python编写程序，输出200以内的所有素数以及素数的总个数？ 用python输出100-200间的素数
python .\u7edf\u8ba1\u5e76\u8f93\u51fa200-700\u4e4b\u95f4\u6240\u6709\u7d20\u6570\u7684\u4e2a\u6570,\u5e76\u5c06\u8fd9\u4e9b\u7d20\u6570\u6253\u5370\u8f93\u51fa\uff1f\u8fed\u4ee3200\u5230700\u4e4b\u95f4\u7684\u6570\u5b57\uff0c\u8bbe\u5b9a\u4e2an,\u8ba9\u6570\u5b57\u53bb\u4f59\u4ed6\u7684\u4e00\u534a\u7684\u7684\u5e8f\u5217\u533a\u95f4\uff0c\u5982\u679c\u6709\u4f590\u7684\u5c31\u52a0\u4e00\u3002\u6700\u7ec8\u6ca1\u6709\u4f590\u7684n\u5c31\u662f0\u3002\u6240\u4ee5n\u4e3a0\u7684\u6570\u5c31\u662f\u7d20\u6570
#coding=utf-8#\u51fd\u6570\u7528\u4e8e\u5224\u65ad\u67d0\u4e00\u4e2a\u6570\u662f\u4e0d\u662f\u7d20\u6570def test(num): list = [] #\u5b9a\u4e49\u5217\u8868\uff0c\u7528\u4e8e\u5b58\u50a8\u8ba1\u7b97 i = num -1#\u53bb\u9664\u672c\u8eab while i > 1:#\u53bb\u96641 if num%i == 0: #\u5224\u65ad\u662f\u5426\u6709\u4f59\u6570 list.append(i)#\u5c06\u6240\u4ee5\u6709\u7684\u80fd\u6574\u9664\u5b83\u6570\u52a0\u5165\u5217\u8868 i -= 1 if len(list) == 0:#\u5982\u679c\u5217\u8868\u4e3a\u7a7a\uff0c\u5c31\u662f\u8868\u793a\u9664\u4e861\u4e2a\u5b83\u672c\u8eab\u80fd\u6574\u9664 print(num,end=" ")#\u6b64\u51fd\u6570\u7528\u4e8e\u5224\u65ad\u8ba1\u7b97\u51fa\u9700\u8981\u5224\u65ad\u7684\u6240\u6709\u6570\u5b57100\uff5e200def test2(star_num,and_num): j = star_num while j < and_num: test(j) j += 1test2(100,200)print("")python \u7528\u51fd\u6570\u5b9e\u73b0\u6c42100-200\u91cc\u9762\u6240\u6709\u7684\u7d20\u6570
采用for循环结合if--elif--else语句判断即可。使用append()方法将符合条件的数值添加到预先创建的列表，最后使用使用len()求列表的长度，该长度即为素数个数。具体如下：
代码示例
大致就是这个思路，如果有其他更简单的判定素数的方法，稍作修改即可。
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不考虑效率的话：
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