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sum of positive integers

P: 4
pls help me to write a program (in c) such that we input an integer x,where x>0. For x, the program has to convert it into the sum of positive integers (all posible!). for e.g. x =8
1. 8
2. 7 + 1
3. 6 + 2
4. 6 + 1 + 1
5. 5 + 3
6. 5 + 2 + 1
7. 5 + 1 + 1 + 1
8. 4 + 4
9. 4 + 3 + 1
10. 4 + 2 + 2
11. 4 + 2 + 1 + 1
12. 4 + 1 + 1 + 1 + 1
13. 3 + 3 + 2
14. 3 + 3 + 1 + 1
15. 3 + 2 + 2 + 1
16. 3 + 2 + 1 + 1 + 1
17. 3 + 1 + 1 + 1 + 1 + 1
18. 2 + 2 + 2 + 2
19. 2 + 2 + 2 + 1 + 1
20. 2 + 2 + 1 + 1 + 1 + 1
21. 2 + 1 + 1 + 1 + 1 + 1 + 1
22. 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
Result should be displayed.

thanx in advance
Nov 27 '07 #1
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7 Replies


Meetee
Expert Mod 100+
P: 931
pls help me to write a program (in c) such that we input an integer x,where x>0. For x, the program has to convert it into the sum of positive integers (all posible!). for e.g. x =8
1. 8
2. 7 + 1
3. 6 + 2
4. 6 + 1 + 1
5. 5 + 3
6. 5 + 2 + 1
7. 5 + 1 + 1 + 1
8. 4 + 4
9. 4 + 3 + 1
10. 4 + 2 + 2
11. 4 + 2 + 1 + 1
12. 4 + 1 + 1 + 1 + 1
13. 3 + 3 + 2
14. 3 + 3 + 1 + 1
15. 3 + 2 + 2 + 1
16. 3 + 2 + 1 + 1 + 1
17. 3 + 1 + 1 + 1 + 1 + 1
18. 2 + 2 + 2 + 2
19. 2 + 2 + 2 + 1 + 1
20. 2 + 2 + 1 + 1 + 1 + 1
21. 2 + 1 + 1 + 1 + 1 + 1 + 1
22. 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
Result should be displayed.

thanx in advance
Help in what sense? What have you tried so far? Please post your efforts here so we can help you. Just apply simple mathematical logic and try to convert it into C conventions.

Please read posting guidelines.

Regards
Nov 27 '07 #2

P: 4
I have problem with converting my idea into C/c++ conventions, because now I begin my adventure with this language.
I know that to split integer into two parts you have to use something like:

b=1;
while(a!=0, a>=b)
{
a--;
printf("%d + %d\n", a, b);
b++;
}

so if we could split number a, and b again it should work, but I think it won't work because it will be problem with order of numbers, and many more.

The second idea I have is to initiate chart with growing size, in with we first put our number( size=1) then size=2 and split our X, then size 3 in which we firstly split number to two first "columns" and in 3th place would be "1". When it's done we add 1 to the 2nd number (-1 in 1st), and than program should try add to the 3th as lond as 3th=<2nd, than we again add 1 to 2nd, and the same with 3th, the same should work with as long as we obtain "1" in everywhere.
I think the second solution should work, but it comlicated and I have problem with how to write it. I thinking about it from a few days, and I have no idea how to do it.
You could help me even by sugestin me other way of solwing this problem. And sorry for my english!
Nov 27 '07 #3

Ganon11
Expert 2.5K+
P: 3,652
I would try using a recursive function to find all the integers adding up to the int value passed. You would call it with 8, and it would first print 8, then 7 and whatever integers added up to 1 (i.e. 1), then 6 and whatever integers added up to 2 (i.e. 2 and 1 + 1), etc etc. Your base condition would be when the number is 1, the sum of integers resulting in it is simply 1.

After this, all that remains is fancy printing.
Nov 27 '07 #4

P: 4
But what with that exemple:
18. 2 + 2 + 2 + 2

in this case we don't have 1.
Nov 28 '07 #5

Expert 100+
P: 671
Hmm, yes, so your algorithm clearly can't rely on there being a 1 at the end, right?

Look, you've just dumped your question here, and walked away. Finding out the algorithm is part of being a programmer and doing the homework assignment. We aren't going to to tell you what the algorithm is. So get to work. Sit down with pencil and paper and come up with ideas.
Nov 28 '07 #6

Ganon11
Expert 2.5K+
P: 3,652
But what with that exemple:
18. 2 + 2 + 2 + 2

in this case we don't have 1.
This is actually included in the recursive solution - I leave it up to you to show that it is. Think of the similarity with 19. 2 + 2 + 2 + 1 + 1 (i.e. (2 + 2 + 2) + 2 and (2 + 2 + 2) + 1 + 1)
Nov 28 '07 #7

P: 4
Ok Ganon11, you are right.
I found this:
http://www.site.uottawa.ca/~ivan/F49-int-part.pdf
and I will try to use it.
Nov 28 '07 #8

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