函数封装2.3 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()

2-100之间共有25个素数

Process finished with exit code 0