Overview

A problem to find the minimum number of coins Takahashi needs to reach a target score from his current score.

Analysis

In this problem, if the current score \(A\) is already greater than or equal to the target score \(B\), nothing needs to be done, so the required number of coins is \(0\).

Otherwise, the required score increase is \(B - A\) points. Each score booster can increase the score by 1 point, and each one costs \(C\) coins. Therefore, the required number of coins is simply \((B - A) \times C\).

This problem is very simple and does not require complex processing such as loops or recursion. Also, since the constraints go up to \(10^9\), a naive approach of increasing the score one point at a time would not finish in time (TLE). However, since the result can be obtained with a single multiplication, it can be solved efficiently.

For example, if \(A = 100\), \(B = 120\), \(C = 5\): - Required points: \(120 - 100 = 20\) - Required coins: \(20 \times 5 = 100\)

Algorithm

1. If the current score \(A\) is greater than or equal to the target score \(B\), the answer is \(0\).
2. Otherwise, calculate the deficit \(B - A\).
3. The number of coins needed to cover that deficit is \((B - A) \times C\).

Complexity

• Time complexity: \(O(1)\)
• Space complexity: \(O(1)\)

Implementation Notes

• Read the input correctly.

• Don’t forget to handle the case where the condition is already satisfied using conditional branching.

• Be careful about the order of multiplication and ensure there is no overflow (this is not a concern in Python).

  Source Code

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

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

This editorial was generated by qwen3-coder-480b.