Birthday Gifts solution codechef

Birthday Gifts solution codechef

It’s Chef’s birthday. He and his twin have received NN gifts in total. The ii-th gift has a price of AiAi. Each twin wants to keep the most expensive gifts for himself.

The twins take KK turns alternately (each has KK turns, for 2K2⋅K turns in total). It is given that 2K+1N2⋅K+1≤N. In a turn, a person may choose one gift. The only exception is the last turn of the twin who moves second, where he gets to choose two gifts instead of one. Assuming they both choose gifts optimally and you can choose who takes the first turn, find the maximum total cost of the gifts that Chef keeps.

Input Birthday Gifts solution codechef

  • The first line contains an integer TT, the number of test cases. Then the test cases follow.
  • Each test case contains two lines of input.
  • The first line contains two space-separated integers NNKK.
  • The second line contains NN space-separated integers A1,A2,,ANA1,A2,…,AN, the price of the gifts.

Output Birthday Gifts solution codechef

For each test case, output in a single line the answer to the problem.

Constraints Birthday Gifts solution codechef

  • 1T1001≤T≤100
  • 3N1033≤N≤103
  • 1KN121≤K≤N−12
  • 1Ai1091≤Ai≤109

Subtasks Birthday Gifts solution codechef

Subtask #1 (100 points): original constraints

Sample Input Birthday Gifts solution codechef

3 1
1 3 2
3 1
3 1 3
5 2
5 1 3 2 4

Sample Output Birthday Gifts solution codechef


Explanation Birthday Gifts solution codechef

Test Case 11: Chef moves first and he chooses the gift with cost 33. His twin then chooses the gifts of costs 11 and 22.

Test Case 22: Chef allows his brother to choose first and his brother chooses a gift with cost 33. Then Chef chooses the remaining gift with cost 33. Since Chef moves second, he is allowed to choose one more gift, so he chooses gift with cost 11. The total cost of Chef’s gifts is 3+1=43+1=4.

Test Case 33: Chef moves first and at the end he will have the gifts with costs 55 and 33. Hence, the total cost of gifts with Chef = 5+3=85+3=8.


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