想象一下，一款精心研發的藥品，其宣傳材料被翻譯成多種語言，準備推向全球市場。這本是令人欣喜的一步，但如果材料中的某個專業術語翻譯不當，或者某個療效描述與當地法規不符，可能會引發誤解，甚至法律風險。這就像是精心準備的宴席，卻因為菜單的介紹詞不達意，而讓賓客望而卻步。藥品多語言宣傳材料的審核，絕非簡單的語言轉換，而是一項關乎患者安全、品牌信譽和法律合規的系統性工程?？得逶陂L期的實踐中深刻認識到，精準、合規、文化適配的審核流程，是藥品全球化傳播的基石。
合規性是藥品多語言宣傳材料審核的首要原則，是所有工作的基礎和前提。這不僅僅是遵循中國的《藥品管理法》、《廣告法》，更是要精準對標目標國家或地區的藥品監管機構的具體要求。例如，美國的食品藥品監督管理局（FDA）、歐洲藥品管理局（EMA）等，對于藥品宣傳的措辭、適應癥表述、風險警示的呈現方式和位置，都有著極為細致和嚴格的規定。任何一點疏忽，都可能導致材料被退回、罰款，甚至影響產品在該市場的準入資格。
康茂峰認為，建立動態的合規知識庫至關重要。審核團隊需要持續追蹤全球主要醫藥市場的法規動態，并將這些變化及時融入到審核標準中。這不僅依賴于法律文本的解讀，更需要理解法規背后的監管意圖。例如，某些地區可能要求將不良反應信息以特殊字體或顏色突出顯示，并且其字號不得小于正文特定比例。這種細節性的要求，必須通過嚴格的校對清單來確保無一遺漏?？梢哉f，合規性審核是為藥品宣傳材料搭建的法律“安全屋”，確保其在全球市場的合法合規流通。
藥品宣傳材料充斥著大量的專業術語，從化學成分、作用機理到臨床數據，每一處都需要精確無誤的翻譯和表述。一個術語的誤譯，可能會完全改變藥品的含義，誤導醫療專業人士和患者。例如，“緩解癥狀”與“治愈疾病”在醫學上是兩個截然不同的概念，其翻譯必須準確反映臨床研究的實際結果，絕不能夸大。
康茂峰的策略是實行“雙軌審核”制度。首先，由具備深厚語言功底和醫學背景的譯員進行翻譯；隨后，審核環節必須引入相關領域的醫學專家或資深藥師進行復審。他們不僅核查術語的準確性，更會審視整個敘述邏輯是否符合醫學常識和臨床實踐。特別是對于臨床試驗數據的呈現，如有效率、生存期等統計指標，其表述方式必須科學、嚴謹，避免產生歧義。專業性確保了宣傳材料的科學內核不被丟失，維護了信息的權威性和可信度。
語言是文化的載體，直譯往往無法實現有效溝通。文化適配性要求審核者超越字面意思，深入理解目標文化的價值觀、信仰、溝通習慣和醫療觀念。例如，在某些文化中，直接展示嚴重疾病的癥狀圖像可能被視為不敬或引發不適，因此需要采用更委婉的視覺表達方式。又如，對“風險”的理解和接受度，不同地區的患者可能存在顯著差異。
康茂峰在審核中特別注重本地化洞察。這包括對色彩、符號、比喻、甚至幽默感的謹慎處理。審核團隊需要思考：這個口號在本地文化中是否會產生負面聯想？這個案例故事是否能引起當地患者的共鳴？通過聘請目標地區的文化顧問或本地醫學撰稿人參與審核，可以有效地識別和化解這些隱形壁壘，使宣傳材料真正“入鄉隨俗”，與目標受眾建立情感連接。
當一種藥品擁有面向幾十個國家的多語言材料時，確保所有版本的核心信息高度一致，是維護全球品牌形象的關鍵。信息不一致不僅會稀釋品牌信息，更可能引發監管機構的質詢，因為不同地區傳遞的矛盾信息可能被視為誤導。核心信息通常包括藥品的通用名# Optimisations
\( score ~~ 1 / \text{price} \times \text{boolean purchased}\)
\( highscore += \text{scores[i]} \times \text{share of purchases}\)
Using a probabilistic interpretation: purchase_i
\[ \text{if purchase_i} = 1 \implies \text{score_i} = 1 / \text{price_i} \\ \text{else score_i} = 0 \]
Additionally, suppose there are 100 customers (not necessarily distinct)
who bought the game, and thus 100 purchase_i variables.
Then naturally, highscore is the sample mean of score_i.
Now, what information does highscore give about purchase_i?
Consider contrarian thinking: What exactly is being purchased here?
The game says purchase equals ownership.
But clearly, if I purchased only 1 title, share_of_purchases = 1 / sum_of_purchases, right?
Actually, let’s rewrite highscore carefully.
Given games indexed by i, each with some price price_i.
Define indicator purchased_i = 1 if purchased, 0 else.
Score for single game i: score_i = (1 / price_i) * purchased_i
Then highscore = sum_i score_i / sum_i purchased_i
Let N = number of games purchased = sum_i purchased_i (which is total # of purchases)
Then highscore = (1/N) * sum_i ( purchased_i / price_i )
So essentially, highscore is the harmonic mean of the prices of purchased games.
Harmonic mean H = N / (sum_i (1/price_i) only for purchased games.
So highscore = 1 / H.
Wait, actually:
Harmonic mean H of prices = N / sum_{purchased} (1/pricei).
But highscore = sum{purchased} (1/price_i) / N = 1/H.
So highscore = reciprocal of harmonic mean of purchased games’ prices.
Thus, to maximize highscore, we want the harmonic mean of purchased prices to be as small as possible.
Harmonic mean is small if there are some very low prices purchased.
But N appears in denominator too.
So strategy: Buy all games with small price, avoid large price?
But if you purchase a large price game, N increases, which might lower highscore.
Unless its 1/price is very small, so it drags harmonic mean up (i.e., makes H larger, so 1/H smaller) → bad.
Actually:
highscore = 1/H, where H = harmonic mean of purchased prices.
We want to maximize highscore = maximize 1/H = minimize H.
So minimize harmonic mean of purchased prices.
Harmonic mean is minimized if the purchased set includes very low prices and excludes high prices.
But harmonic mean is more sensitive to small values than arithmetic mean.
Example: If we purchase a game with price 1, H will be low. If we purchase a game with price 1000, H increases.
Thus optimal strategy: Purchase all games with price below some threshold T, and none above.
Why? Because adding a higher-priced game always increases H (harmonic mean increases when you add a larger number to the set), thus decreases highscore.
Check: For set S, H = |S| / sum_{i in S} 1/price_i.
Add a new game with price p_new > old H? Not exactly straightforward.
Better: The harmonic mean is a mean, so if you add a value greater than the current harmonic mean, the new harmonic mean increases.
Proof: For positive numbers, if x > H(S), then H(S ∪ {x}) > H(S).
Because harmonic mean is increasing function when adding a larger element (by the convexity of 1/x).
So indeed, if current harmonic mean of purchased set is H, adding a game with price > H will increase H, thus decrease highscore. Adding a game with price < H will decrease H, thus increase highscore.
So dynamically:
Start with empty set: harmonic mean undefined. Add the smallest price game first: H = that price.
Then add next smallest price if it’s < H? Actually, if next price p < H, then adding it reduces H.
Continue until the next smallest price p is > H, then adding it would increase H → don’t add.
So optimal set: purchase all games with price ≤ H_final, where H_final is the harmonic mean of that set.
This is a fixed-point condition: If you purchase all games with price ≤ H, then H is the harmonic mean of those prices.
So find H such that H = harmonic mean of {prices ≤ H}.
We can solve this by iterating:
Let S = {prices sorted ascending}.
Start with H = price of first game.
Check if next price ≤ H, add it, recompute H = harmonic mean of all so far.
Keep adding until next price > H.
Algorithm:
Sort prices ascending: p1 ≤ p2 ≤ … ≤ pn.
Initialize:
purchased = [p1]
H = p1 (since harmonic mean of one number is that number)
For i = 2 to n:
if p_i ≤ H:
add p_i to purchased
recalc H = |purchased| / sum(1/p for p in purchased)
else:
stop
Final highscore = 1/H.
Wait, but highscore = 1/H by definition.
So final answer = 1 / H_final.
Let’s test with example.
Example: Prices = [1, 2, 3, 4].
Start: purchased = [1], H = 1.
i=2: p2=2, is 2 ≤ 1? No → stop.
So purchased = [1], H=1, highscore=1. That’s 1.
But maybe we can do better? Try [1,2]:
H = 2 / (1⁄1 + 1⁄2) = 2 / (1 + 0.5) = 2 / 1.5 = 4⁄3 ≈ 1.333.
Then highscore = 1/H = 3⁄4 = 0.75 — worse.
So indeed adding 2 reduced highscore.
So algorithm correct.
But wait, in example, highscore = 1/H? Let’s check formula:
highscore = sum(purchased) of (1/price * purchased_i) / N = sum of (1/p_i) / N = (1.5)/2 = 0.75. Yes.
So it’s correct.
Thus solution: Sort prices, keep adding while price <= current H, then stop. Compute final H, answer = 1/H.
But careful: When we add a new price p, new H = (k+1) / (sum_reciprocals + 1/p).
We can compute incrementally.
Code plan:
sort prices
sum_recip = 0
count = 0
H = 0
for p in sorted_prices:
if p <= H or count == 0: # for first element, count=0, we add.
count += 1
sum_recip += 1/p
H = count / sum_recip
else:
break
highscore = 1 / H
But wait, condition: when count==0, we add. Then after adding first, H computed.
Then for next p, if p <= H, add.
Yes.
Let’s test with [1,2,3,4]:
p=1: count=0 → add: count=1, sum_recip=1, H=1.
p=2: 2<=1? No → stop.
highscore=1⁄1=1.
Test with [1, 1, 2]:
p=1: add → count=1, sum_recip=1, H=1.
p=1: 1<=1 → add: count=2, sum_recip=2, H=2⁄2=1.
p=2: 2<=1? No → stop.
highscore=1⁄1=1.
Test with [2, 3, 4]:
p=2: add → H=2.
p=3: 3<=2? No → stop.
highscore=1⁄2=0.5.
Test with [1, 3, 3]:
p=1: add → H=1.
p=3: 3<=1? No → stop.
highscore=1.
Seems consistent.
What if [1, 2, 4]:
p=1: H=1.
p=2: 2>1 → stop.
highscore=1.
But what if we try [1,4]?
H = 2 / (1 + 1⁄4) = 2 / 1.25 = 1.6, highscore=0.625 — worse.
So correct.
Thus solution is straightforward.
We just need to implement this iteration.
But note: prices may have duplicates, but our loop handles duplicates because we add if p<=H, and if p=H, we add.
Also, if all prices are large, we still buy at least one (the smallest) because for first element condition is count==0.
So code:
Read input, sort, then apply algorithm.
Let’s write Python code.
Wait, but the problem says: “Given the list of prices of all available games, write a function that returns the highest highscore that can be achieved.”
So we are allowed to choose any subset. Our algorithm finds the optimal subset.
We must buy at least one? Yes, because if none purchased, highscore undefined, but problem assumes purchase at least one.
Our loop buys at least one.
So code:
def highestHighscore(prices):
prices.sort()
count = 0
sum_recip = 0.0
H = 0.0
for p in prices:
if count == 0 or p <= H:
count += 1
sum_recip += 1.0 / p
H = count / sum_recip
else:
break
return 1.0 / H
Test with given example: prices = [1, 2, 3, 4] → 1.0.
Test with [1, 1, 2] → 1.0.
Test with [2, 3, 4] → 0.5.
Test with [1, 3, 3] → 1.0.
Test with [1, 2, 4] → 1.0.
Test with [1, 1, 1] →
p=1: count=1, sum_recip=1, H=1.
p=1: add: count=2, sum_recip=2, H=1.
p=1: add: count=3, sum_recip=3, H=1.
highscore=1.
Test with [1, 10, 100]:
p=1: H=1.
p=10: 10>1 → stop.
highscore=1.
Test with [5, 5, 5]:
all added: H=5, highscore=0.2.
Test with [1, 2]:
p=1: H=1.
p=2: 2>1 → stop.
highscore=1.
But if we try both: H=2/(1 + 0.5)=1.333, highscore=0.75. So indeed buying only first is better.
So algorithm correct.
Let’s implement.
But use float? Prices are integers, but reciprocals are fractions. Floating point should be fine.
We can use exact fractions if needed, but numbers are small enough.
Code:
We’ll write the function.
Note: We must sort.
Let’s code.
