C / C++ / MFC

// CWzdThread class (class B)
myDlg* pDlg;

int value1 = input; // store value we will decrement

for (int value2 = 0; value2 <= input; ++value2, --value1)  // You can put mutiple statements into a for loop by using commans, this is true of any of the three parts involved.  You can also just put a ; if, for example, value2 was already set it could be for(;value2<=input;++value2, --value1)
{
    cout << value1 << " + " value2 << " = ";
}

cout << input << endl;

#include "stdafx.h"

#include <iostream>
#include <math.h>

using std::cout;
using std::cin;
using std::endl;

int main(int argc, char* argv[])
{
	
	// It's a good idea to declare variables as you need them, not all at the top like this
	int input;

	cout <<"please enter a number ";
	cin >> input;

	for (int i = 2; i < 10; ++i)
	{
// This works because, for example, if input can be created by three consecutive numbers, then 
//  their values will be input/value, input/value+1 and input/value+2, therefore input/3 will have 
//  a remainder of 2
		if (input%i == i-1) break;
	}

// We check to see if the loop fell through.  Substitute any number to make that the maximum count
// of numbers in the sequence

	if (i == 10)
	{
		cout << "No solution found\n";
		return 0;
	}

// Find the lowest value in the sequence.  Floor guarentees a round down, otherwise we might be out by 1 at times.

	int lowest = floor(input/i);
// Just use the same vale of i for a new loop, we don't use the value anyhow, we just need that many iterations

	for (;i>=0;--i)
	{
		cout << lowest++;
// make sure we don't put a '+' at the end of the sequence
		if (i > 0) cout << " + ";
	}
// And we're done.  LMK if there is anything here that needs further explanation.

	cout << " = " << input << endl;

	return 0;
}

#include "stdafx.h"

#include <iostream>
#include <math.h>

using std::cout;
using std::cin;
using std::endl;

int main(int argc, char* argv[])
{
	
	// It's a good idea to declare variables as you need them, not all at the top like this
	int input;

	cout <<"please enter a number ";
	cin >> input;

	for (int i = 2; i < 10; ++i)
	{
		if ((input%i) == (i-1)) break;
	}

	if (i == 10)
	{
		cout << "No solution found\n";
		return 0;
	}

	int lowest = floor(input/i);
// Added to make sure we never provide a solution of 0 + 1 + ....
	if (lowest < 1) lowest = 1;

// Keep track of the sequence because sequences starting from 1 seem to overrun right now, this fixes that.
	int sequence = 0;

	for (;i>0;--i)
	{
// Make sure we never overrun our value
		sequence += lowest;
		if (sequence > input) break;
		
		cout << lowest++;
		if (i > 1 && sequence + lowest <= input) cout << " + ";
	}
	cout << " = " << input << endl;

	return 0;
}

#include "stdafx.h"

#include <iostream>
#include <math.h>

using std::cout;
using std::cin;
using std::endl;

int main(int argc, char* argv[])
{
	
	// It's a good idea to declare variables as you need them, not all at the top like this
	int input;

	cout <<"please enter a number ";
	cin >> input;

	int result = 1;

	int low = 1, high = 1;

// Figure out a sequence starting at 1 that is >= input

	while (result < input) 
	{ 
		result += ++high;
	}

// Until the sequence is equal to input we subtract from the bottom until it's less than input,
// then add at the top until its greater than or equal
// We use while loops because they will run however many times they need to in order to fulfill the condition
// We could equally have put for (;result != input;) but that would have been pointless.

	while (result != input)
	{
		while (result > input)
			result -=low++;

		while (result < input)
			result += ++high;
	}

// output the result

	for (int i = low; i < high; ++ i)
	{
		cout << i << " + ";
	}

	cout << high << " = " << result << endl;


	return 0;
}