Here is a simple solution to the fibonacci problem. The case of element 0 and element 1 are special cases, i.e. they get the whole series going and so they are dealt with in a special if. Then the while loop procedes and generates all elemnets in the series.

int sum=1, term, term1=0, term2=1;

cin >>term;
if (term < 2)
   cout<<term;
else
{
  //term--;
  while (term >1 )
  {
    sum = term1 + term2;
    term1 = term2;
    term2 = sum;
    term--;
  }

cout << sum;
}

This is a slightly different solution which solves the problem without an if. See if you can figure how it improves on the first solution.

int main()
unsigned long current = 0, prev=0, prevprev=1;
int term;
cout << "enter term you want :";
cin >> term;

while (term)
{
current=prev+prevprev;
prevprev=prev;
prev=current;
term--;
}
cout<<current;

Here is a clever solution to the factorial problem. Note that there is no such thing as a negative factorial, but neverthless I have made this solution to solve both negative and possitive factorial questions and I did it without using an if. And it also works for zero. How would you solve this problem had you not seen this solution?

int number, result =1;
cin >> number;
while (number>1)
{
   result*=number;
   number--;
}
while (number<0)
{
   result*=number;
   number++;
}  
cout << result;

Here is a solution to finding unique pythagorean triples. I think you will find that this is the most efficient solution to the problem, short of using fancy things like square roots and powers or for that matter Platos theorum.

#define MAX 200
int main(int argc, char *argv[])
{
   unsigned long a, b, c, e, _2sides;
   for (a=1; a<MAX; a++)
   {
      for (b=a+1; b<MAX; b++)
         {
	        _2sides=a+b;//No side of any triangle can ever be larger than the sum of 2 other sides
                for (c=b+1; c<_2sides; c++)
                {
                       if ((c*c)==(a*a)+(b*b))//found a triple
                       {
                          for (e=2; e<a; e++)
                          {
                              if ((a%e)==0&&(b%e)==0&&(c%e)==0) // found reject
                              {
                                 break;
                              }
                          }
                          if (e==a) // i have gone through entire loop and found no mults
                          {
                              cout<<a<<", "<<b<<", "<<c<<endl;
                              break;//don't bother searching for any more c's since I already found one
                          }
                       }
                }
         }   
   }
}