[{"createTime":1735734952000,"id":1,"img":"hwy_ms_500_252.jpeg","link":"https://activity.huaweicloud.com/cps.html?fromacct=261f35b6-af54-4511-a2ca-910fa15905d1&utm_source=V1g3MDY4NTY=&utm_medium=cps&utm_campaign=201905","name":"华为云秒杀","status":9,"txt":"华为云38元秒杀","type":1,"updateTime":1735747411000,"userId":3},{"createTime":1736173885000,"id":2,"img":"txy_480_300.png","link":"https://cloud.tencent.com/act/cps/redirect?redirect=1077&cps_key=edb15096bfff75effaaa8c8bb66138bd&from=console","name":"腾讯云秒杀","status":9,"txt":"腾讯云限量秒杀","type":1,"updateTime":1736173885000,"userId":3},{"createTime":1736177492000,"id":3,"img":"aly_251_140.png","link":"https://www.aliyun.com/minisite/goods?userCode=pwp8kmv3","memo":"","name":"阿里云","status":9,"txt":"阿里云2折起","type":1,"updateTime":1736177492000,"userId":3},{"createTime":1735660800000,"id":4,"img":"vultr_560_300.png","link":"https://www.vultr.com/?ref=9603742-8H","name":"Vultr","status":9,"txt":"Vultr送$100","type":1,"updateTime":1735660800000,"userId":3},{"createTime":1735660800000,"id":5,"img":"jdy_663_320.jpg","link":"https://3.cn/2ay1-e5t","name":"京东云","status":9,"txt":"京东云特惠专区","type":1,"updateTime":1735660800000,"userId":3},{"createTime":1735660800000,"id":6,"img":"new_ads.png","link":"https://www.iodraw.com/ads","name":"发布广告","status":9,"txt":"发布广告","type":1,"updateTime":1735660800000,"userId":3},{"createTime":1735660800000,"id":7,"img":"yun_910_50.png","link":"https://activity.huaweicloud.com/discount_area_v5/index.html?fromacct=261f35b6-af54-4511-a2ca-910fa15905d1&utm_source=aXhpYW95YW5nOA===&utm_medium=cps&utm_campaign=201905","name":"底部","status":9,"txt":"高性能云服务器2折起","type":2,"updateTime":1735660800000,"userId":3}]
<>爱因斯坦阶梯问题
<>解释：设有台阶若干，每跨2阶，最后余1阶，跨3阶余2阶，跨5阶余4阶，跨6阶余5阶，跨7阶正好到达阶梯顶部。
<>分析：先假设有i，余2等于1，余3等于2，以此类推。（代码1）
i%2 == 1 and i%3 == 2 and i%5 == 4 and i%6 == 5 and i%7 == 0
符合以上代码即可。
可以使用while和for语句进行书写
<>第一种（while判定）（代码2）
#爱因斯坦阶梯问题 i = 1 while i % 2 != 1 or i % 3 != 2 or i % 5 != 4 or i % 6 != 5 or i
% 7 != 0: i+=1 print(i)
<>先进行判定如果，如果和全部不符合条件则为爱因斯坦阶梯数。
<>第二种（while判定）（代码3）
i=1 while True: #无限循环 i+=1 if(i%2==1 and i%3==2 and i%5==4 and i%6==5 and i%7==
0): print(i) break #跳出该循环
<>while无限循环再用if进行判定，符合条件输出。（如果不用break退出循环可以判定出所有的爱因斯坦阶梯数，但是是无限循环！）
<>第三种（for判定）（代码4）
i = 1 for i in range(10000): if (i % 2 == 1 and i % 3 == 2 and i % 5 == 4 and
r i% 6 == 5 and i % 7 == 0): print(i) break
<>最后的结果为 119
小白上路，如有不同意见，希望大家多多讨论！谢谢大家！！！
