#1
1
(由 Microsoft 翻譯) 我認為大約 25% 的時間沒有 「假」 硬幣。 要麼那個, 要麼重量差是如此之小, 可以忽略不計地有很多硬幣在規模上。
(原文) Coin Weighing Game
I think that about 25% of the time there is no "fake" coin. Either that or the weight differential is so small as to be negligible with lots of coins on the scale.
I think that about 25% of the time there is no "fake" coin. Either that or the weight differential is so small as to be negligible with lots of coins on the scale.
作者 DingleBerries
2006-08-24 02:56:33
讚好
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#2
1
(由 Microsoft 翻譯) 經過進一步審查 - 我只是不太擅長比賽。
(原文) Upon further review - I'm just not that good at the game.
作者 DingleBerries
2006-08-24 05:10:56
讚好
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#3
(由 Microsoft 翻譯) 我們已經發佈了更新。
這應該。
1) 解決您提到的問題。
2) 確保如果移動錯誤,遊戲將丟失。
3) 如果做出正確的移動,使結果更加隨機。
這應該。
1) 解決您提到的問題。
2) 確保如果移動錯誤,遊戲將丟失。
3) 如果做出正確的移動,使結果更加隨機。
(原文) We've posted an update.
This should
1) solve the problem you mentioned
2) ensure that the game will be lost if a wrong move is made
3) make the result more random if a correct move is made
This should
1) solve the problem you mentioned
2) ensure that the game will be lost if a wrong move is made
3) make the result more random if a correct move is made
作者 Novel Games
2006-08-24 17:11:31
讚好
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#4
2601
(由 Microsoft 翻譯) 我同意 DingleBerries 的觀點，儘管這條資訊是 2006 年發佈的。
(原文) I AGREE with DingleBerries, even though this message is from 2006.
作者 Alex Gordillo
2025-05-06 12:44:34
讚好
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#5
-997308
(由 Microsoft 翻譯) 在 39 個硬幣中，最初有 78 種可能性（39 個硬幣中的一個更重或更輕）。 @Everyone 在稱重結果為 <、= 或 >的情況下，您必須確保第一次稱重將可能性分成不超過 27 個可能性。然後 @Everyone 您必須確保第二次權衡將可能性分成不超過 9 個可能性（如果稱重結果為 <、= 或 >）。再說一次， @Everyone ，您必須確保第三個稱量在稱量結果為 <、= 或 >的情況下將可能性分成不超過 3 種可能性。 最後， @Everyone， 您必須確保在稱量結果<的情況下，第四個稱量將可能性分為不超過 1 種。 = 或 >。縮小範圍后，您就可以選擇答案。
(原文) In 39 coins, there are initially 78 possibilities (one of 39 coins is either heavier or lighter). @Everyone You must make sure that the first weigh splits the possibilities into no more than 27 each in cases of weigh result being <, =, or >. Then @Everyone you must make sure that the second weigh splits the possibilities into no more than 9 each in cases of weigh result being <, =, or >. Then again, @Everyone you must make sure that the third weigh splits the possibilities into no more than 3 each in cases of weigh result being <, =, or >. Then finally, @Everyone, you must make sure that the fourth weigh splits the possibilities into no more than 1 each in cases of weigh result being <, =, or >. And after it's been narrowed down then you can pick the answer.
作者 Piotr Grochowski
2025-05-06 19:57:01
讚好
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