 # How to solve Xor sum problem from Hackerearth?

I am trying to solve Xor sum problem from Hackerearth Practice.

I am not getting the idea on how to apply the merge operation in the construction of segment tree for this problem. Any hints with code is appreciated.

Thank you You need the count of each bit, Let a_i denote the number of times bit i is set, and b_i denote the number of times it is not set in the given range. The answer is \sum (\binom{a_i}{1}\binom{b_i}{2} + \binom{a_i}{3}\binom{b_i}{0} )\times 2^i

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I got it, so for a triple xor to be 1 you need to have 1 set bit and 2 non-set bits or 3 set bits. Number of ways of doing that is what given right?

Thank you )

I guess segment tree is used to get those number of set and non-set bits in a range.
Implemeting it using fenwick tree would be quite easier

#include<bits/stdc++.h>
using namespace std;
#define ll long long int
#define MOD 1000000007
ll n,q,a,seg,l,r,po;
/*void build(ll node,ll st,ll en)
{
if(st==en)
{
for(ll i=0;i<=42;i++)
{
if(a[st]&(1<<i))
seg[node][i]=1;
}
}
else
{
ll mid=(st+en)/2;
build(2*node,st,mid);
build(2*node+1,mid+1,en);
for(ll i=0;i<=42;i++)
{
seg[node][i]=seg[2*node]+seg[2*node+1];
}
}
}
ll qry(ll node,ll st,ll en,ll l,ll r,ll idx)
{
if(st>en||st>r||en<l||l>r)
return 0;
else if(st>=l&&en<=r)
return seg[node][idx];
else
{
ll mid=(st+en)/2;
return qry(2*node,st,mid,l,r,idx)+qry(2*node+1,mid+1,en,l,r,idx);
}
}*/
void create()
{
ll i,j;
for(i=0;i<=42;i++)
{
for(j=1;j<=n;j++)
{
seg[j][i]=seg[j-1][i];
if(a[j]&(1LL<<i))
seg[j][i]++;
}
}
}
int main()
{
//ios::sync_with_stdio(0);
//cin.tie(0);
freopen("in05.txt","r",stdin);
freopen("out05.txt","w",stdout);
ll i,j,k;
cin>>n;
po=1;
for(i=1;i<=100002;i++)
{
po[i]=2*po[i-1];
po[i]%=MOD;
}
for(i=1;i<=n;i++)
{
cin>>a[i];
}
//build(1,1,n);
create();
cin>>q>>j;
while(q--)
{
cin>>l>>r;
ll cnt1,cnt0,ans=0,ans1=0;
for(i=0;i<=42;i++)
{
cnt1=seg[r][i]-seg[l-1][i];
cnt0=r-l+1-cnt1;
ans=cnt1*(cnt0*(cnt0-1))/2;
ans+=(cnt1*(cnt1-1)*(cnt1-2))/6;
ans%=MOD;
ans=ans*po[i];
ans%=MOD;
ans1+=ans;
ans1%=MOD;
}
cout<<ans1<<"\n";
}
return 0;
}


Explanation

- cnt1 stores no. of 1s occurring in i-th place of binary representation of numbers in the array from range L to R
- cnt1=seg[r][i]-seg[l-1][i];
- cnt0 stores no. of 0s occurring in i-th place of binary representation of numbers in the array from range L to R
- cnt0=r-l+1-cnt1;

- (cnt0*(cnt0-1))/2 (nC2) calculates no. of distinct pairs of 0s that can be formed from 0s present in i-th place of binary representation of numbers in the array from range L to R
- multiply no. of 1s with the no of 0-pairs calculated
- it will result in no of trios formed whose xor will give 1 (because 1 xor 0 xor 0 is 1)
ans=cnt1*(cnt0*(cnt0-1))/2;

- (cnt1*(cnt1-1)*(cnt1-2))/6 (nC3) calculates no. of distinct trios of 1s that can be formed from 0s present in i-th place of binary representation of numbers in the array from range L to R
- it will also result in no of trios formed whose xor will give 1 (because 1 xor 1 xor 1 is 1)
ans += (cnt1*(cnt1-1)*(cnt1-2))/6;

ans%=MOD;
ans=ans*po[i];
ans%=MOD;
ans1+=ans;
ans1%=MOD;