Approach for better solution - Sum of medians

Question

Here is the question Spoj-WEIRDFN

Problem:

Let us define :

F[1] = 1
F[i] = (a*M[i] + b*i + c)%1000000007 for i > 1

where M[i] is the median of the array {F[1],F[2],..,F[i-1]} Given a,b,c and n, calculate the sum F[1] + F[2] + .. + F[n].

Constraints:

0 <= a,b,c < 1000000007
1 <= n <= 200000

I came up with a solution which is not so efficient

MY SOLUTION::--

#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define mod 1000000007
int main() {
    // your code goes here
    int t;
    scanf("%d",&t);
    while(t--)
    {
      ll a,b,c,sum=0;
      int n;
      scanf("%lld%lld%lld%d",&a,&b,&c,&n);
      ll f[n+1];
      f[1]=1;
      f[0]=0;
      for(int i=2;i<=n;i++)
      {  
          ll temp;
          sort(&f[1],&f[i]);
          temp=f[i/2];
          f[i]=((a*(temp)%mod)+((b*i)%mod)+(c%mod))%mod;
          sum+=f[i];
      }
      printf("%lld\n",sum+f[1]);
    }
    return 0;
}

Can anybody give me hint for for better algorithm or data structure for this task

Can you add some descriptions for the problem, not just the link :) ? — Pham Trung, Mar 03 '15 at 07:17
I can add description but it will make this question long long....that's why i just added link — doodle doodle, Mar 03 '15 at 07:23
Uhm, I can help you, if you don't mind :) this is one of this site's requirement, which help you to get better response and help other also :) — Pham Trung, Mar 03 '15 at 07:25
No,problem sir..you already did much for me(giving me the best approach) — doodle doodle, Mar 03 '15 at 07:26

score 1 · Accepted Answer · edited May 23 '17 at 10:29

1

For each test case, you can maintain a binary search tree, thus you can find the median of n elements in O(log n) time, and you only need O(log n) time to add a new element into the tree.

Thus, we have an O(T*nlogn) algorithm, with T is number of test case, and n is number of elements, which should be enough to pass.

edited May 23 '17 at 10:29

Community

1
1

answered Mar 03 '15 at 07:16

Pham Trung

11,204
2
24
43

I once thought of implementing this using stl(priority_queue)..but than i thought i will still lead me to TLE..BST is the best i can do.Thanks for suggestion – doodle doodle Mar 03 '15 at 07:22
It's important to keep the BST *balanced*, so that you only need to consider the root and possibly its children to find the median. Otherwise (e.g. if all inputs are in increasing order) your BST could wind up looking like a linked list, and the per-test-case running time will balloon out to O(n^2). – j_random_hacker Mar 03 '15 at 12:48

Approach for better solution - Sum of medians

1 Answers1

Linked