[英]Program works but does not produce correct output?
我制作了一個程序,可以找到低於 n 數的素數。
這是代碼:-
import time
n = 10000
start = time.time()
primes = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,
61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139,
149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227,
229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311,
313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401,
409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491,
499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599,
601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683,
691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797,
809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887,
907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997]
if n > 1000000:
for j in range(1,n+1):
for i in range(0,len(primes)):
if j % primes[i] == 0:
break
else:
primes[:25].append(j)
print primes
else:
for j in range(1,n+1):
for i in range(0,len(primes[:25])):
if j % primes[i] == 0:
break
else:
primes[:25].append(j)
print primes
end = time.time() - start
print end
我知道該代碼的工作原理是找到前 1000 個所需的時間,而 100 萬個素數則大不相同。
但它只打印最多 1000 的素數,即它不打印附加列表。 我哪里錯了?
復制和更改您的代碼來測試它似乎是一個壞主意,因為這兩個版本會分開。 相反,修改初始條件並運行相同的代碼:
import time
start = time.time()
n = 10000000
primes = [ \
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, \
61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, \
149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, \
229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, \
313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397, 401, \
409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, \
499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, \
601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, \
691, 701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797, \
809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, \
907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997, \
]
# for testing:
n = 10000
primes = primes[:25]
for j in range(3, n + 1, 2):
for prime in primes:
if j % prime == 0:
break # skip composites and primes in list
else: # no break
primes.append(j)
print primes
end = time.time() - start
print end
即使這不僅僅是為了測試,您也可以將其轉換為條件:
if n <= 1000000:
primes = primes[:25]
並且仍然避免代碼重復。
我只想在最短的時間內找到質數
如果是這種情況,難道您不應該只檢查列表中的素數直到j
平方根而不再進一步檢查嗎? 當然,您還需要更改主要列表管理邏輯。
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