Project Euler #2: Even Fibonacci numbers Discussions |

8 years ago+ 2 comments

I used Binet's formula, changed to long in C++ and I still get a few timeouts. Any recommendations?

8 years ago+ 0 comments

well this won't give any timeouts but one test case is failing , i think it's because of inaccuracy we get in golden ratio while calculating it for very large term for e.g. the following code will generate 27th term as 37889062373143904 instead of 37889062373143906. rectify it and you will get the correct code :)

int main(){
    int t; 
    scanf("%d",&t);
    for(int a0 = 0; a0 < t; a0++){
        long n, sum = 0; 
        scanf("%ld",&n);
        long curr_num = 2;
        while(curr_num<=n){
            sum += curr_num;
            curr_num = (long)floor((pow(((1 + pow(5 , 0.5)) / 2),3) * curr_num + 0.5));
            //sum += curr_num;
        }
        printf("%ld \n", sum);
    }
    return 0;
}

7 years ago+ 1 comment

sorry for late advice - store the first fibbonachi numbers in memory and go thrue tests $-)

import functools
@functools.lru_cache(maxsize=None)
def fib(n):
    if n<=2:
        return 1
    else:
        return fib(n-1) + fib(n-2)



n = int(input().strip())
s=0
for i in range(3,n,3):
    _ = fib(i)
    if _<=n:
        s+=_
    else:
        break
        
print(_)

all works on slow but wise Python

6 years ago+ 0 comments

for i in range(3,n,3) I see what you did there, smart.

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