Subtask 1 is a special case. Iterate all possible ~k \le n~, and try to take exactly ~k~ molecules.

Subtask 2 is a special case. Iterate all possible ~k_1, k_2 : k_1+k_2 \le n~, try to take exactly ~k_1~ molecules of minimal weight and exactly ~k_2~ molecules of maximal weight.

Subtasks 3 and 4 may be solved using dynamic programming. The task is a classic knapsack problem. Use ~\mathcal O(nu)~ of time, and ~\mathcal O(u)~ of space.

You may optimize the dynamic programming, use bitset and you'll pass Subtask 5.

We suggested, a contestant who solves Subtasks 5 and 6 will invent a greedy approach. If you implement greedy in ~\mathcal O(n^2)~, you'll pass only Subtask 5. Good time to pass Subtask 6 is ~\mathcal O(n \log n)~.

There are 3 correct greedy solutions. All of them start with sorting the array of weights in nondecreasing order.

Greedy 1. Let fix ~k~, number of molecules to take. We can choose set of size ~k~ with sum in ~[l \dots r]~ iff ~minSum[k] \le u~ and ~maxSum[k] \le l~. Where ~minSum[]~ is partial minimums on prefixes and ~maxSum[]~ is partial maximums on suffixes. Both of them may be precalculated in ~\mathcal O(n)~.

Proof. Let's take minimal possible ~k~ molecules, its summary weight does not exceed ~l~. Let's change the set smoothly from "~k~ minimal molecules" to "~k~ maximal molecules". One step: drop any one molecule, and take any other. Each step changes the sum by at most ~u-l~. The last value of the sum is at least ~l~. So one of the intermediate steps gives ~l \le sum \le u~. ■

Greedy 2. There exists an answer which forms a segment. Use two pointers to find it in ~\mathcal O(n)~.

Proof. Let's fix ~k~ – number of molecules in the answer. The smallest ~k~ molecules form the leftmost segment, and the biggest ~k~ form the rightmost segment. Let's change the set smoothly from "the leftmost segment" to "the rightmost segment". One step: drop the leftmost molecule, and add a new one at the right. ■

Greedy 3. There exists an answer which forms a union of prefix and suffix. Use two pointers to find it in ~\mathcal O(n)~.

Proof. Let's fix ~k~ – number of molecules in the answer. The smallest ~k~ molecules form prefix, the biggest ~k~ form suffix. Let's change the set smoothly from "prefix" to "suffix". One step: make prefix shorter by one, make suffix longer by ~k~. ■