求素数 list=[]i=2 for i in range (2,100):j=2 for j in range(2,i):if(i%j==0):break else:list.append(i)print(list)
python中如何编程求1到100之间的素数
#!/usr/bin/python27# coding:utf8'''100以内的质数(素数)'''l = []for n in range(1,101):if n == 1:continueelif n == 2:l.append(2)else:if 0 not in [n%i for i in range(2,n)]:l.append(n)print l运行结果:[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]2014-12-29
#！/usr/bin/python
#-*- coding:UTF-8 -*-
#求素数
list=[]
i=2
for i in range (2,100):
j=2
for j in range(2,i):
if(i%j==0):
break
else:
list.append(i)
print(list)
扩展资料：
python：for语句的使用方法
for循环的语法格式：
for i in range(n):#从数据类型中拿一个值赋值给i
print(i)#打印i
例如：
#for
for i in range (1,6,2):#从一开始到六之前每隔上2个数字
print(i)#结果为1,3,5
# ------------------------------------------------
s = ["man", "woman", "girl", "boy", "sister"]
for i in s:#列表s中的每个元素给i
print(i)
#-------------------------------------------------
for i in range(5):
print(i)#结果为：0,1,2,3,4
for循环实例：数字0,1,2组成一个百位数，并且数字不重复！
#for
for i in range(0,3):
for j in range(0,3):
for k in range(0,3):
if (i != 0) and (i != j) and (i != k) and (j != k):
print (i,j,k)
2020-04-20
看你要不要考虑效率了，不考虑效率最简单的就是双层循环试呗，100以为的话，这笨办法也不慢，哈哈2014-12-29
def primes(x): # prepair data space plist = [0, 0] + range(2,x+1) for i in xrange(2, x): if plist[i]: plist[i+i::i] = [0] * len(plist[i+i::i]) return filter(None, plist)print primes(100)筛选法是高效的素数列表计算算法, python的列表切片赋值可以极大地简化代码
2015-07-15
num=[]for i in range(2,101): if 0 not in [i%j for j in range(2,i)]: num.append(i)print(num)2019-05-23
[p for p in range(2, 101) if 0 not in [p% d for d in range(2, int(p**0.5)+1)]]2015-06-28
