我需要找到 10000 以下的最長的非素?cái)?shù)行。如何僅打印輸出中的最后一行？沒有 max 函數(shù)，也沒有簡單的數(shù)值約束。我認(rèn)為它只需要一些小的調(diào)整，只是不知道在哪里或如何調(diào)整。priemen = []for x in range(2,10000): #prime generator tel = 0 for deler in range(1,x): if x % deler == 0: tel += x % deler == 0 if tel <2: priemen.append(x) a = priemen[0]b = priemen[1]maxrow = 0for next in priemen[2:]: a = b b = next row = b - a - 1 if row > maxrow: maxrow = row print("The longest row starts at", a+1, "and stops at", b-1, "and is", maxrow, "long.")------------------Output: The longest row starts at 8 and stops at 10 and is 3 long.The longest row starts at 24 and stops at 28 and is 5 long.The longest row starts at 90 and stops at 96 and is 7 long.The longest row starts at 114 and stops at 126 and is 13 long.The longest row starts at 524 and stops at 540 and is 17 long.The longest row starts at 888 and stops at 906 and is 19 long.The longest row starts at 1130 and stops at 1150 and is 21 long.The longest row starts at 1328 and stops at 1360 and is 33 long.The longest row starts at 9552 and stops at 9586 and is 35 long.我只需要它來打印最后一張
2 回答
繁星點(diǎn)點(diǎn)滴滴
TA貢獻(xiàn)1803條經(jīng)驗(yàn) 獲得超3個(gè)贊
您需要將a和的值保存b到單獨(dú)的變量中，以便可以在循環(huán)后打印它們。
b = priemen[1]
maxrow = 0
for n in priemen[2:]:
a = b
b = n
row = b - a - 1
if row > maxrow:
maxrow = row
a_max = a
b_max = b
if maxrow != 0:
print("The longest row starts at", a_max + 1, "and stops at", b_max - 1, "and is", maxrow, "long.")
其他注意事項(xiàng)：
a_max我還沒有初始化b_max- 但最終的if測試是為了防止任何尚未設(shè)置的情況我已重命名
next為n，因?yàn)?code>next這是內(nèi)置名稱該
a = priemen[0]行毫無意義，所以我已將其刪除
動(dòng)漫人物
TA貢獻(xiàn)1815條經(jīng)驗(yàn) 獲得超10個(gè)贊
我發(fā)現(xiàn)這段代碼效率低下并且存在問題。首先，它的效率很低，因?yàn)樗鼫y試從 1 到x
for deler in range(1, x):
當(dāng)它只需要測試從 3 到的平方根的奇數(shù)除數(shù)（處理偶數(shù)之后）時(shí)。但即使這樣也是低效的，因?yàn)樗趧?chuàng)建一個(gè)素?cái)?shù)列表，它可以用作除數(shù)來進(jìn)一步加快速度！最后，就效率而言，我相信它可以一次性完成：x
TARGET = 10_000 # below this number
primes = [2]
start = end = primes[-1]
for number in range(3, TARGET, 2):
def is_prime(number):
for divisor in primes:
if divisor * divisor > number:
return True
if number % divisor == 0:
return False
return True
if is_prime(number):
primes.append(number)
if primes[-1] - primes[-2] > end - start:
start, end = primes[-2:]
print("The longest run starts at", start + 1, "and stops at", end - 1, "and is", end - start - 1, "long.\n")
最后，就目標(biāo)而言，問題未明確說明，并且解決方案可能是錯(cuò)誤的?？紤]目標(biāo)為 9586 而不是 10000。編寫的代碼將打?。?/p>
The longest run starts at 1328 and stops at 1360 and is 33 long.
但是通過在主循環(huán)之后添加以下代碼：
if TARGET - primes[-1] > end - start:
start, end = primes[-1], TARGET
我們得到正確答案：
The longest run starts at 9552 and stops at 9585 and is 34 long.
如果目標(biāo)更大，跑步會(huì)更長，但這仍然是最長的跑步。
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