Sifid and Strange Subsequences(思维)

该博客讨论了一个编程挑战,题目要求找到一个整数数组中最长的奇怪子序列,这里的奇怪定义为序列中任意两个元素的绝对差大于等于序列的最大值。博主通过分析得出结论,最长的奇怪子序列中最多只有一个正数,并给出了一种解决方案,即检查序列中差值最小值是否大于等于最小正数。博主提供了AC代码实现,通过排序和遍历数组来找到满足条件的子序列长度。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

A sequence (b1,b2,…,bk) is called strange, if the absolute difference between any pair of its elements is greater than or equal to the maximum element in the sequence. Formally speaking, it’s strange if for every pair (i,j) with 1≤i<j≤k, we have |ai−aj|≥MAX, where MAX is the largest element of the sequence. In particular, any sequence of length at most 1 is strange.

For example, the sequences (−2021,−1,−1,−1) and (−1,0,1) are strange, but (3,0,1) is not, because |0−1|<3.

Sifid has an array a of n integers. Sifid likes everything big, so among all the strange subsequences of a, he wants to find the length of the longest one. Can you help him?

A sequence c is a subsequence of an array d if c can be obtained from d by deletion of several (possibly, zero or all) elements.

Input

The first line contains an integer t (1≤t≤104) — the number of test cases. The description of the test cases follows.

The first line of each test case contains an integer n (1≤n≤105) — the length of the array a.

The second line of each test case contains n integers a1,a2,…,an (−109≤ai≤109) — the elements of the array a.

It is guaranteed that the sum of n over all test cases doesn’t exceed 105.

Output

For each test case output a single integer — the length of the longest strange subsequence of a.

Example Input

6
4
-1 -2 0 0
7
-3 4 -2 0 -4 6 1
5
0 5 -3 2 -5
3
2 3 1
4
-3 0 2 0
6
-3 -2 -1 1 1 1

Output

4
5
4
1
3
4

Note

In the first test case, one of the longest strange subsequences is (a1,a2,a3,a4)
In the second test case, one of the longest strange subsequences is (a1,a3,a4,a5,a7).

In the third test case, one of the longest strange subsequences is (a1,a3,a4,a5).

In the fourth test case, one of the longest strange subsequences is (a2).

In the fifth test case, one of the longest strange subsequences is (a1,a2,a4).

题意:给你n个数,让你从这n个数中找m个数,保证这m个数中任意两个数的差的绝对值大于等于这m个数中最大值。求一个最大的m。

题解:这m个数中最多只能有一个正数,因为如果有两个正数,这两个正数的差的绝对值一定小于最大值。所以我们只需要判断当有正数的时候,我们判断这个序列中的差的绝对值的最小值是否大于等于这个最小的正数,符合就加一。

AC代码:

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;
const ll inf=0x7fffffff;//ll中的最大值
const int inf1=0x3f3f3f3f;//int中的最大值
int a[maxn];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n,i,j,k;
		scanf("%d",&n);
		for(i=1; i<=n; i++)
			scanf("%d",&a[i]);
		sort(a+1,a+1+n);
		int minn=inf1;
		k=0;
		int f=0;
		for(i=1; i<=n; i++)
		{
			if(a[i]>0)
			{
				f=1;
				break;
			}
			k++;
			minn=min(minn,abs(a[i]-a[i+1]));
		}
		if(minn>=a[i]&&f)
			k++;
		printf("%d\n",k);
	}
	return 0;
}
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值