python判断素数代码
@滕申1112：python编写一个函数 判断是否为素数 -
蔡陆13943553786…… 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
@滕申1112：求Python 代码:实现质数的判断 -
蔡陆13943553786…… def isPrime(n): if n <= 1: return False i = 2 while i*i <= n: if n % i == 0: return False i += 1 return True
@滕申1112：python编程素数判断 -
蔡陆13943553786…… 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...
@滕申1112：如何用python语言判断i是否为素数 -
蔡陆13943553786…… 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
@滕申1112：如何使用Python检测素数实例说明 -
蔡陆13943553786…… 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)
@滕申1112：求助用python编写“判断101 - 200之间有多少个素数.”的代码? -
蔡陆13943553786…… 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))
@滕申1112：python,不用for 和 while循环写一个判断素数的代码 -
蔡陆13943553786…… 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
@滕申1112：求助python 查询2到100之间的素数: 代码: i =2 j =1 c=0 while -
蔡陆13943553786…… 这是3-100的素数,供参考!def is_prime(n): i = 2 while(i < n): if n % i == 0: break i += 1 if n == i: return True else: return False p = 0 for i in range(3,101): if is_prime(i): print i, p += 1 if p % 10 == 0: print '\n'
@滕申1112：python 求素数用什么方法达到时间最快
蔡陆13943553786…… 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
蔡陆13943553786…… 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
@滕申1112：求Python 代码:实现质数的判断 -
蔡陆13943553786…… def isPrime(n): if n <= 1: return False i = 2 while i*i <= n: if n % i == 0: return False i += 1 return True
@滕申1112：python编程素数判断 -
蔡陆13943553786…… 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...
@滕申1112：如何用python语言判断i是否为素数 -
蔡陆13943553786…… 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
@滕申1112：如何使用Python检测素数实例说明 -
蔡陆13943553786…… 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)
@滕申1112：求助用python编写“判断101 - 200之间有多少个素数.”的代码? -
蔡陆13943553786…… 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))
@滕申1112：python,不用for 和 while循环写一个判断素数的代码 -
蔡陆13943553786…… 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
@滕申1112：求助python 查询2到100之间的素数: 代码: i =2 j =1 c=0 while -
蔡陆13943553786…… 这是3-100的素数,供参考!def is_prime(n): i = 2 while(i < n): if n % i == 0: break i += 1 if n == i: return True else: return False p = 0 for i in range(3,101): if is_prime(i): print i, p += 1 if p % 10 == 0: print '\n'
@滕申1112：python 求素数用什么方法达到时间最快
蔡陆13943553786…… 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
