Finding the fibonacci number of large number

问题

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];
        }
    }
}

回答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.

来源：https://stackoverflow.com/questions/38554793/finding-the-fibonacci-number-of-large-number