[{"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}]
内容目录
思路解析
运用知识点
代码实现
1.思路解析
先接受存入数组的个数以及数组，然后将原序数组折中交换：
即将数组对半，第一个与最后一个交换，第二个与倒数第二个交换，以此类推。
此过程涉及双指针移动。
2.运用知识点
数组，指针（双指针移动，指针传址调用）
3.代码实现
#include <stdio.h> //逆序存放函数： void store(int *b,int x) { int *i,*j,*p,t,m;
m=(x-1)/2; i=b;j=b+x-1;p=b+m; for(;i<=p;i++,j--) { t=*i; *i=*j; *j=t; } }
//主函数: int main() { int i,n; int a[999]; printf("请输入需要储存数的个数：");
scanf("%d",&n); printf("\n请存入%d个数：",n); for(i=0;i<n;i++) { scanf("%d",&a[i]); }
store(a,n); printf("\n逆序存放为："); for(i=0;i<n;i++) { printf("%d ",a[i]); }
printf("\n"); return 0; }
此方法较容易理解，不是最好的方法，以作借鉴，也希望大家多多支持。
