Thread: Just learning, problem with an excercise.

  1. #1
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    Just learning, problem with an excercise.

    I have to do the following:
    Read a number and calculate the following sequence:
    1-1/2+1/3-1/4+1/5...1/n

    This is the way I'm doing it, but it doesn't work:

    Code:
    #include <stdio.h>
    
    int main()
    {
    	int a=0, n, suma=0;
    	float den;
    	
    	printf ("\n Ingrese el número : ");
    	scanf ("%i", &n);
    	
    	while (a<n)
    	{
    		if (a%2==0){
    			a++;
    			den=1/a;
    			suma-=den;}
    		
    		else{
    			a++;
    			den=1/a;
    			suma+=den;}
    	}
    	
    	
    	printf ("\n Resultado secuencia = %i", suma);
    	
    	return 0;
    }

  2. #2
    Banned master5001's Avatar
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    I personally would (either in my head or on a sheet of paper) see what can be constructively noted about this function.

    1. So it looks like the denominator simply gets incremented by one each time.
    2. Even denominators get subtracted and odds added

    So couldn't you make a very simple loop sheerly from this alone?

    Example:
    Code:
    $denominator = 1
    $sum = 0.0;
    
    while $denominator < $n+1
      if iseven($denominator)
        sum -= 1.0 / $denominator
      else
        sum += 1.0 / $denominator
    loop

  3. #3
    and the Hat of Guessing tabstop's Avatar
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    Note also that:
    When a==0 1/a is NaN
    When a==1 1/a is 1
    Every other a 1/a is 0

    Maybe use 1.0/a instead.

  4. #4
    Banned master5001's Avatar
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    Yeah but the variable a should surely be 1 to begin with. Oooooooooh ok, nevermind. I just saw that your post was totally directed toward the OP.

  5. #5
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    This is how it looks now:

    Code:
    #include <stdio.h>
    
    int main()
    {
    	int a=0, n;
    	float den, suma=0.0;
    	
    	printf ("\n Ingrese el n&#250;mero : ");
    	scanf ("&#37;i", &n);
    	
    	while (a<n)
    	{
    		if (a%2==0){
    			a++;
    			den=1.0/a;
    			suma-=den;}
    		
    		else{
    			a++;
    			den=1.0/a;
    			suma+=den;}
    	}
    	
    	printf("%f", den);
    	printf ("\n Resultado secuencia = %f", suma);
    	
    	return 0;
    }
    But this is what I get:

    Ingrese el n&#250;mero : 2
    Resultado secuencia = -0.500000

    Why the negative number?
    Last edited by medeshago; 10-15-2008 at 06:59 PM.

  6. #6
    Banned master5001's Avatar
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    Read my code (even if it is pseudo code) and read what tabstop suggested. I see a pattern...

    Example:
    Code:
    #include <stdio.h>
    
    int main()
    {
    	int a=1, n;
    	float suma=0.0f;
    	
    	printf ("\n Ingrese el n&#250;mero : ");
    	scanf ("&#37;i", &n);
    	
    	while (a<n)
    	{
    		if (a%2==0)
    			suma-=1.0f/a;
    		else
    			suma +=1.0f/a;}
    		a++;
    	}
    	
    	// printf("%d", den); <- why? Who cares
    	printf ("\n Resultado secuencia = %d", suma);
    	
    	return 0;
    }
    Last edited by master5001; 10-15-2008 at 07:01 PM.

  7. #7
    and the Hat of Guessing tabstop's Avatar
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    Quote Originally Posted by medeshago View Post
    This is how it looks now:

    Code:
    #include <stdio.h>
    
    int main()
    {
    	int a=0, n;
    	float den, suma=0.0;
    	
    	printf ("\n Ingrese el número : ");
    	scanf ("%i", &n);
    	
    	while (a<n)
    	{
    		if (a%2==0){
    			a++;
    			den=1.0/a;
    			suma-=den;}
    		
    		else{
    			a++;
    			den=1.0/a;
    			suma+=den;}
    	}
    	
    	printf("%f", den);
    	printf ("\n Resultado secuencia = %f", suma);
    	
    	return 0;
    }
    But this is what I get:

    Ingrese el número : 2
    Resultado secuencia = -0.500000

    Why the negative number?
    Because you start by subtracting? You're currently doing -1 + 1/2 - 1/3 + 1/4 - ...

  8. #8
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    Thanks, now it works great.

  9. #9
    Banned master5001's Avatar
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    You need to start placing your brackets a little more cleanly. But at least you did use [code] tags. So I will simply say: you're welcome.

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