我正在嘗試解決以下問(wèn)題：A store sells large individual wooden letters for signs to put on houses. The letters are priced individually.The total cost of letters in LOAN is 17 dollars.The total cost of letters in SAM is 18 dollars.The total cost of letters in ANNA is 20 dollars.The total cost of letters in ROLLO is 21 dollars.The total cost of letters in DAMAGES is 30 dollars.The total cost of letters in SALMON is 33 dollars.How much would the letters in the name GARDNER cost?我使用以下 python 代碼強(qiáng)制執(zhí)行字母成本，但需要幾個(gè)小時(shí)才能收斂，因?yàn)樗鼈冇?33^10 個(gè)可能的組合來(lái)測(cè)試。我使用 n=33，因?yàn)樗敲Q(chēng)的最大成本，但確實(shí)，n 可以減少到 15 甚至 10，但如果不保證它會(huì)收斂。def func(letters):    print letters    if letters['L']+letters['O']+letters['A']+letters['N'] != 17:        return False    elif letters['S']+letters['A']+letters['M'] != 18:        return False    elif 2*letters['A']+2*letters['N'] != 20:        return False    elif letters['R']+2*letters['O']+2*letters['L'] != 21:        return False    elif letters['D']+2*letters['A']+letters['M']+letters['G']+letters['E']+letters['S'] != 30:        return False    elif letters['S']+letters['A']+letters['L']+letters['M']+letters['O']+letters['N'] != 33:        return False    return Truedef run(letters, n, forbidden_letters):    for letter in letters.keys():        if letter not in forbidden_letters:            for i in range(1, n):                letters[letter] = i                if not func(letters):                    if letter not in forbidden_letters:                        forbidden_letters+=letter                    if run(letters, n, forbidden_letters):                        return letters                else:                    return lettersLETTERS = {    "L":1,    "O":1,    "A":1,    "N":1,    "S":1,    "M":1,    "R":1,    "D":1,    "G":1,    "E":1,}n=33print run(LETTERS, n, "")暴力破解最終會(huì)奏效，但它占用的 CPU 資源如此之多，以至于它肯定不是最佳解決方案。有沒(méi)有人有更好的解決方案來(lái)解決這個(gè)問(wèn)題？要么通過(guò)減少計(jì)算時(shí)間，要么通過(guò)良好的數(shù)學(xué)方法。謝謝大家。