Huawei machine test question -- hj76 nikochus theorem

describe

To test the Nicolas Theorem , namely ： Any integer m All the cubes can be written as m The sum of consecutive odd numbers .

for example ：

1^3=1

2^3=3+5

3^3=7+9+11

4^3=13+15+17+19

Enter a positive integer m（m≤100）, take m The cube of is written as m Output in the form of the sum of consecutive odd numbers .

Data range ：1\le m\le 100\1≤m≤100 

Advanced ： Time complexity ：O(m)\O(m) , Spatial complexity ：O(1)\O(1) 

Input description ：

Enter a int Integers

Output description ：

Output after decomposition string

Example 1

Input ：

6

Copy output ：

31+33+35+37+39+41

#include <iostream>
#include <vector>
#include <sstream>
std::string itoa(int num)
{
    std::stringstream oss;
    oss <<  num;
    return oss.str();
}

void nikochus(int n)
{
    int coub = n * n * n;
    int tempi = n * n - n + 1;          // Record the coordinates at the beginning of each accumulation 
    if(tempi % 2 == 0)
    {
        tempi -= 1;
    }
    std::vector<std::string> rtnVect;
    while(1)
    {
        int sum = 0;
        int start = tempi;
        while(start < coub / 2)
        {
            sum += start;
            start += 2;
            int end = start;
            if(sum == coub)
            {
                for(int k = tempi; k < end; k += 2)
                {
                    rtnVect.push_back(itoa(k));
                    rtnVect.push_back("+");
                }
                
                std::vector<std::string>::iterator iter = rtnVect.begin();
                while(iter != rtnVect.end()-1)
                {
                    std::cout << *iter++;
                }
                std::cout << std::endl;
                return;
            }
            if(sum > coub)
            {
                break;
            }
        }
        tempi += 2;
        sum = 0;
    }

    // print(rtnVect);
}

//6         31+33+35+37+39+41
int main()
{
    int n;
    while(std::cin >> n)
    {
        nikochus(n);
    }

    return 0;
}