Decomposition of positive integers into the sum of several contiguous natural numbers

Title: Enter a positive integer that outputs all possible positive integer sequences if the number can be represented by a sum of several consecutive positive integers.

A positive integer may be represented as a sum of N (n>=2) consecutive positive integers, such as:
15=1+2+3+4+5
15=4+5+6
15=7+8

Some numbers can be written as the sum of successive n (>1) natural numbers, such as 14=2+3+4+5; some cannot, say, 8. So how do you tell if a number can be written as the sum of successive n natural numbers?

If a number M can be written as the sum of successive n natural numbers starting with a, then m=a+ (a+1) + (a+2) +...+ (a+n-1) =n*a+n* (n-1)/2, a!=0 is required, otherwise a+1 is the continuous n-1 integer, that is, a requirement (m-n* (n-1)/2 %n==0, so it's easy to tell if a number can be written in the form of a continuous n natural number, traversing n=2...sqrt (M), and outputting all the solutions.

void divide (int num)  {      int i,j,a;      for (i=2; i<=sqrt (float) num) (++i)      {          if (num-i* (i-1)/2)%i==0)          {a= (num-i*              (i-1)/2)/I;              if (a>0)              {for                  (j=0; j<i; ++j)                      cout<<a+j<< "";              }              cout<<endl;          }}}     

The second question is what number can be written as the sum of successive n natural numbers, and what number cannot?

The programming experiment found that all the other numbers except 2^n could be written in that form. The following explains why.
If the number of M meets the conditions, then there is m=a+ (a+1) + (a+2) +...+ (a+n-1) = (2*a+n-1) *N/2, and 2*a+n-1 and N must be an odd-numbered one even, that is, M must have an odd factor, and all 2^n have no odd factor, and therefore certainly do not meet the criteria.
It is proved that only m has an odd factor, that is, M!=2^n,m can be written as the sum of successive n natural numbers. Suppose that M has an odd factor a, then m=a*b.

1. If B is also an odd number, as long as A-1/2>0,m can be written as B (A-1)/2, a continuous a natural numbers, the conclusion of A and B exchange, still set up. 15=3*5=1+2+3+4+5=4+5+6.
2. If B is an even number, then we have an odd number A and an even number B.

• 2.1 B (A-1)/2>0,m can be written as a continuous a natural number starting with B (A-1)/2. 24=3*8=7+8+9.
• 2.2 if (a+1)/2-b>0,m can be written in (a+1)/2-b the beginning of a continuous 2*b a natural number. 38=19*2=8+9+10+11.

The above two inequalities must have at least one set up, so it can be proved that as long as M has an odd factor, it can be written as the sum of successive n natural numbers.

Another algorithm for the decomposition of positive integers:
SUM (i,j) for I summation to J and
Make I=1 j=2
If sum (i,j) >n i++
else if sum (i,j) <n J + +
else cout I...J

#include <iostream>   using namespace std;    int add (int m,int n)  {      int sum=0;      for (int i=m;i<=n;i++)          sum+=i;      return sum;  }    void divide (int num)  {      int i=1,j=2,flag;      int sum=0;      while (I<=NUM/2)      {       sum=add (i,j);       while (Sum!=num)       {          if (sum>num)              i++;          else              j + +;          Sum=add (I,J);       }       for (int k=i;k<=j;k++)          cout<<k<< "";       ++i;       cout<<endl;      }  }    int main ()  {      int num;      cout<< "Please input your number:" <<endl;      cin>>num;      Divide (num);      return 0;  }  

Decomposition of positive integers into the sum of several contiguous natural numbers