ODTPIC | Editorial by puns

**Odd Topic** is a classic **bitset** implementation problem.
If you need basic info about bitsets then you should visit https://codeforces.com/blog/entry/73558 .
We maintain arrays of bitsets here for inputs and return count of bitset ans as answer here.
Basically , you save a bitset for every position from 0 to n or 0 to m and toggle the bit for every occurence of the same number in the indexes of the particular bitset. 
For example, in a bitset of size 5 : input 1 : 3           bitset : 01000 (count bits from right,beginning from 0) 
                                                             input 2 : 2           bitset : 01100 
                                                             input 3 : 3           bitset : 00100  (notice the bit at index 3 from right toggles to become 0 again at even occurence of 3)
                                                             input 4 : 4           bitset : 10100 ...and so on.
Now for both bitsets a & b,you can simply xor out the bitsets at positions (r-1) and (l-2) for a & (rt-1) and (lt-2) for b.
Finally ,the two bitsets obtained from above steps ,we xor them to obtain our "ans" bitset.
The count of 1's in this bitset is our answer.
```
 
#include <bits/stdc++.h>

using namespace std;
#define fin for(int i = 0; i < n; i++)
 
#define rep(i, n) for(int i = 0; i < n; i++)
 
 
 
int main()
{
 
	 int n,m,q,l,r,lt,rt;
	 cin>>n>>m>>q;
	 int a[n],b[m];
	 bitset<4005> aa[n],bb[m],ans,ansa,ansb;

fin{
	 	cin>>a[i];
	 	if(i==0){
	 		aa[0][a[i]]=true;
	 	}
	 	if(i>0){
	 	if(aa[i-1][a[i]]){
	 		aa[i]|=aa[i-1];
	 		aa[i][a[i]]=false;
	 
	 	}else{
	 		aa[i][a[i]]=true;
	 		aa[i]|=aa[i-1];
	  
	 	}}
	 }
	 rep(i,m){
	 	cin>>b[i];
	 	if(i==0){
	 		bb[0][b[i]]=true;
	 	}
	 	if(i>0){
	 	if(bb[i-1][b[i]]){
	 		bb[i]|=bb[i-1];
	 		bb[i][b[i]]=false;
	  
	 	}else{
	 		bb[i][b[i]]=true;
	 		bb[i]|=bb[i-1];
	 
	 	}}
	 }

rep(i,q){
	 	cin>>l>>r>>lt>>rt;
 
	 	if(l==1){
	 		ansa=aa[r-1];
	 	}
	 	else{
	 		ansa= (aa[r-1]^(aa[l-2]));
	 	}

if(lt==1){
	 		ansb=bb[rt-1];
	 	}else{
	 		ansb= (bb[rt-1]^(bb[lt-2]));
	 	} 
	 
 
	 	ans=((ansa^(ansb)));
	 	 
	 	cout<<ans.count()<<endl;
	 }

return 0;
}
```

ODTPIC | Editorial by puns

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