# HackerEarth Problem B Pikachu vs Team Meowstic and Helping Hand solution

In this HackerEarth ( Problem B) Pikachu vs Team Meowstic and Helping Hand problem solution Pikachu loves battling with other Pokemon. This time he has a team of N Meowstic to fight, ith of which has strength ai. He wants to fight with all of them K times. Team Meowstic came to know about this and now they have devised a strategy to battle against the mighty Pikachu.

All the N Meowstic stand in a straight line numbered from 1 to N. Before every round of battle, they simultaneously use a move, called Helping Hand. It changes the attacking power of Team Meowstic as follows:

The attacking power of first Meowstic remains ai.
The attacking power of remaining Meowstic changes as ai = ai | ai - 1 where 2 <= i <= N and A|B represents the bitwise OR of A and B
For example, if the current attacking powers are [2,1,4,6,3], after using the Helping Hand, the powers change to [2,1|2,4|1,6|4,3|6], or [2,3,5,6,7].

Help Pikachu by finding the attacking powers of all Meowstic when he fights each of them for the last time, that is, for the Kth round.

## HackerEarth ( Problem B ) Pikachu vs Team Meowstic and Helping Hand problem solution.

`#include <bits/stdc++.h>using namespace std;int main(){    int N,K;    cin>>N>>K;    vector<int>x(N);    for(int i=0;i<N;i++){        cin>>x[i];    }    vector<int>last_set(30,INT_MIN);    vector<int>ans(N);    for(int i=0;i<N;i++){        for(int j=0;j<30;j++){            if((x[i]>>j)&1)last_set[j]=i;            if(i<=K+last_set[j])ans[i]|=(1<<j);        }    }    for(auto i:ans)cout<<i<<" ";}`

### Second solution

`n,k = map(int, raw_input().split())arr = map(int, raw_input().split())ans = [0 for __ in xrange(r)]for b in xrange(30):    ds = -1    for i in xrange(n):        if arr[i]&1: ds = i+k        arr[i] >>= 1        ans[i] |= (i<=ds)<<bprint " ".join(map(str, ans))`