Candy|Candy Box (easy version) CodeForces - 1183D 题解 CandyBox(easyversion)CodeForces-1183

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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.
【Candy|Candy Box (easy version) CodeForces - 1183D 题解】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 a(n+1); for(int i=0; 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; }