CODEWITHSUNDEEP
javafibonaccibiginteger
user8076955
user8076955

Reputation:

find nth fibonacci number where n can vary till 10^9

// i was trying to find (n^n)%1000000007 where n is the nth fibonacci term of the series. example : n=3 ans 4 (2^2) 

 public class Sol_Big {
 public static void main(String args[])
{  
       int n =10000000;
       BigInteger a = BigInteger.ZERO;
       BigInteger b = BigInteger.ONE;
       BigInteger c = BigInteger.valueOf(1);
       BigInteger MOD = BigInteger.valueOf(1000000007);

       for (int j=2 ; j<=n ; j++)
       {
           c =  a.add(b);
           c=c.mod( MOD);
           a = b.mod(MOD);
           b = c.mod(MOD);
       }
        System.out.println(c);
        BigInteger result = BigInteger.valueOf(1);
        result = c.modPow(c, MOD);
        int intval = result.intValue();
        System.out.println( intval);

      }
}

// my code is working for 10^7 input but for 10^8 and onward it is exceeding time limit....

Upvotes: 0

Views: 244

Answers (1)

Johannes Kuhn
Johannes Kuhn

Reputation: 15173

The trick is to notice that you can calculate the Fibonacci numbers using matrix multiplication:

| 0 1 |   | a |   | b     |
| 1 1 | * | b | = | a + b |

With this knowledge, we can calulate the n-th Fibonacci number:

| 0 1 |^n   | 0 |
| 1 1 |   * | 1 |

Because matrix multiplication is associative, we can efficiently calculate m^n.

  • If n is even, m^n = m^(n/2) * m^(n/2)
  • If n is odd, m^n = m^((n-1)/2)) * m^((n-1)/2)) * m

Note that we need the same thing twice, but we only have to compute it once.
This makes it possible to calculate the n-th Fibonacci number in O(log n).

I leave it to you to write the code.

Upvotes: 2

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