i am trying to solve this problem but not getting idea can anybody help please with this problem
Most probably I overkilled the solution so feel free to ignore it, I’d appreciate if someone can share a neater solution.
So for any given N we have to calculate the following S,
Now we will use the standard trick of bit manipulation. let c[i] be the number of terms where bit i is set. So final answer would just be
So all we need to do now is to find c[i] for all i's.
For any arbitrary i consider these 2 subcases.
CASE 1 (EVEN & ODD) :
In binary, any even number would look like E=100\underbrace{1}_\text{i}.....0 (some prefix of bits ending with a 0)
Then the next odd number just after E odd would look like O=100\underbrace{1}_\text{i}.....1 (same prefix ending with a 1)
Hence it’s pretty trivial to see that E\&O will have i th bit set if it’s set in E.
Observation 1: If any even number has i th bit set then its bitwise AND with next odd will contribute 1 to the bit count of i i.e. increase c[i] by 1
CASE 2(ODD & EVEN):
Similarly say, O=100\underbrace{1}_\text{i}..\underbrace{0}_\text{k}...1
Here k is the lowest unset bit of O

CASE A: k <i :
E=100\underbrace{1}_\text{i}..\underbrace{1}_\text{k}...0. It’s pretty clear that k would become the lowest set bit and bit i will be set in E if it was set in O. Hence O\&E will also contribute 1 to c[i] as in EVEN & ODD case. 
CASE B: k>i:
O=10\underbrace{0}_\text{k}\underbrace{1}_\text{i}.....1
E=10\underbrace{1}_\text{k}\underbrace{0}_\text{i}.....0
Now in this case i th bit will become unset and hence won’t contribute to the answer.
OBSERVATION 2: If any odd number has i th bit set then its bitwise AND with next even will contribute 1 to the bit count of i i.e. increase c[i] by 1 except for the case when all bits after i (including i ofc) are set..
So the final answer can be deduced by the following equation.
F(N, i)  This returns the count of numbers that are less than equal to N and have their i th bit set. Note that we’ve added the contributions from both even and odd numbers in F(N,i).
G(N,i) This returns the count of numbers that are less than N and have all bits set after i (including i )
And these functions are very trivial to compute. For F(N,i) you can refer this and G(N,i) would just be equal to N right shifted i+1 times .
Adding the code just for the sake of completeness.
AC_CODE
#include "bits/stdc++.h"
#define int long long
using namespace std ;
const int M = 1e9+7 ;
int pM(int b,int p){
int r=1 ;
for(;p;b=(b*b)%M,p/=2)
if(p&1)
r=(r*b)%M ;
return r ;
}
int F(int n, int i){
int ans = (n >> (i + 1LL)) << i;
if (((n >> i) & 1LL) != 0)
ans += n & ((1LL << i)  1LL);
return ans%M;
}
int G(int n,int i){
return (n>>(i+1)) ;
}
void solve(){
int n,ans=0;cin >> n ;
vector<int>c(61) ;
for(int i=0;i<61;i++)
c[i]=F(n,i)G(n,i);
for(int i=0;i<61;i++)
ans=((c[i]*pM(2,i))%M+ans)%M ;
cout << ans << '\n' ;
}
signed main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int TC ;
cin >>TC ;
while(TC)
solve() ;
}
thank you very much bro
i think you should add this as editorial in problem so that people will get directly from their
request : could you explain the next question ( CodeChef: Practical coding for everyone) you can do that whenever you get free time and thanks again
My approach was that for each bit we can calculate the no of valid pairs in which it is set.
So for any ith bit a valid pairs starts from its every odd multiple and its next 2^i1 pairs are valid
For example for 4th bit, 2^4 = 16, so its first pair is (16,17 ) ,(17,18 ) … (30,31)…
then we next valid pair occurs at (48,49)(49,50)…(62,63) and then next occurs at (80,81).
So we can calulate easily no of odd factors of 2^i less than n and then for each of those we get (2^i1) pairs and each of which adds 2^i to the answer
we get (n/2^i)/2 valid factors
but if (n/2^i) is odd then we get (n/2^i)/2 such points from which we get complete (2^i1) pairs.
Apart from this we have to add some pairs seprately as we won’t get complete (2^i 1) pairs, but some of them.
E.g if n = 85, for 4th bit, 85/16 = 5, we get (5/2 = 2) complete ones, we also have to add contribution of (n80) = (5) more pairs, which are from (80,81)…(84,85)