[{"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}]
subject ： Yes 5 Arrays of different lengths a1,a2,a3,a4,a5, Now take a number from the array , Form a new array , Request from a1 The number taken out must be less than from a2 Number taken , From the same principle a2 The number taken out must be less than from a3 Number taken , and so on , Lists all arrays that meet the criteria .
The title is quite simple , Take a look at it , And then I used a bunch of them for the first time for loop , The script is as follows ：
a1 = [1,2,4] a2 = [2,3,4,5] a3 = [1,4,7] a4 = [3,6,7,8] a5 = [3,6,8,9,10] for i
in a1: for j in a2: for k in a3: for l in a4: for m in a5: if i<j<k<l<m: print(i
,j,k,l,m)
The output is right , But look at this pile for loop , I always feel uncomfortable , After a search , Found out itertools.product(A,
B) This function , This function returns A,B Tuple of Cartesian product of elements in , Seems to meet the conditions , So the script above changes to the following
import itertools for i in itertools.product(a1,a2,a3,a4,a5): if i[0]<i[1]<i[2]<
i[3]<i[4]: print(i)