New Year Transportation | Editorial by destroyerguy

Explanation:-
        This problem is a basic implementation problem which can be solved in O(n) time complexity.The user can jump to specific indices  only,i.e. if we consider 0 based indexing user is initially at index 0,from here he can only go to a[0],then from a[0] to a[0] + a[a[0]] and so on.
        We can run a while loop and fetch the position of the user at each iteration .If we get the required position t-1 (for 0 based indexing),we set flag=1 and break from the while loop.Also we break from the loop when the current position of the user becomes more than n-2 (since there are only n-1 positions).
        We print YES when flag is 1 or NO when it is 0.
 Code:-
```
                    /* 
                        Coded with love by :- Ashutosh Singh
                        Indian Institute of Information Technology Sricity
                    */
                    #include <bits/stdc++.h>
                    //#include<boost/multiprecision/cpp_ll.hpp>
                    //using namespace boost::multiprecision;
                    using namespace std;
                    typedef long long int ll;
                    #define mod 1000000007
                    #define f(i, n) for (ll i = 0; i < n; i++)
                    #define fp(i, k, n) for (ll i = k; i <= n; i++)
                    #define fr(i, k, n) for (ll i = k; i >= n; i--)
                    typedef map<string, ll> msi;
                    typedef map<string, string> mss;
                    int main()
                    {
                      int n, t, x, flag = 0;
                          cin >> n >> t;
                      int a[n-1];
                      f(i, n - 1)
                              cin >> a[i];
                      x = a[0];
                      while (1)
                      {
                        if (x == t - 1)
                        {
                          flag = 1;
                          break;
                        }
                        if (x > n - 2)
                          break;
                        x = x + a[x];
                      }
                      if (flag)
                        cout << "YES\n";
                      else
                        cout << "NO\n";
                    }
```

New Year Transportation | Editorial by destroyerguy

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