hdu 2192 MagicBuilding

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 

屠龙宝刀点击就送

#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 ;
}