Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2418    Accepted Submission(s):
1088

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
 
Recommend
 
题目大意 :
高度不同的可以合并为一个楼,问最少能合并成多少楼
简单模拟 
#include <algorithm>
#include <cstdio> using namespace std;
int maxn,ans,t,n,buil[];
int main()
{
scanf("%d",&t);
for(;t--;)
{
maxn=ans=;
scanf("%d",&n);
for(int i=;i<=n;i++)
scanf("%d",&buil[i]);
sort(buil+,buil++n);
for(int i=;i<=n;i++)
{
if(buil[i]==buil[i-])
ans++;
else
{
if(ans>maxn) maxn=ans;
ans=;
}
}
if(ans>maxn) maxn=ans;
printf("%d\n",maxn);
}
return ;
}

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