Candy Box (easy version) CodeForces - 1183D 题解

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题目

CodeForces - 1183D
This problem is actually a subproblem of problem G from the same contest.

There are n candies in a candy box. The type of the i-th candy is ai (1≤ai≤n).

You have to prepare a gift using some of these candies with the following restriction: the numbers of candies of each type presented in a gift should be all distinct (i. e. for example, a gift having two candies of type 1 and two candies of type 2 is bad).

It is possible that multiple types of candies are completely absent from the gift. It is also possible that not all candies of some types will be taken to a gift.

Your task is to find out the maximum possible size of the single gift you can prepare using the candies you have.

You have to answer q independent queries.

If you are Python programmer, consider using PyPy instead of Python when you submit your code.

Input
The first line of the input contains one integer q (1≤q≤2⋅105) — the number of queries. Each query is represented by two lines.

The first line of each query contains one integer n (1≤n≤2⋅105) — the number of candies.

The second line of each query contains n integers a1,a2,…,an (1≤ai≤n), where ai is the type of the i-th candy in the box.

It is guaranteed that the sum of n over all queries does not exceed 2⋅105.

Output
For each query print one integer — the maximum possible size of the single gift you can compose using candies you got in this query with the restriction described in the problem statement.

Example
Input

3
8
1 4 8 4 5 6 3 8
16
2 1 3 3 4 3 4 4 1 3 2 2 2 4 1 1
9
2 2 4 4 4 7 7 7 7

Output

3
10
9

Note
In the first query, you can prepare a gift with two candies of type 8 and one candy of type 5, totalling to 3 candies.

Note that this is not the only possible solution — taking two candies of type 4 and one candy of type 6 is also valid.

分析

题目大意

给定一个序列，统计出每个数字出现的个数，然后求解个数相加后的最大值，这里的相加的个数的值不能重复，如果重复可以减小该值。如1212，则和为2+1=3。

解题思路

用vector存储每种数字的个数，数组下标代表该数字。之后利用sort进行升序排序，用一个变量ans存储当前最大的个数，遍历vector，求解ans的和，若数值大于等于ans则说明该值已被取过，故ans-1，否则说明数值与ans间还有间隔，取ans=数值。

代码

#include 
#include 
#include 
#include 
const int maxn=200010;
using namespace std;

int main()
{
    int q,n;
    scanf("%d",&q);
    while(q--)
    {
        scanf("%d",&n);
        vector<int> a(n+1); 
        for(int i=0;i<n;i++)
        {
            int x;
            scanf("%d",&x);
            a[x]++; //记录x的出现次数
        }
        sort(a.rbegin(),a.rend()); //反响迭代器，这里作用是升序排列
        int ans=a[0]; //取得最大值
        int sum=a[0];
        for(int i=1;i<n;i++)
        {
            if(ans==0) //代表不同的值已被取尽
                break;
            if(a[i]>=ans) //由于ans初始值即为最大值，故若a[i]>=ans，表明a[i]的值已被取过，只需加上ans-1.
            {
                sum+=(ans-1); 
                ans--;
            } else //这种情况可能是ans值偏大，与a[i]间的值无法被取到
            {
                sum+=a[i];
                ans=a[i];
            }
        }
        printf("%d\n",sum);
    }
    return 0;
}