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**HackerEarth Special Bit Numbers problem solution**, A number is known as a special bit number if its binary representation contains at least two consecutive 1's or set bits. For example 7 with binary representation 111 is a special bit number. Similarly, 3(11) is also a special bit number as it contains at least two consecutive set bits or ones.Now the problem is, You are given an array of N integers and Q queries. Each query is defined by two integers L, R. You have to output the count of special bit numbers in the range L to R.

## HackerEarth Special Bit Numbers problem solution.

`#include <bits/stdc++.h>`

using namespace std;

int main()

{

int n, q;

cin >> n >> q;

int pre[n + 1] = {};

for(int i = 1; i <= n; i++){

int x;

cin >> x;

pre[i] = pre[i - 1];

if(x & (x << 1))

pre[i]++;

}

while(q--){

int l, r;

cin >> l >> r;

cout << pre[r] - pre[l - 1] << endl;

}

return 0;

}

### Second solution

`#include<bits/stdc++.h>`

#define LL long long int

#define M 1000000007

#define endl "\n"

#define eps 0.00000001

#define check(a , b , c) assert(a >= b && a <= c)

LL pow(LL a,LL b,LL m){LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}

LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}

LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}

using namespace std;

int a[100001];

int main()

{

ios_base::sync_with_stdio(0);

int n , q;

cin >> n >> q;

for(int i = 1; i <= n; i++)

{

cin >> a[i];

int pre = 0 , cur = 0;

while(a[i])

{

cur = a[i] % 2;

if(pre == 1 && cur == 1)

{

a[i] = 1;

break;

}

pre = cur;

a[i] /= 2;

}

a[i] = a[i - 1] + a[i];

}

while(q--)

{

int l , r;

cin >> l >> r;

cout << a[r] - a[l - 1] << endl;

}

}

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