我是编程新手。请帮助我更正我的代码。
n = int(input("Enter the number: "))
s = set()
while True:
for i in range(2, n):
if n%i==0:
break
else:
s.add(i)
print(s)
答案 0 :(得分：0)
第一和基本筛分算法
T = int(input("Limit").rstrip())
def genp(n): # General algo.
for i in range(2, n):
var1 = True
for j in range(2, i):
if i%j==0:
var1 = False
break
if var1:
yield i
# if T = 300
>>> list(genp(T))
[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]
使用平方根策略的第二个改进算法（比第一个更快）
def genp2(n):
upper = int(n**0.5) + 1
if upper <= 2:
yield 2
return
for i in range(upper ,n):
var1 = True
for j in range(2, upper):
if j <= i:
if i%j==0:
var1 = False
break
if var1:
yield i
t = genp2(upper)
for i in t: yield i
>>> sorted(list(genp2(T))) # sorting the unorganized list
[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]
在质数直到第n位数字之前的最终和最佳算法（其中） 更有效地使用平方根方法 （最快）
def root(n):
Root = []
upper = int(n**0.5) + 1
while int(upper)> 2:
Root = [upper] + Root
upper = int(upper**0.5) + 1
return Root
Root = root(T)
def genp3(n):
for i in range(2, n):
var1 = True
for j in primes:
if j>= i: continue
if j<i:
if i%j==0:
var1 = 0
break
if var1: yield i
primes = [2, 3]
for i in Root+[T]:
primes += list(genp3(i))
primes = list(set(primes))
>>> print(sorted(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]
按时间进行比较：
from timeit import timeit
s1 = '''
def genp1(n):
for i in range(2,n):
var1 = 1
for j in range(2, i):
if i%j==0:
var1 = 0
break
if var1:
yield i
l1 = list(genp1({}))
'''.format(T)
s2 = '''
def genp2(n):
upper = int(n**0.5) + 1
if upper <= 2:
yield 2
return
for i in range(upper ,n):
var1 = True
for j in range(2, upper):
if j <= i:
if i%j==0:
var1 = False
break
if var1:
yield i
t = genp2(upper)
for i in t: yield i
l2 = sorted(list(genp2({})))
'''.format(T) # Better algo
s3 = '''
def root(n):
Root = []
upper = int(n**0.5) + 1
while int(upper)> 2:
Root = [upper] + Root
upper = int(upper**0.5) + 1
return Root
# Root
Root = root({})
def genp3(n):
for i in range(2, n):
var1 = True
for j in primes:
if j>= i: continue
if j<i:
if i%j==0:
var1 = 0
break
if var1: yield i
primes = [2, 3]
for i in Root+[{}]:
primes += list(genp3(i))
#print(primes)
'''.format(T, T)
# for T = 200,000;
>>> print(timeit(stmt = s1, number = 1))
121.92417089999799
>>> print(timeit(stmt = s2, number = 1))
0.8547031289999723
>>> print(timeit(stmt = s3, number = 1))
0.28995667900016997
# for T = 104730; (measuring speed for 10,000 primes; 10,000th prime is 104729)
>>> print(timeit(stmt = s1, number = 1))
35.426819173000695 #Good
>>> print(timeit(stmt = s2, number = 1))
0.35784729400256765 #Better
>>> print(timeit(stmt = s3, number = 1))
0.13226230800137273 #Best
答案 1 :(得分：0)
尝试一下：
n = int(input("Enter the number: "))
s = set()
for i in range(2, n+1):
flag = 1
for j in range(2, i):
if i%j == 0:
flag = 0
break
if flag == 1:
s.add(i)
print(s)
答案 2 :(得分：0)
这里有两个问题。 while True循环永无止境； break语句只是中断内部循环，因此它将永远运行。您也不会迭代最多n个数字，因此实际上您只是在for循环中检查n是否为质数。
请尝试以下操作：
n = int(input("Enter the number: "))
primes = set()
for i in range(2, n): # Check each number i up to n
for j in range (2, n): # Check if i is prime
if j>i/2: # if j>i/2, i is prime
primes.add(i)
break
if i%j==0:
break # Not prime
print(primes)
请注意，python中的set没有（也不能被排序）。如果要按顺序填入素数，请尝试使用list。