All In One Coding

INTRODUCTORY PROBLEMS 1 . WEIRD ALGORITHM SOLUTION: #include < bits / stdc ++. h > using namespace std ; int main (){ long long n ; //beacuse n can be greater than max size of int. cin >> n ; while ( n != 1 ){ // loop will break when n become 1. cout << n << " " ; if ( n & 1 ) n = n * 3 + 1 ; // when n is odd. else n = n / 2 ; // when n is even. } cout << "1" ; } 2 . MISSING NUMBER SOLUTION : LOGIC: just sum all the inputs and subtract from total sum(n*(n+1)/2). #include < bits / stdc ++. h > using namespace std ; int main (){         long long n , s , sum = 0 ;                cin >> n ;           for ( long long i = 0 ; i < n - 1 ; i ++){              cin >> s ; sum += s ; //sum all the inputs.           }    ...

JUNE LONG CHALLENGE                                             Problem code : Contain /* Algorithm :-   --> We will create concentric hulls as many as possible.        This can be done using a tracking array to keep track of points which are still unused and passing.        it again in convex hull function We will do this until our track array becomes empty . Algorithm used :-        Finding convex hull using graham scan in O(nlogn) ( For AC )        Finding convex hull using jarvis march in O(n^2) ( For 35 points )  --> Then we will memoize all these convex hulls.        Now we cant create hulls for each query as it will result in  O(q*n^2) ...

JUNE LONG CHALLENGE:                   Problem code: EOEO The Tom and Jerry Game!  logic : For odd case: you just have to find all the values which are even. For even case : Jerry only wins when Jerry strength is even and tom strength is odd so we have to find only those values of JS which can divided by 2 more times than times who in simpler terms (or in bit terms ) we have to to find all that values which have mores zeroes in the end than TS CODE:  #include<bits/stdc++.h> using namespace std; int main(){     long long t;     cin>>t;     while(t--){         long long tom,jerry,count=0;         cin>>tom;         if(tom%2!=0)cout<<(tom-1)/2<<"\n"; /* if tom strength is odd  */     ...

Who am I?.   I am Manish Gupta,a Computer Science Engineering undergrad pursuing my B.Tech in NIT Srinagar and a competitive programmer.I am well versed in C++,Python,JavaScript,C and Bash and as programmer I am eager to learn new things.I'll make sure to post each and every line of elegant code that I write and also ensure to share solutions of some beautiful problems that I encounter.I'll also share some of the projects that I happen to work on. I am also interested in Ethical hacking and am looking forward to pursue a side career in it so I might share some of the "shady" code here :)