c ++将向量中出现n次的所有元素作为向量返回

Question

Goal: Return all elements in vector A that appear N times and put results in vector B .

Expected Result:

--Begin With---

Vector A=(10,20,30,30,40,50,100,50,20,100,10,10,200,300)

Do some code to return the name of elements that appear in Vector A
when N=3

Result should be Vector B=(10) //because only 10 is in vector A N=3 times.

My Attempt: I got the counts of all the elements placed into another vector but I don't have the part that can give back all of the elements that appear N times. I'm very flexible with how it can be done if it means a speed increase .

#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <iterator>     // std::back_inserter
#include <algorithm>    // std::copy

int main()
{
    std::vector<int> v{ 1, 2, 3, 4, 4, 3, 7, 8, 9, 10 };

    std::vector<std::pair<int, int> > shows;
    int target1;
    int num_items1;
    int size = static_cast<int>(v.size());

    for(int x=0; x<size; x++)
    {
        target1 = v[x];

        num_items1 = std::count(v.begin(), v.end(), target1);

        shows.push_back(std::make_pair(target1, num_items1));

        std::cout << "number: " << target1 << " count: " << num_items1 << '\n';
    } 
}

ACCEPTED SOLUTION TO QUESTION

#include <iostream>
#include <string>
#include <sstream>
#include <vector>
#include <iterator>     // std::back_inserter
#include <algorithm>    // std::copy
#include <set>
#include <map>

using namespace std;

int main()
{
    std::vector<int> v{ 1, 2, 3, 4, 4, 3, 7, 8, 9, 10 };

    std::vector<int> shows;
    std::map<int, int> freqOfv;

    for(const auto& i: v)
    {
        freqOfv[i]++;
    }

    std::set<int> s(v.begin(), v.end());

    int N = 2; //Can be read from stdin as well...

    for ( auto it = s.begin(); it != s.end(); it++ )

    {
        if(freqOfv[*it] ==N)
        {

            shows.push_back(*it);
        }
    }

    for (std::vector<int>::const_iterator i = shows.begin(); i != shows.end(); ++i)
    {
        std::cout << *i << ' ';
    }
    return 0;
    }

Answer 1

As suggested in the comments, std::map will simplify the code:

int main()
{
    std::vector<int> v{ 1, 2, 3, 4, 4, 3, 7, 8, 9, 10 };    
    std::map<int, int> freqOfv; 
    for(const auto& i: v)
        freqOfv[i]++;

    int N = 2; //Can be read from stdin as well...

    for(const auto& i: freqOfv)
    {
        if(N == i.second)
            std::cout << "The value " << i.first << " occurs " << N << " times." << std::endl;
    }   
}

This produces the following output:

The value 3 occurs 2 times.
The value 4 occurs 2 times.

Of course, you need #include <map> at the beginning to use maps in your code.