Codeforces - Next Round

Problem solving with programming: Next Round - A problem from Codeforces
http://codeforces.com/problemset/problem/158/A
Given a list of scores of contestants in a competition, arranged in decreasing order. We have to calculate the number of people qualified for the next round.
This is determined as follows, consider the score at a given index, Whoever scored more than this score, qualifies for the next round as long as they have a score of greater than zero.

Case 1: score[index] > 0
    In this case, we have to find the right most occurrence of the cutoff, which will give the required answer

Case 2: score[index[ <= 0
    In this case, we have to find the left most occurrence of zero, and count the remaining elements from the beginning.

 cin >> n >> k;


 vector<int> scores(n);


 int i;


 for( i = 0; i < n; i++ )


 {


  cin >> scores[i];


 }


 


 int cutoff = scores[k-1];


 if( cutoff < 0 )


  cutoff = 0;


 if( cutoff > 0 )


 {


  int ind = rightMost(scores,cutoff);


  cout << ind+1 << endl;


 }


 else


 {


  vector<int>::iterator loIt;


  int ind = leftMost(scores,cutoff);


  if( ind <= 0 )


  {


   cout << "0" << endl;


  }


  else


  {


   cout << ind << endl;


  }


 }

int leftMost( vector<int>& nums, int s )


{


 int l = 0, h = nums.size()-1;


 while ( l <= h )


 {


  int mid = l + (h-l)/2;


  if( nums[mid] == s )


  {


   if( mid > 0 && nums[mid] == nums[mid-1] )


   {


    h = mid-1;


   }


   else


    return mid;


  }


  else if( nums[mid] > s )


  {


   h = mid-1;


  }


  else


  {


   l = mid+1;


  }


 }


 return -1;


}


int rightMost( vector<int>& nums, int s)


{


 int l = 0, h = nums.size()-1;


 while ( l <= h )


 {


  int mid = l + (h-l)/2;


  if( nums[mid] == s )


  {


   if( mid < nums.size()-1 && nums[mid] == nums[mid+1] )


   {


    l = mid+1;


   }


   else


    return mid;


  }


  else if( nums[mid] > s )


  {


   h = mid-1;


  }


  else


  {


   l = mid+1;


  }


 }


 return -1;


}

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