結果
| 問題 |
No.1352 Three Coins
|
| コンテスト | |
| ユーザー |
lam6er
|
| 提出日時 | 2025-03-20 21:06:29 |
| 言語 | PyPy3 (7.3.15) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 1,055 bytes |
| コンパイル時間 | 193 ms |
| コンパイル使用メモリ | 82,336 KB |
| 実行使用メモリ | 71,536 KB |
| 最終ジャッジ日時 | 2025-03-20 21:06:36 |
| 合計ジャッジ時間 | 3,021 ms |
|
ジャッジサーバーID (参考情報) |
judge3 / judge5 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 2 |
| other | AC * 32 WA * 2 |
ソースコード
import math
def compute_gcd(a, b, c):
return math.gcd(a, math.gcd(b, c))
A, B, C = map(int, input().split())
d = compute_gcd(A, B, C)
if d != 1:
print("INF")
else:
m = min(A, B, C)
# Use a maximum size to avoid memory issues. 1e6 is chosen based on problem constraints
max_n = 10**6
dp = [False] * (max_n + 1)
dp[0] = True # base case: 0 can be formed
consecutive = 0
result = 0
for i in range(1, max_n + 1):
# Check if current i can be formed by any of the coins
can_form = False
if i >= A and dp[i - A]:
can_form = True
if not can_form and i >= B and dp[i - B]:
can_form = True
if not can_form and i >= C and dp[i - C]:
can_form = True
dp[i] = can_form
if can_form:
consecutive += 1
if consecutive >= m:
break # All subsequent numbers can be formed
else:
consecutive = 0
result += 1 # Count numbers that cannot be formed
print(result)
lam6er