1. Problem Statement & Logic

Problem: Chef wants to give 1 candy to each of $N$ children. He currently has $X$ candies. He can buy packets containing 4 candies each. Find the minimum number of packets he needs to buy.

Logic:

• If $X \ge N$, Chef already has enough candies. He needs 0 packets.
• If $X < N$, he needs $N - X$ more candies.
• Since each packet has 4 candies, matching the demand requires $\lceil (N-X)/4 \rceil$ packets.
• Integer math for ceiling division of $A/B$ is $(A + B - 1) / B$.

1.1 Dry Run

Example: $N=20, X=12$

• Candies needed = $20 - 12 = 8$.
• Packets = $8 / 4 = 2$. Output: 2.

Example: $N=10, X=5$

• Candies needed = $10 - 5 = 5$.
• Packets = $\lceil 5 / 4 \rceil = 2$. (1 packet gives 4, remaining 1 candy needs another packet).

1.2 Pseudocode Solution

READ T
FOR i FROM 1 TO T:
    READ N, X
    
    IF X >= N:
        PRINT 0
    ELSE:
        needed = N - X
        # Ceiling division (needed / 4)
        packets = (needed + 3) / 4 (integer division)
        PRINT packets

• Time Complexity: $O(1)$.
• Space Complexity: $O(1)$.