Actually it does. One can skip few steps Using this formula one will get only the even fibinoacci numbers. Traditional algo will be :

f1=1;

f2=2;

f=0;

while(f1 less than n)

{

if(f%2==0) sum+=f;

f=f1+f2;

f2=f1;

f1=f;

}

print -> sum

However you can skip the step of verifing odd-even. Also you are skiping 2 odd number and directly getting only the even numbers.

f1=2;

f2=0;

f=0;

while(f1 less than n) {

sum+=f1;

f=4*f1+f2;

f2=f1;

f1=f;

}

print -> sum

PS-Sorry for bad english