寻找大数的斐波那契数

Question

I wrote the following program for finding the modulus of large Fibonacci's number. This can solve large numbers but fails to compute in cases like fibo_dynamic(509618737,460201239,229176339) where a = 509618737 , b = 460201239 and N = 229176339 . Please help me to make this work.

long long  fibo_dynamic(long long  x,long long  y,long long  n, long long a[]){
    if(a[n]!=-1){
         return a[n];
    }else{
        if(n==0){
            a[n]=x;
            return x;
        }else if(n==1){
            a[n]=y;
            return y;
        }else {
            a[n]=fibo_dynamic(x,y,n-1,a)+fibo_dynamic(x,y,n-2,a);
            return a[n];
        }
    }
}

Answer 1

The values will overflow because Fibonacci numbers increase very rapidly. Even for the original fibonacci series (where f(0) = 0 and f(1) = 1 ), the value of f(90) is more than 20 digits long which cannot be stored in any primitive data type in C++. You should probably use modulus operator (since you mentioned it in your question) to keep values within range like this:

a[n] = (fibo_dynamic(x,y,n-1,a) + fibo_dynamic(x,y,n-2,a)) % MOD;

It is safe to mod the value at every stage because mod operator has the following rule:

if a = b + c, then:
a % n = ((b % n) + (c % n)) % n

Also, you have employed the recursive version to calculate fibonacci numbers (though you have memoized the results of smaller sub-problems). This means there will be lots of recursive calls which adds extra overhead. Better to employ an iterative version if possible.

Next, you are indexing the array with variable n . So, I am assuming that the size of array a is atleast n . The value of n that is mentioned in the question is very large. You probably cannot declare an array of such large size in a local machine (considering an integer to be of size 4 bytes , the size of array a will be approximately 874 MB ).

Finally, the complexity of your program is O(n) . There is a technique to calculate n_th fibonacci number in O(log(n)) time. It is "Solving Recurrence relations using Matrix Exponentiation." Fibonacci numbers follow the following linear recurrence relation:

f(n) = f(n-1) + f(n-2)   for n >= 2

Read this to understand the technique.