Zeckendorf’s Theorem (Non-Neighbouring Fibonacci Representation), Find nth Fibonacci number using Golden ratio, n’th multiple of a number in Fibonacci Series, Space efficient iterative method to Fibonacci number, Factorial of each element in Fibonacci series, Fibonomial coefficient and Fibonomial triangle, An efficient way to check whether n-th Fibonacci number is multiple of 10, Find Index of given fibonacci number in constant time, Finding number of digits in n’th Fibonacci number, Count Possible Decodings of a given Digit Sequence, Program to print first n Fibonacci Numbers | Set 1, Modular Exponentiation (Power in Modular Arithmetic), Find Square Root under Modulo p | Set 1 (When p is in form of 4*i + 3), Find Square Root under Modulo p | Set 2 (Shanks Tonelli algorithm), Euler’s criterion (Check if square root under modulo p exists), Multiply large integers under large modulo, Find sum of modulo K of first N natural number. Given a positive integer n, generate all possible unique ways to represent n as sum of positive integers. we first print p[] and then update p[] to store the next partition. Input Description: An integer \(n\). How to check if a given number is Fibonacci number? It has algorithms for many combinatorial computations including generating all partitions of an integer. They generate partitions in lexicographic and antilexicographic order, respectively. Recursive sum of digits of a number formed by repeated appends, Find value of y mod (2 raised to power x), Modular multiplicative inverse from 1 to n, Given two numbers a and b find all x such that a % x = b, Exponential Squaring (Fast Modulo Multiplication), Subsequences of size three in an array whose sum is divisible by m, Distributing M items in a circle of size N starting from K-th position, Discrete logarithm (Find an integer k such that a^k is congruent modulo b), Finding ‘k’ such that its modulus with each array element is same, Trick for modular division ( (x1 * x2 ….
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We are given current partition in p[] and its size.
(If order matters, the sum becomes a composition. Steps to get next partition from current partition:

Now there may be two cases: Copy p[k] to next position, increment k and reduce count by p[k] while p[k] is less than rem_val. It's recursive and to generate the solution it calls the breakup function 46 times (thanks to the magic of Memoize)..

Given a positive integer n, generate all possible unique ways to represent n as sum of positive integers. Lots of good stuff here, but beware the sometimes hard to understand fortran.


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