MagicBuilding

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1314    Accepted Submission(s): 599

Problem Description

As the increase of population, the living space for people is becoming smaller and smaller. In MagicStar the problem is much worse. Dr. Mathematica is trying to save land by clustering buildings and then
we call the set of buildings MagicBuilding. Now we can treat the buildings as a square of size d, and the height doesn‘t matter. Buildings of d1,d2,d3....dn can be clustered into one MagicBuilding if they satisfy di != dj(i != j).

Given a series of buildings size , you need to calculate the minimal numbers of MagicBuildings that can be made. Note that one building can also be considered as a MagicBuilding.

Suppose there are five buildings : 1, 2, 2, 3, 3. We make three MagicBuildings (1,3), (2,3), (2) .And we can also make two MagicBuilding :(1,2,3), (2,3). There is at least two MagicBuildings obviously.

Input

The first line of the input is a single number t, indicating the number of test cases.

Each test case starts by n (1≤n≤10^4) in a line indicating the number of buildings. Next n positive numbers (less than 2^31) will be the size of the buildings.

Output

For each test case , output a number perline, meaning the minimal number of the MagicBuilding that can be made.

Sample Input

2
1
2
5
1 2 2 3 3

Sample Output

1
2

Author

scnu

题意：

就是给定几组数据，把这些数据进行合并，并且合并的时候，一个组里面的没有两个数值是相同的，判断最少是几组数据？

难点：

如何判定是几组数据？

解决办法：

就是找出相同的数最多的即可。

代码如下 ：

<span style="font-size:14px;">#include<stdio.h>
#include<algorithm>
using namespace std;
int a[10010];
int main()
{
	int t,i,n;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d",&n);
		for(i=0;i<n;i++)
		{
			scanf("%d",&a[i]);
		}
		sort(a,a+n);
		int sum=1;
		int Max=1;
		int m=a[0];
		for(i=1;i<n;i++)
		{
			if(a[i]==m)
			{
				++sum;
			}
			else
			{
				m=a[i];
				sum=1;
			}
			if(Max<sum)
			Max=sum;
		}
		printf("%d\n",Max);
	}
	return 0;
}</span>