HackerEarth Special Bit Numbers problem solution

In this 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;            }    }`