怎么判断素数python
@酆陈6584：python编写一个函数 判断是否为素数 -
司史18259943531…… def Is_Prime(n): for i in range(2,int(n**(1/2))+1): if n % i == 0: return False break else: return True 程序缩进如图所示
@酆陈6584：求Python 代码:实现质数的判断 -
司史18259943531…… def isPrime(n): if n <= 1: return False i = 2 while i*i <= n: if n % i == 0: return False i += 1 return True
@酆陈6584：如何用python语言判断i是否为素数 -
司史18259943531…… a = 0 list = [] for i in range(101,200,2): flag=0 for j in range(2,i/2+1): if (i%j == 0): flag = 0 break else: flag = 1 continue if flag == 1: list.append(i) a+=1 print list print a
@酆陈6584：python编程素数判断 -
司史18259943531…… def getprimelist(n): length=(n-3)//2+1 primelist=[True]*length for i in range(length): if primelist[i] is True: number=2*i+3 for j in range(i+number,length,number): primelist[j]=False return [2]+[2*i+3 for i,sign in enumerate(primelist) if sign] def isprime...
@酆陈6584：如何使用Python检测素数实例说明 -
司史18259943531…… num = int(input(＂please enter the number:＂)) for i in range(2, num):if num % i == 0:print(＂ %d is not a prime number!＂ % num) break else:print(＂ %d is a prime number!＂ % num)
@酆陈6584：求助用python编写“判断101 - 200之间有多少个素数.”的代码? -
司史18259943531…… s=set(( n for n in range (101,201) for k in range (2,n) if n%k==0 )) t=set((n for n in range(101,201))) print(sorted(t-s))
@酆陈6584：python 求素数用什么方法达到时间最快
司史18259943531…… 判断是否素数:def isprime(n): if n in [2,3,5,7,11]: return True for i in range(3, int(n**0.5)+1): if n%i == 0: return False return True
@酆陈6584：python,不用for 和 while循环写一个判断素数的代码 -
司史18259943531…… 1 2 3 4 5 6 7 8 9 10 11 12 defisPrime(n, i=2): ifn <=1: returnFalseifi *i <=n: ifn %i ==0: returnFalse else: i +=1 returnisPrime(n, i)returnTrue
@酆陈6584：python如何算质数和 -
司史18259943531…… 你好的!import math def is_prime(n): # 简化问题,先利用函数判断是否为质数 if n == 1: # =是赋值,==才是判断,切记切记 return False for i in range(2, int(math.sqrt(n))+1): # 质数判断条件,注意+1 if n % i == 0: return False return True ...
司史18259943531…… def Is_Prime(n): for i in range(2,int(n**(1/2))+1): if n % i == 0: return False break else: return True 程序缩进如图所示
@酆陈6584：求Python 代码:实现质数的判断 -
司史18259943531…… def isPrime(n): if n <= 1: return False i = 2 while i*i <= n: if n % i == 0: return False i += 1 return True
@酆陈6584：如何用python语言判断i是否为素数 -
司史18259943531…… a = 0 list = [] for i in range(101,200,2): flag=0 for j in range(2,i/2+1): if (i%j == 0): flag = 0 break else: flag = 1 continue if flag == 1: list.append(i) a+=1 print list print a
@酆陈6584：python编程素数判断 -
司史18259943531…… def getprimelist(n): length=(n-3)//2+1 primelist=[True]*length for i in range(length): if primelist[i] is True: number=2*i+3 for j in range(i+number,length,number): primelist[j]=False return [2]+[2*i+3 for i,sign in enumerate(primelist) if sign] def isprime...
@酆陈6584：如何使用Python检测素数实例说明 -
司史18259943531…… num = int(input(＂please enter the number:＂)) for i in range(2, num):if num % i == 0:print(＂ %d is not a prime number!＂ % num) break else:print(＂ %d is a prime number!＂ % num)
@酆陈6584：求助用python编写“判断101 - 200之间有多少个素数.”的代码? -
司史18259943531…… s=set(( n for n in range (101,201) for k in range (2,n) if n%k==0 )) t=set((n for n in range(101,201))) print(sorted(t-s))
@酆陈6584：python 求素数用什么方法达到时间最快
司史18259943531…… 判断是否素数:def isprime(n): if n in [2,3,5,7,11]: return True for i in range(3, int(n**0.5)+1): if n%i == 0: return False return True
@酆陈6584：python,不用for 和 while循环写一个判断素数的代码 -
司史18259943531…… 1 2 3 4 5 6 7 8 9 10 11 12 defisPrime(n, i=2): ifn <=1: returnFalseifi *i <=n: ifn %i ==0: returnFalse else: i +=1 returnisPrime(n, i)returnTrue
@酆陈6584：python如何算质数和 -
司史18259943531…… 你好的!import math def is_prime(n): # 简化问题,先利用函数判断是否为质数 if n == 1: # =是赋值,==才是判断,切记切记 return False for i in range(2, int(math.sqrt(n))+1): # 质数判断条件,注意+1 if n % i == 0: return False return True ...
