Реши задачу на пайтоне

創建於：2025年12月6日

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問題

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Реши задачу на пайтоне

思考

Cross-checking rating equation logic

I’ll zoom in on a small section to verify, but the important part is the equation: he participates in n games, gaining a points for a win and losing b points for a loss. The task is to find how many wins he can have such that his rating is back to the initial value. If it’s impossible, output -1. Ok, let's sort this out.

Evaluating win count with equation constraints

I need to compute the maximum number of wins, w, where 0 ≤ w ≤ n. The equation simplifies to w = b*n / (a + b). To make sure w is valid, it must be an integer and satisfy the condition w ≤ n. Let me check if this condition holds for different values of n, a, and b.

Analyzing the edge cases and constraints

I’m confirming that the condition $bn \leq n(a+b)$ holds true except for when $n = 0$ . For $n = 0$ , the solution is clear, $w = 0$ . The challenge is when $a = 0$ , as wins won’t change the rating. For these cases, $w = n$ , and it holds based on the equation.

Handling the special case of a = 0 and b = 0

If $b = 0$ , losses don't affect the rating, and the formula still works with $w = 0$ . However, if both $a$ and $b$ are 0, the formula becomes undefined due to the $0/0$ division. In this case, no rating change means any combination of wins and losses is valid, but since he wants the maximum wins, $w = n$ works as the solution.

Verifying edge cases and conditions

When both $a = 0$ and $b = 0$ , the rating remains constant, and any combination of wins and losses works. If $n = 0$ , the win count must be 0. Otherwise, I just need to check for divisibility using the formula and ensure results fit in the constraints. The solution checks divisibility and handles up to $1e9$ .

Finalizing Python code solution

The problem is straightforward in Python. Essentially, when both $a$ and $b$ are zero, the solution is $w = n$ . Otherwise, I check if the equation can be satisfied by confirming if the product $b \times n$ is divisible evenly by $a + b$ . If it’s not, the solution is “-1.”

Here’s the final code:

python
n,a,b = map(int, input().split())
if a==0 and b==0:
    print(n)
else:
    s=a+b
    x = b*n
    if x % s != 0:
        print(-1)
    else:
        print(x//s)

Thought for 1m 1s

回答

Нужно, чтобы после всех игр рейтинг не изменился.

Пусть

$n$ — количество партий,
$a$ — очки за победу,
$b$ — очки, снимаемые за поражение,
$w$ — число побед, тогда поражений $n-w$ .

Суммарное изменение рейтинга:

\Delta = w \cdot a - (n-w) \cdot b

Требуется $\Delta = 0$ :

w a - (n-w) b = 0 \\ (a + b) w = b n \\ w = \frac{b n}{a + b}

То есть:

если $a+b = 0$ (то есть $a = b = 0$ ), рейтинг не меняется от любых результатов, значит можно выиграть все партии: ответ $w = n$ ;
иначе нужно, чтобы $b n$ делилось на $a+b$ . Если не делится — ответа нет, выводим $-1$ .
Если делится, то это единственное возможное число побед; оно всегда $\le n$ .

Код на Python:

python
n, a, b = map(int, input().split())

# Особый случай: ни победы, ни поражения не меняют рейтинг
if a == 0 and b == 0:
    print(n)
else:
    s = a + b
    total = b * n
    if total % s != 0:
        print(-1)
    else:
        w = total // s
        print(w)

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