[{"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}]
<>前言
最近刷题的时候遇到了一个问题，是关于变量在运算过程（并不是最终结果）中超出其变量范围引起的错误，故写此文章。
一、什么时候需要1LL
这里举个例子，比如我一个int类型的数据需要进行一个运算
100000*100000/1000，虽然我最后算出的结果不超过int（只有16位编译器的unsigned int
是65535，现在是4294967295），但是在我运算的过程里（进行除法之前）超过了int类型的范围，这时候我们就需要强制类型转换来消除错误，但一直强制转换会有些麻烦，这个时候用1LL就方便了。
<>二、如何使用1LL
#include <stdio.h> #include <stdlib.h> int main() { int a = 100000 * 100000 /
1000; printf("%d", a); system("pause"); return 0; }
这里明显是答案错误，说明这句话是会超出范围的，当然我们可以通过强制转换类型达到效果，只不过最后还要转回成int类型过于麻烦，这里我们就使用1LL，具体看下面。
#include <stdio.h> #include <stdlib.h> int main() { int a = 1LL * 100000 *
100000 / 1000; printf("%d", a); system("pause"); return 0; }
这时候就运行正确了，1LL会在运算时把后面的临时数据扩容成long long类型，再在赋值给左边时转回int类型。
