Just Arrange the Icons
BerPhone X is almost ready for release with nn n applications being preinstalled on the phone. A category of an application characterizes a genre or a theme of this application (like "game", "business", or "education"). The categories are given as integers between 11 1 and nn n , inclusive; the ii i -th application has category c**ici c_i .
You can choose mm m — the number of screens and ss s — the size of each screen. You need to fit all nn n icons of the applications (one icon representing one application) meeting the following requirements:
-
On each screen, all the icons must belong to applications of the same category (but different screens can contain icons of applications of the same category);
-
Each screen must be either completely filled with icons (the number of icons on the screen is equal to
ss
) or almost filled with icons (the number of icons is equal to
s−1s−1
).
Your task is to find the minimal possible number of screens mm m .
Input
The first line contains an integer tt t (1≤t≤100001≤t≤10000 1 \le t \le 10,000 ) — the number of test cases in the input. Then tt t test cases follow.
The first line of each test case contains an integer nn n (1≤n≤2⋅1061≤n≤2⋅106 1 \le n \le 2\cdot10^6 ) — the number of the icons. The second line contains nn n integers c1,c2,…,c**nc1,c2,…,cn c_1, c_2, \dots, c_n (1≤c**i≤n1≤ci≤n 1 \le c_i \le n ), where c**ici c_i is the category of the ii i -th application.
It is guaranteed that the sum of the values of nn n for all test cases in the input does not exceed 2⋅1062⋅106 2\cdot10^6 .
Output
Print tt t integers — the answers to the given test cases in the order they follow in the input. The answer to a test case is an integer mm m — the minimum number of screens on which all nn n icons can be placed satisfying the given requirements.
Example
Input
Copy
3
11
1 5 1 5 1 5 1 1 1 1 5
6
1 2 2 2 2 1
5
4 3 3 1 2
Output
Copy
3
3
4
Note
In the first test case of the example, all the icons can be placed on three screens of size 44 4 : a screen with 44 4 icons of the category 11 1 , a screen with 33 3 icons of the category 11 1 , and a screen with 44 4 icons of the category 55 5 .
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f;
const int N=1e6+5;
//int num[2*N];
vector<int>vt;
int x[2*N];
int main()
{
int t,n;
//cin>>t;
scanf("%d",&t);
while(t--)
{
// cin>>n;
vt.clear();
int mi=INF,mx=-1,p;
//int cnt=0;
// cnt=0;
scanf("%d",&n);
// memset(num,0,sizeof(num));
memset(x,0,sizeof(x));
for(int i=0;i<n;i++)
{
//cin>>num[i];
scanf("%d",&p);
x[p]++;
mx=max(mx,p);
}
for(int i=0;i<=mx;i++)
{
if(x[i]==0) continue;
else
{
//num[cnt++]=x[i];
vt.push_back(x[i]); //vector速度比数组快 ，用于统计数量
mi=min(mi,x[i]);
}
}
int tot,r=0;
int minn=INF;
int ans=0;
for(int i=1;i<=mi+1;i++)
{
r=0;
ans=0;
//for(int j=0;j<cnt;j++)
for(auto X :vt)
{
tot=(X-1)/i+1;
if(X%i==0) ans+=X/i;
else if(X-tot*(i-1)>=0&&X-tot*i<=0) ans+=tot;
else
{
r=-1;
break;
}
}
if(r!=-1)
minn=min(minn,ans);
}
//cout<<ans<<endl;
printf("%d\n",minn);
}
return 0;
}
思路：
有
