Overview

This problem asks you to calculate the number of coins needed to raise your current score to a target score. You need to purchase items to make up for the score deficit and find the total cost.

Analysis

To solve this problem, we focus on the following points.

Key Observations

1. When the goal is already achieved: If the current score \(A\) is greater than or equal to the target score \(B\), there’s no need to buy any items (cost is \(0\))
2. When the score is insufficient: The score increase needed to reach the target score \(B\) is \(B - A\) points
3. Number of items needed: Since each score booster gives \(1\) point, the number of items needed is \((B - A)\)
4. Total cost: Since each item costs \(C\) coins, the total cost is \((B - A) \times C\) coins

Verification with a Concrete Example

For example, if \(A = 30\) (current score), \(B = 50\) (target score), \(C = 10\) (item price): - Required score increase: \(50 - 30 = 20\) points - Required number of items: \(20\) - Total cost: \(20 \times 10 = 200\) coins

Why This Problem is Simple

In this problem, the naive approach directly gives the optimal solution. Since the only way to increase the score is through score boosters, we simply need to buy the minimum necessary number, resulting in a straightforward structure.

Algorithm

1. Read \(A\), \(B\), \(C\) as input
2. If \(A \geq B\) (goal already achieved):
   • Output \(0\)
3. If \(A < B\) (score is insufficient):
   • Output \((B - A) \times C\)

if A >= B:
    answer = 0
else:
    answer = (B - A) * C

Complexity

• Time complexity: \(O(1)\)
  • Only requires reading input and a constant number of arithmetic operations
• Space complexity: \(O(1)\)
  • Uses only a few variables and does not depend on input size

Implementation Notes

• Watch out for overflow: Based on the constraints, \(B - A\) and \(C\) can each be up to approximately \(10^9\). Their product can be up to approximately \(10^{18}\), but since Python does not have integer overflow, no special handling is needed. If using languages like C++, you need to use the long long type.
• Don’t forget the case distinction: If you forget to output \(0\) when \(A \geq B\), you will end up outputting a negative value.

Source Code

A, B, C = map(int, input().split())

if A >= B:
    print(0)
else:
    print((B - A) * C)

This editorial was generated by claude4.5opus.