I have come up with the solution but still this is not the for the
general case. I would like to make my problem clear once again.
The problem is to find total number of solutions for the following
expression.
if k=1, the expression is like this: 1*x1=1, so number of solution is
1, i.e. x1=1.
if k=2, the expression is like this: 1*x1 + 2*x2=2, so number of
solution is 2: (x1=2,x2=0) and (x1=0,x2=1)
so the general expression is like this: 1*x1 + 2*x2 + 3*x3
+4*x4+........+k*xk = k, where xn adn k are non-negative integers. we
need to find how many possible solution are there for this expression.
I can code for the specific value of k but i couldn't figure out for
general case: please help if you can find me for general case. here is
my logic.
if you are given k=3
then
for(x1=0;x1<=k;x1++)
{
sum=x1;
s1=sum;
for(x2=0;x2<=k;x2++)
{
sum=sum+2*x2;
s2=sum;
for(x3=0;x3<=k;x3++)
{
sum=sum+3*x3;
if(sum==k)
count=count+1;
sum=s2;
}
sum=s1;
}
}
print(count);
here you can see that loop increases as k increases i.e. number of
loops=k and we check for the condition at most inner loop. that means
check every possible combination. can any one help me for general
case?
thanks
--~--~---------~--~----~------------~-------~--~----~
You received this message because you are subscribed to the Google Groups
"Algorithm Geeks" group.
To post to this group, send email to [email protected]
To unsubscribe from this group, send email to [EMAIL PROTECTED]
For more options, visit this group at http://groups.google.com/group/algogeeks
-~----------~----~----~----~------~----~------~--~---
