# Integers: Recreation One

### [Integers: Recreation One](https://www.codewars.com/kata/55aa075506463dac6600010d)

Divisors of 42 are : 1, 2, 3, 6, 7, 14, 21, 42. These divisors squared are: 1, 4, 9, 36, 49, 196, 441, 1764. The sum of the squared divisors is 2500 which is 50 \* 50, a square!

Given two integers m, n (1 <= m <= n) we want to find all integers between m and n whose sum of squared divisors is itself a square. 42 is such a number.

The result will be an array of arrays or of tuples (in C an array of Pair) or a string, each subarray having two elements, first the number whose squared divisors is a square and then the sum of the squared divisors.

## Examples:

```
list_squared(1, 250) --> [[1, 1], [42, 2500], [246, 84100]]
list_squared(42, 250) --> [[42, 2500], [246, 84100]]
```

The form of the examples may change according to the language, see `Example Tests:` for more details.

**Note**

In Fortran - as in any other language - the returned string is not permitted to contain any redundant trailing whitespace: you can use dynamically allocated character strings.

### Solutions

#### 🧠 C++

```cpp
#include <vector>
#include <numeric>
#include <sstream>
#include <cmath>
#include <utility>
#include <string>

using ll = long long;

class SumSquaredDivisors
{
public:
    static std::string listSquared(ll m, ll n)
  {
      std::vector<std::pair<ll,ll>> res;
      for(ll i = m; i <=n; i++)
      {
        //find divisors
        std::vector<ll> divisors;

        for(ll j = 1; j<=i; ++j)
          if(i%j == 0)
            divisors.push_back(j);

        //power all divisors
        std::for_each( divisors.begin(), divisors.end(), [](ll & x){x=std::pow(x,2);});

        ll sum_of_elems = std::accumulate(divisors.begin(), divisors.end(), 0);

        double divisors_sqrt = std::sqrt(sum_of_elems);
        if ( divisors_sqrt-(int)divisors_sqrt == 0.0)
          res.push_back({i, sum_of_elems});
      }

      //if we found nothing
      if(!res.size())
        return "{}";

      //format vector of pairs to string via stringstream
      std::stringstream ss_resstr;
      ss_resstr << "{";
      for (auto k : res)
          ss_resstr<<"{"<<k.first<<", " << k.second << "}, ";

      std::string resstr = ss_resstr.str();
      resstr.replace(resstr.end()-2, resstr.end(), "}");

      return resstr;
    }

};
```


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://anton-veselskyi.gitbook.io/codding-problems-solutions/codewars/5-kyu/integers-recreation-one.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.