Overview

This problem asks you to maximize the number of students who can participate in crafts, given a rope of length \(L\) that can be cut at most \(K\) times to create at most \(K+1\) pieces, where each student receives one piece of rope that is at least as long as they need.

Analysis

Key Observations

1. At most \(K+1\) pieces of rope: Cutting \(K\) times produces at most \(K+1\) pieces. This means at most \(K+1\) students can receive rope.

2. Greedy approach works: If we want to distribute rope to as many students as possible, we should prioritize students who need shorter rope.

Why distribute to students needing shorter lengths first?

Let’s consider a concrete example.

• Rope length \(L = 10\), maximum cuts \(K = 2\) (at most 3 pieces)
• Lengths needed by students: \(A = [2, 3, 6]\)

Distributing from shortest first: - Give length 2 to student 1 → remaining 8 - Give length 3 to student 2 → remaining 5 - Student 3 needs length 6 but only 5 remains → cannot distribute - Result: Can distribute to 2 students

Distributing from longest first: - Give length 6 to student 3 → remaining 4 - Student 2 needs length 3 → remaining 1 - Student 1 needs length 2 but only 1 remains → cannot distribute - Result: Can distribute to 2 students

In this example the results are the same, but for \(A = [1, 1, 9]\), \(L = 10\): - From shortest: 1 + 1 + 9 = 11 > 10, so 2 students - From longest: 9 + 1 = 10, so 2 students

They actually turn out the same, but taking from the shortest has the advantage of “being able to consider more students”.

Correctness of the Algorithm

By distributing to students in order of increasing required length, we can serve more students with the same rope consumption. This is a typical pattern of the greedy method.

Algorithm

1. Sort array \(A\) in ascending order
2. Assign rope to students in order, starting from those needing the shortest lengths
3. Continue distributing rope to students as long as the following conditions are met:
   • The number of students served so far is less than \(K+1\)
   • The total length of rope distributed so far + the current student’s required length \(\leq L\)
4. Output the number of students who received rope

total_length = 0, count = 0
for each student i (sorted by A[i]):
    if count >= K+1: break
    if total_length + A[i] <= L:
        total_length += A[i]
        count += 1
    else:
        break
return count

Complexity

• Time complexity: \(O(N \log N)\)
  • Sorting takes \(O(N \log N)\)
  • Loop takes \(O(N)\)
• Space complexity: \(O(N)\)
  • Storing array \(A\) requires \(O(N)\)

Implementation Notes

1. Sort direction: It is important to sort in ascending order and process students starting from those with the smallest requirements.

2. Loop termination conditions: Exit when either of these two conditions is met:

   • The number of students served reaches \(K+1\) (cannot create more pieces)
   • The remaining rope cannot satisfy the next student’s requirement

3. Early termination optimization: Since the array is sorted, once we encounter a case where “the length is insufficient”, all subsequent students will also have insufficient rope. Therefore, we can immediately exit the loop with break.

   Source Code

def solve():
    N, L, K = map(int, input().split())
    A = list(map(int, input().split()))
    
    # Sort A in ascending order to prioritize students who need less rope
    A.sort()
    
    # We can make at most K cuts, resulting in at most K+1 pieces
    # We want to maximize the number of students who can participate
    
    # Greedy approach: Give rope to students who need the least amount first
    # This way we can maximize the number of students
    
    # We can give rope to at most min(K+1, N) students
    max_pieces = K + 1
    
    # Try to give rope to as many students as possible
    # Starting from students who need the least rope
    
    total_length = 0
    count = 0
    
    for i in range(N):
        if count >= max_pieces:
            break
        if total_length + A[i] <= L:
            total_length += A[i]
            count += 1
        else:
            break
    
    print(count)

solve()

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