int query(int id, int idx)
{
int ret=0;
for(;idx;idx-=idx&amp;-idx)
ret+=tree[id][idx];
return ret;
}

int main()
{
//    freopen(&quot;in.txt&quot;,&quot;r&quot;,stdin);
//	  freopen(&quot;out.txt&quot;,&quot;w&quot;,stdout);

int n;
sf(n);
for(int i=1; i&lt;n; i++)
{
int a,b;
sff(a,b);
g[a].pb(b);
g[b].pb(a);
}

dfs(0,0);

int qq;
sf(qq);

for(int i=0;i&lt;n;i++)
{
node_cnt[times[i].ff]=1;
node_cnt[times[i].ss]=-1;
}

partial_sum(node_cnt,node_cnt+(2*mx),node_cnt);

ms(dp,-1);

while(qq--)
{
int a,b,c;
sfff(a,b,c);
if(a==1)
{
update(c,times[b].ff,1);
update(c,times[b].ss,-1);
}
else
{
if(times[b].ff&gt;times.ff)
swap(b,c);

bool ans=0;

int lca=lca_query(b,c);

for(int i=1;i&lt;=30;i++)
{
int x=query(i,times.ff);
int y=query(i,times[b].ff);
int z=query(i,times[lca].ff);
x=(x+y-(2*z));
x+=query(i,times[lca].ff)-query(i,times[lca].ff-1);
y=node_cnt[times[b].ff]+node_cnt[times.ff]-(2*node_cnt[times[lca].ff])+(node_cnt[times[lca].ff]-node_cnt[times[lca].ff-1]);
if(x)
{
if(x==y)
ans=1;
break;
}
}

if(ans==0)
printf(&quot;NO\n&quot;);
else
printf(&quot;YES\n&quot;);

}
}
return 0;
}

```

# SPOJ: SUMSUM – Enjoy Sum with Operations

0

Solution Idea:

This one is a interesting problem. The main idea behind the problem is combinatorics. At first we need to count the number of 1 bit and number of 0 bit in the i_th position of a given number. Then we need to do some combinatorial calculation.

Now think about an array of 4 number A=[1,0,1,0]. If we perform OR operation in the range 1 to 4 what will we get? We get 5 as answer. If we form pair from this number and perform OR operation between then then we will get the answer. Now as here is 4 element so we can perform pair in 4*(4-1)/2 = 6 ways. Now comes the tricky part. For OR operation how many pair will contribute nothing ?? Here is 2 zero and we can form 1 pair from this two element who will not contribute anything. So for i_th bit position the contribution for OR operation is number_of_active_pair*(2^i).

For AND operation we need to consider the number of pair can be form by 1 bit and for XOR the number of pair will be number_of_one_bit*number_of_zero_bit.

We can consider a number as an array of 30 element whose value is either 0 or 1 and perform above operation.

```
#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

int tree[28][mx];

void update(int id, int idx, int val)
{
for(;idx&lt;mx &amp;&amp; idx;idx+=idx&amp;-idx)
tree[id][idx]+=val;
}

ll query(int id, int idx)
{
ll ret=0;
for(;idx;idx-=idx&amp;-idx)
ret+=tree[id][idx];
return ret;
}

int ara[mx];

int main()
{
//    freopen(&quot;in.txt&quot;,&quot;r&quot;,stdin);
//	  freopen(&quot;out.txt&quot;,&quot;w&quot;,stdout);

int n,q;
sff(n,q);
for(int i=1;i&lt;=n;i++) sf(ara[i]);

for(int i=1;i&lt;=n;i++)
{
for(int j=0;j&lt;28;j++)
{
if(check(ara[i],j))
update(j,i,1);
}
}

while(q--)
{
int a,b,c;
sf(a);
if(a==1)
{
sff(b,c);
for(int i=0;i&lt;28;i++)
{
if(check(ara,i))
update(i,c,-1);
}
ara=b;
for(int i=0;i&lt;28;i++)
{
if(check(ara,i))
update(i,c,1);
}
}
else
{
char ss[10];
scanf(&quot; %s&quot;,&amp;ss);
string str=string(ss);
sff(b,c);
ll temp[30];
for(int i=0;i&lt;28;i++)
{
temp[i]=query(i,c);
}
for(int i=0;i&lt;28;i++)
{
temp[i]-=query(i,b-1);
}

ll ans=0;

for(int i=0;i&lt;28;i++)
{
ll one=temp[i];
ll zero=(c-b+1)-one;
ll pairs=0;
if(str==&quot;OR&quot;)
{
ll total=one+zero;
pairs=(total*(total-1))/2;
pairs-=(zero*(zero-1))/2;
}
else if(str==&quot;AND&quot;)
{
pairs=(one*(one-1))/2;
}
else if(str==&quot;XOR&quot;)
{
pairs=one*zero;
}
ans+=(1LL&lt;&lt;i)*pairs;
}

printf(&quot;%lld\n&quot;,ans);
}
}

return 0;
}

```

# SPOJ: FREQUENT – Frequent values

0

Solution Idea:

In this problem an array of non-decreasing order is given. In each query a range l and r is given. You have to tell the maximum frequency of a number in the range l to r.

Now as the given array is in non-decreasing order. So you can replace array a = {1 1 1 2 2 3 3 3 3} by another array b = {3 2 4}. After this compression the problem is converted to a Range Maximum query problem which can be easily solved by segment tree or sparse table. Now for each query determine the index in the compress array and perform range maximum query on that range.

Think about the first and index of the query range. You need to handle them separately.

```
#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

pii block[mx];

int dp[mx][20];

int ara[mx],id[mx],cnt=0,table[mx];

int func(int idx, int p)
{
if(idx&gt;cnt) return 0;
if(p==0)
{
return dp[idx][p]=table[idx];
}
int &amp;ret=dp[idx][p];
if(ret!=-1) return ret;
ret=max(func(idx,p-1),func(idx+(1&lt;&lt;(p-1)),p-1));
return ret;
}

int query(int l, int r)
{
if(r&lt;l) return 0;
if(l==r) return table[l];
int log=(int)log2(r-l);
return max(func(l,log),func(r-(1&lt;&lt;log)+1,log));
}

int main()
{

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

int n,q;
while(sf(n))
{
if(n==0) break;
sf(q);
for(int i=1; i&lt;=n; i++)
{
sf(ara[i]);
}
cnt=0;
ms(table,0);
ms(dp,-1);
//        ms(block,0);
//        ms(id,0);
for(int i=1; i&lt;=n; i++)
{
int j=i;
cnt++;
block[cnt].ff=i;
while(ara[j]==ara[i] &amp;&amp; j&lt;=n)
{
id[j]=cnt;
table[cnt]++;
j++;
}
j--;
block[cnt].ss=j;
i=j;
}

while(q--)
{
int a,b;
sff(a,b);
if(id[a]==id[b])
{
printf(&quot;%d\n&quot;,b-a+1);
}
else
{
int l=id[a];
int r=id[b];
int ans=0;
ans=max(ans,block[l].ss-a+1);
l++;
ans=max(ans,b-block[r].ff+1);
r--;
ans=max(ans,query(l,r));
printf(&quot;%d\n&quot;,ans);
}
}

}

return 0;
}

```