[{"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}]
试题详情:
将2个按升序排列的整数数列,仍然按升序合并,存放到另一个数组中,要求每个数依次插入恰当的位置,也就是一次到位,不得在新数组中重新排序。
输入提示:
先输入两个整数,分别为以下两行升序数列的长度,然后输入两行升序整数数列。
输出提示:
输出仅一行升序数列。
输入样本:
4 3
1 3 4 7
2 3 6
输出样本:
1 2 3 3 4 6 7
#include <stdio.h> int main() { int n1,n2,a[100],b[100],c[200],i,j,k;
scanf("%d%d",&n1,&n2); for(i=0;i<n1;i++) scanf("%d",&a[i]); for(i=0;i<n2;i++)
scanf("%d",&b[i]); i=0; j=0; k=0;while(i<n1&&j<n2) { if(a[i]<b[j]) { c[k]=a[i];
k++; i++; } else { c[k]=b[j]; k++; j++; } } while(i<n1) { c[k]=a[i]; k++; i++;
} while(j<n2) { c[k]=b[j]; k++; i++; } for(i=0;i<(n1+n2);i++) printf("%d
",c[i]); return 0; }