WOJ1040-Magic Building

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 sqaure 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. 

输入格式

The first line of the input is a single number t (0<t<10), 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.

输出格式

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

样例输入

212 51 2 2 3 3

样例输出

1
2

#include<iostream>#include<map>using namespace std;int main(){int t,n,i,ans=0;map<int,int> count;cin>>t;while(t-->0){cin>>n;count.clear();ans=0;while(n-->0){cin>>i; ++count[i]; }map<int,int>::iterator it=count.begin();for(;it!=count.end();++it){if(it->second>ans)ans=it->second;}cout<<ans<<endl;}return 0;}