A. Yet Another Dividing into Teams

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are a coach of a group consisting of nn students. The ii-th student has programming skill aiai. All students have distinct programming skills. You want to divide them into teams in such a way that:

  • No two students ii and jj such that |ai−aj|=1|ai−aj|=1 belong to the same team (i.e. skills of each pair of students in the same team have the difference strictly greater than 11);
  • the number of teams is the minimum possible.

You have to answer qq independent queries.

Input

The first line of the input contains one integer qq (1≤q≤1001≤q≤100) — the number of queries. Then qq queries follow.

The first line of the query contains one integer nn (1≤n≤1001≤n≤100) — the number of students in the query. The second line of the query contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1001≤ai≤100, all aiai are distinct), where aiai is the programming skill of the ii-th student.

Output

For each query, print the answer on it — the minimum number of teams you can form if no two students ii and jj such that |ai−aj|=1|ai−aj|=1 may belong to the same team (i.e. skills of each pair of students in the same team has the difference strictly greater than 11)

Example

input

Copy

4
4
2 10 1 20
2
3 6
5
2 3 4 99 100
1
42

output

Copy

2
1
2
1

Note

In the first query of the example, there are n=4n=4 students with the skills a=[2,10,1,20]a=[2,10,1,20]. There is only one restriction here: the 11-st and the 33-th students can't be in the same team (because of |a1−a3|=|2−1|=1|a1−a3|=|2−1|=1). It is possible to divide them into 22 teams: for example, students 11, 22 and 44 are in the first team and the student 33 in the second team.

In the second query of the example, there are n=2n=2 students with the skills a=[3,6]a=[3,6]. It is possible to compose just a single team containing both students.

解题说明:水题,遍历判断即可。

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>

int main()
{
	int q, n, a[101], i, j, flag;
	scanf("%d", &q);
	while (q)
	{
		flag = 1;
		scanf("%d", &n);
		for (i = 1; i <= n; i++)
		{
			scanf("%d", &a[i]);
		}
		for (i = 1; i <= n && flag == 1; i++)
		{
			for (j = i + 1; j <= n && flag == 1; j++)
			{
				if (fabs(a[j] - a[i]) == 1)
				{
					flag = 2;
				}
			}
		}
		printf("%d\n", flag);
		q--;
	}
	return 0;
}
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转载自blog.csdn.net/jj12345jj198999/article/details/102750933