计算 2-100 之间素数的个数, 返回结果
- def primeNum(f):
- def f1():
- sum_pri = 0
- for i in range(2,101):
- for j in range(2,i):
- if i % j == 0:
- break
- elif j == i - 1:
- sum_pri += 1
- return f(sum_pri+1)
- return f1
- @primeNum
- def f(p):
- print("2-100 之间共有 {} 个素数".format(p))
- f()
输出结果:
2-100 之间共有 25 个素数
Process finished with exit code 0
来源: http://www.bubuko.com/infodetail-3320054.html