某企业招聘题目：获取最小的10000个素因子只有2、3、5的数

import time p=time.time() s=set({2,3,5}) t=set({}) i=0 j=1 while i<10000: if j in s: t.add(j) s.add(j2) s.add(j3) s.add(j*5) i+=1 if i%100==0: print(i) j+=1 t=list(t) t.sort() q=time.time() print(q-p) print(t)

```python #! /usr/bin/env python def mul_nums(nums,size): curnums = [] retnums = [] minnum = nums[0] for c in nums: curnums.append(c) retnums.append(c) if c < minnum: minnum = c i = len(retnums) while i < size: minnum = curnums[0] minidx = 0 idx = 0 for c in curnums: if c < minnum: minnum = c minidx = idx idx += 1 minmulnum = curnums[0] * nums[0] for c in curnums: for d in nums: if (c * d) < minmulnum: minmulnum = (c * d) retnums.append(curnums[minidx]) curnums[minidx] = minmulnum i += 1 return retnums def main(): retnums = mul_nums([2,3,5],10000) for c in retnums: print('[%d]\n'%(c)) main() ```

``` #!/usr/bin/env python # -*- coding:utf-8 -*- """ 10000以内 素因子只有2,3,5的数 只需要log2(10000) * log3(10000)*log5(10000)次循环 520次 最终的数据个数小于这个数 """ import math def findOnly235(): max = 10000 maxL2 = int(math.log(max, 2)) maxL3 = int(math.log(max, 3)) maxL5 = int(math.log(max, 5)) print maxL2, maxL3, maxL5 L = [] for i in range(maxL2): for j in range(maxL3): for k in range(maxL5): number = pow(2, i) * pow(3, j) * pow(5, k) if number <= max and number > 1: L.append(number) return L if __name__ == '__main__': l = findOnly235() l.sort() print l ```