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Problem  :  TSHOW2 - Fibonacci brain Solution : We can easily calculate nth Fibonacci number by dynamic programming using the recurrence relation - \[ { f(x) } = \left\{ \begin{array}{ll} { f(x-1) + f(x-2) } & \mbox{if } x > 2 \\ 1 & \mbox{if } x = 2 \\ 0 & \mbox{if } x = 1 \end{array} \right. \] f[1] = 0 f[2] = 1 for i in { 3 ... n } f[i] = f[i-1] + f[i-2] But we cannot calculate Fibonacci number by dynamic programming when n is large like 10^18. We can easily calculate large 10^18 Fibonacci number by  matrix exponentiation . We can write that, \[ \begin{bmatrix} {1} & {1} \\ {1} & {0} \end{bmatrix} ^ {n} = \begin{bmatrix} { f(n+1) } & { f(n) } \\ { f(n) } & { f(n-1) } \end{bmatrix} \] n is large , using binary exponentiation we can calculate nth Fibonacci number by O(log n) complexity.    

Infinite fractions \[ \text{When is }\ x \mid y \text{ ?} \ \] x can be represent as \[ {x} = \prod_{i=1}^n p_{i}^{k} \] \[ \text{where}\ p_{i} =\text{ is a prime factor of } {x} \text{ and } {x} \text{ has } {n} \text{ prime factors.} \] \[ \text{Example: }\ {x} = {12} = {2}^{2} \times {3} \] We can represent the y by the same way. x divides y ( x is a divisor of y ) if and only if all prime factors of x are available in y prime factorization and each prime factor (p) power (p^a) in x factorization is less than equal to prime factor (p) power (p^b) in y factorization (a <= b). Or we can say that when GCD(x, y) = x  , x divides y. \[ \text{When is }\ x\nmid y \text{ ?} \ \] y = q * x + r [ where r < x ] When reminder (r) is not equal to 0, y is not divisible by x.  Only when r is not equal to zero there are numbers after the decimal point of fraction p/q. How to get numbers after the decimal point of a fraction ? We recursively do it until MOD = 0. initially MOD