need to solve simple c problem

This is a discussion on need to solve simple c problem within the C Programming forums, part of the General Programming Boards category; Problem Name: Y = MX Everyone know that the equation of a straight line is Y = MX + C. ...

  1. #1
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    Exclamation need to solve simple c problem

    Problem Name: Y = MX
    Everyone know that the equation of a straight line is Y = MX + C. But if C = 0, then the straight line is passing through the origin. Here I always put C = 0. So our equation will be always Y = MX. Straight line is a geometrical term but my problem is not geometrical. I just use the equation Y = MX. Since Y = MX, so Y is must be divisible by M and X, and for a specific value of Y and various value of M and X, the equation must be satisfy.
    But in this problem we find (N) the number of M and X that satisfy the equation with condition M must be less than X (M16 = 1 * 16 (1<16) so it is countable.
    16 = 2 * 8 (2<8) so it is countable.
    16 = 4 * 4 (4 = 4) so it is not countable.
    16 = 8 * 2 (8>2) so it is not countable.
    16 = 16 * 1 (16>1) so it is not countable.
    So our answer N=2.

    So your duty is to find out how many way to find M and X (M
    Input: You have to provide an integer Y (1<=Y<=10000).

    Output: You just print the value of N.

    Sample input:
    16

    Sample output:
    2

    what output i'll get for following those input 128,500,666,1000,10000?
    i need code and solution

    i tried many time but can't get logic. how i improve my logic?
    help to clera this program logic concept. ... Shift+R improves the quality of this image. Shift+A improves the quality of all images on this page.
    Code:
    #include<stdio.h>
    int main()
    {
        
        int a,b,c=1,i;
        scanf("%d",&a);
        for(i=2;i<=a;i+=2)
        {
            if((a%i)==0)
            {
            printf("%d ",i);
            b=(a%i);
            }
    
            if(b>i)
             c++;
    
        }
    
        printf("\n%d",c);
    }
    Last edited by raihan004; 09-15-2012 at 02:00 PM.

  2. #2
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    i need code and solution
    Are you joking? please read the forum FAQ.

  3. #3
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    Quote Originally Posted by root4 View Post
    Are you joking? please read the forum FAQ.
    sorry am new. can't i ask any problem solution code ? what is my fault?

  4. #4
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    can't i ask any problem solution code ?
    Indeed. You can get some help, here, with programming issues, but nobody is going to solve any assignment.
    Work on your problem first, show us some code and possible problems you face, then we'll help.

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    I have been feverishly looking for a silver platter with a gold rim - something suitable for the delivery of this code and solution.

    All in vain.

  6. #6
    and the hat of wrongness Salem's Avatar
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    You can only get so far by looking at code.
    The real learning starts when you actually try and write some yourself.

    Make an actual effort, and then we'll help you.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
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    now i inlcude my tried attetmp code . now make me my code fullfill and make clear concept

  8. #8
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    1. Your assignment gives you explicit names for the objects so use them in your code, that gives a readability bonus (e.g. what you named "a" is "y").
    2. Indent correctly your code.

    You have all the pieces, but not composed correctly: to increment your counter (c), Y has to be divisible by your current M (Y%M==0) and that M must be less than (Y/M) so you have to merge your two conditionals.
    I don't know why you increment M by 2.
    Last edited by root4; 09-15-2012 at 03:04 PM.

  9. #9
    Algorithm Dissector iMalc's Avatar
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    The problem comes from asking the equivalent of "do it for me" when you should be asking the equivalent of "teach me".
    My homepage
    Advice: Take only as directed - If symptoms persist, please see your debugger

    Linus Torvalds: "But it clearly is the only right way. The fact that everybody else does it some other way only means that they are wrong"

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    Cool [solved] problem solve

    i solve this problem . if any wrong about my logic please repaly or notify me thanks all
    Code:
      
    
    #include<stdio.h>
    int main()
    {
    
        int y,m,n=0,i,x=0;
        scanf("%d",&y);
        for(i=1;i<=y;i++)
        {
            if((y%i)==0)
            {
    
    
    x=y/i;
    
            if(x>i)
            {
                printf("%4d",i);
             n++;
            }
            }
        }
    
        printf("\ntotal %d step need to do this",n);
    }
    
    :redface:

  11. #11
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    Looks ok.
    Indent your code properly.
    You can merge the two conditionals with a logical 'and' operator (&&).
    Your variable 'm' is unused so far, simply replace 'i' with 'm'.

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