the intervals

寻找最优区间算法

最新推荐文章于 2022-07-07 15:54:32 发布

原创最新推荐文章于 2022-07-07 15:54:32 发布 · 313 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#intervals #numbers #output #each #arrays #search

本文介绍了一个寻找最优区间的算法问题，对于给定的两个数组，需要找到包含特定元素的最小区间。通过排序和二分查找的方法实现了高效的解决方案。

The Intervals

Time Limit: 1 Second Memory Limit: 32768 KB

Given two arrays of numbers {A(n)} and {B(m)}. For each B(i) in {B(m)}, find 2 numbers a and b from {A(n)}, such that B(i) is in [a,b) and b-a<=|b'-a'| for all a' and b' from {A(n)} such that [a',b') contains B(i).

Input

There are several test cases.

In each test case, the first line gives n and m.

The second line contains n numbers, which are the elements of {A(n)}.

The third line contains m nubmers, which are the elements of {B(m)}.

Output

For each B(i) in {B(m)}, output a line containing the interval [a,b).

If there is no such interval, output "no such interval" instead.

Print a blank line after each test case.

Sample Input

3 3
10 20 30
15 25 35

Sample Output

[10,20)
[20,30)
no such interval

/*题目要求找中间值，所以先排序比较好解，二分查找比顺序查找快，要灵活运用还要有创造力，把二分查找返回条件改写一下就能完成这道题了,若值不在最大或最小范围内则输出nosuch intervals,注意条件每个案例后面输出一个空行*/

#include<stdio.h>
#include<stdlib.h>
int a[1000],mid,low,high;
int comp(const void *a,const void *b)
{
          return *(int *)a-*(int *)b;
}
int search(int n,int k)
{
low=0,high=n-1,mid;
while(low<=high)
{
  mid=(low+high)/2;
  if(a[mid]<=k&&a[mid+1]>k) return(mid);
  if(a[mid]>k)
   high=mid-1;
  else low=mid+1;
}
return -1;
}

int main()
{   int b;
register int i;
int m,n;
while(scanf("%d%d%*c",&m,&n)!=EOF)
{
  for(i=0;i<m;i++)
   scanf("%d",&a[i]);
  qsort(a,m,sizeof(int),comp);
     for(i=0;i<n;i++)
  {
   scanf("%d",&b);
   if(b<a[0]||b>=a[m-1]||search(m,b)==-1) {printf("no such interval/n");continue;}
   printf("[%d,%d)/n",a[mid],a[mid+1]);

  }
printf("/n");
}
return 0;
}