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Problems with code
I create a program that calculates the products of a series of numbers."k" we type from the keyboard.I also do not understand how to find factorial in my incident.Help me please with the implementation My code #include <stdio.h> #include <stdlib.h> #include <math.h> int main() { int n,k; float z,prd=1; printf("Enter k: \n"); scanf("%d",&k); for(n=-3;n<=k;n++) { z=((n+2)*(abs(n-4))/(n+3); // (n+3)!(factorial) prd*=z //product of numbers } printf("5.2f \n",prd); return 0; }
6 Respuestas
+ 3
Alright. That product is 0 as soon as n is -2.
But still, to calculate the factorial you can either write a simple algorithm yourself. Can look like this:
int factorial(int n)
{
int f = 1;
while(n) f *= n--;
return f;
}
But that means to repeat a lot of computations. It'd be smarter to produce the factorial as a side-effect of the loop.
A thing to remember: When dividing integers it will be integer division (division with remainder). To get floating points, you first need to convert one element to a floating point type (explicit type cast). So, the loop could look like this:
double prod = 1.0;
int f = 1;
for(int n=-3; n<=k; n++) {
prod *= ((double)(n+2) * abs(n-4)) / f;
f *= n+4;
}
Edit:
Just in case the product is supposed to be a sum (which would make more sense to me...):
double z = 0;
int f = 1;
for(int k=-3; k<n; k++) {
z += ((double)(k+2) * abs(k-4)) / f;
f *= k+4;
}
+ 2
Could you write out the series you want to calculate without using "code" ? Right now, as soon as n is -2 one factor becomes zero. That means your product will become zero. [I understand that the denominator is supposed to be factorial, right? Otherwise you run into trouble there as well, dividing by 0...]
+ 2
I am sorry, that makes even less sense than before. Now, you are trying to take the factorial of a negative number which isn't defined.
The k-th term is
(k+2) |k-4| / (k-4)! or
(k+2) |k-4| / (k+3)! ?
And the product runs from k=-3 up to n ?
+ 1
thank you,Leif!
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z=П k up,n=-3 down below (n+2)|n-4|/(n-4)!