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Openai/6955e0f9-71f0-8005-b29b-45f8f5be7e41
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=== Assistant: 明白了。你現在的規則其實是: === 明白了。你現在的規則其實是: * 每個模組 = 7 個 step 的字串(只用 a/b/c) * 第 7 個一定是 c * 排除缺少 a/b/c 其中之一(所以每條都必須同時有 a、b、c) * 分類規則 #1:主導字母 = 出現最多的那個(必須是「唯一最多」,不能打平) * 分類規則 #2(你之前 highlight 的):主導落點 step = 主導字母「第一次出現」的 step 在這一套限制下: ✅ 不可能有 21 種(7×3) 因為「第 7 位固定是 c」會讓很多「晚出現才主導」的情況根本做不到。 我用程式把所有符合條件的字串掃過一遍後,結論是: ===== ## ===== 格式:Step + 主導字母 → 例子(第7位固定c,且含a/b/c) ===== - 1A:aaaaabc ===== * 2A:baaaaac * 3A:bbaaaac * 4A:bbcaaac ===== - 1B:baabbbc ===== * 2B:ababbbc * 3B:aabbbbc * 4B:aacbbbc ===== - 1C:caabbcc ===== * 2C:acabbcc * 3C:aacbbcc * 4C:aabcbcc * 5C:aabbccc ==== 因為你把 Step7 鎖死為 c,所以: ==== * A / B 不可能在 Step5–7 才開始主導(後面只剩 2 個位置能放 A 或 B,怎樣都不夠「最多」) * C 也不可能在 Step6–7 才開始主導(Step7 固定是 c,但只靠最後一位沒辦法從落後翻成最多) 所以自然只剩 13 種。 ==== 最少改一個就行(你選一個): ==== # 第7位不固定(就回到 21 種) 或 # 允許主導字母打平(例如 a=b>c 也算主導) 或 # 主導落點不是“第一次出現”,改成“開始連續主導/最後翻盤點”(也能湊回 21) 你要我照哪一種改,才能給你「真正 21 種」?只回我:1 / 2 / 3 就行。
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