# HackerEarth The Great Kian problem solution

In this HackerEarth The Great Kian problem solution The great Kian is looking for a smart prime minister. He's looking for a guy who can solve the OLP (Old Legendary Problem). OLP is an old problem (obviously) that no one was able to solve it yet (like P=NP).

But still, you want to be the prime minister really bad. So here's the problem:

Given the sequence a1, a2, ..., an find the three values a1 + a4 + a7 + ..., a2 + a5 + a8 + ... and a3 + a6 + a9 + ... (these summations go on while the indexes are valid).

## HackerEarth The Great Kian problem solution.

`#include <bits/stdc++.h>#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/tree_policy.hpp>using namespace __gnu_pbds;using namespace std;#define Foreach(i, c) for(__typeof((c).begin()) i = (c).begin(); i != (c).end(); ++i)#define For(i,a,b) for(int (i)=(a);(i) < (b); ++(i))#define rof(i,a,b) for(int (i)=(a);(i) > (b); --(i))#define rep(i, c) for(auto &(i) : (c))#define x first#define y second#define pb push_back#define PB pop_back()#define iOS ios_base::sync_with_stdio(false)#define sqr(a) (((a) * (a)))#define all(a) a.begin() , a.end()#define error(x) cerr << #x << " = " << (x) <<endl#define Error(a,b) cerr<<"( "<<#a<<" , "<<#b<<" ) = ( "<<(a)<<" , "<<(b)<<" )\n";#define errop(a) cerr<<#a<<" = ( "<<((a).x)<<" , "<<((a).y)<<" )\n";#define coud(a,b) cout<<fixed << setprecision((b)) << (a)#define L(x) ((x)<<1)#define R(x) (((x)<<1)+1)#define umap unordered_map#define double long doubletypedef long long ll;typedef pair<int,int>pii;typedef vector<int> vi;typedef complex<double> point;template <typename T> using os =  tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;template <class T>  inline void smax(T &x,T y){ x = max((x), (y));}template <class T>  inline void smin(T &x,T y){ x = min((x), (y));}ll x[3];int main(){  iOS;  int n, a;  cin >> n;  For(i,0,n){    cin >> a;    x[i % 3] += (ll)a;  }  For(i,0,3)    cout << x[i] << ' ';  cout << endl;  return 0;}`

### Second solution

`#include<bits/stdc++.h>const int N = 100031;using namespace std;int n;int ar[N];long long res[10];int main(){  ios_base::sync_with_stdio(0);  //cin.tie(0);  cin >> n;  for (int i = 0; i < n; i++)  {    cin >> ar[i];  }  for (int i = 0; i < 3; i++)  {    int cur = i;    do    {      res[i] += ar[cur];      cur += 3;    } while (cur < n);  }  for (int i = 0; i < 3; i++)  {    if (i)      cout << " ";    cout << res[i];  }  cout << endl;  return 0;}`