Equalize the Array（前缀和）

题目链接

题目描述：

Polycarp was gifted an array aa of length nn . Polycarp considers an array beautiful if there exists a number CC , such that each number in the array occurs either zero or CC times. Polycarp wants to remove some elements from the array aa to make it beautiful.

For example, if n=6n=6 and a = [1, 3, 2, 1, 4, 2]a=[1,3,2,1,4,2] , then the following options are possible to make the array aa array beautiful:

• Polycarp removes elements at positions 22 and 55 , array aa becomes equal to [1, 2, 1, 2][1,2,1,2] ;
• Polycarp removes elements at positions 11 and 66 , array aa becomes equal to [3, 2, 1, 4][3,2,1,4] ;
• Polycarp removes elements at positions 1, 21,2 and 66 , array aa becomes equal to [2, 1, 4][2,1,4] ;

Help Polycarp determine the minimum number of elements to remove from the array aa to make it beautiful.

 

Input：

follow.

The first line of each test case consists of one integer nn ( 1 \le n \le 2 \cdot 10^51≤n≤2⋅105 ) — the length of the array aa .

The second line of each test case contains nn integers a_1, a_2, \ldots, a_na1​,a2​,…,an​ ( 1 \le a_i \le 10^91≤ai​≤109 ) — array aa .

It is guaranteed that the sum of nn over all test cases does not exceed 2 \cdot 10^52⋅105 .

Output：

For each test case, output one integer — the minimum number of elements that Polycarp has to remove from the array aa to make it beautiful.

 

Example：

input：

3
6
1 3 2 1 4 2
4
100 100 4 100
8
1 2 3 3 3 2 6 6

output：

2
1
2

题意：有 t 组数据。每组数据给出一个长度为 n 的序列 a_1,a_2,...,a_n，问最少需要删除多少个元素，才可以使得剩余的序列中，所有不同数字出现的次数均相等。

思路：每个数都出现刚好要么 0 次要么 c次。考虑枚举这个 c。首先处理出每个数的出现次数然后把这些出现次数从小到大排序。假定我们在考虑第 i个出现次数 b_i，如果所有的数的出现次数都要变为 b_i，

说明多了的次数要减掉，比 b_i 少的出现次数要全部删掉。sum_i−1​ 为比 b_i 出现少的，要全部删掉，s_n - s_i 表示所有大于 b_i 的 b_k之和，减掉不需要删除的 (cnt-i)*b_i 即可。

#include<bits/stdc++.h>
#include<iostream>
#include<cctype>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
using namespace std;
typedef unsigned long long ull;
typedef  long long ll;
typedef pair<ll,ll> pi;
#define IOS std::ios::sync_with_stdio(false)
#define ls p<<1
#define rs p<<1|1
#define mod 1000000000 + 7
#define PI acos(-1.0)
#define INF   1e18
#define N 200000 + 5
/*********************Code*********************/
int t,n,a[N],ans,cnt,sum[N],b[N];
map<int,int>mp;
int main(void){
    IOS;
    cin>>t;
    while(t--){
        cin>>n;
        mp.clear();
        ans = 1e9,cnt = 0;
        for(int i = 0;i < n;i++){
            cin>>a[i];
            mp[a[i]]++;
        }
        for(auto i: mp){
            b[++cnt] = i.second;
        }
        sort(b,b+cnt+1);
        for(int i =1;i <=cnt;i++ )
            sum[i] =sum[i-1]+b[i];
        for(int i = 1;i <=cnt;i++){
            ans = min(ans,sum[i-1]+sum[cnt]-sum[i]-(cnt-i)*b[i]);
        }
        cout<<ans<<endl;
    }
    return 0;
}