This is a fast solution in C++ to Huge Fibonacci number modulo m question asked in the Algorithms course in UC San Diego Coursera course on Algorithms. It first calculates the Pisano period, then utilizes it to reduce the problem to a smaller size. The code also contains iterative solver to calculate Fibonacci number modulo m.
#include <iostream> #include <vector> using namespace std; // This iteratively calculates the mod m of Fn, i.e. Fn % m [cost: O(n)] template <typename T> T calc_fibModm(T n, T m) { T fn=0, fn_minus1=1,fn_minus2=1; if (n==0) return 0; else if (n==1 || n==2) return 1; else { for (T i=2; i<n; i++){ fn =( fn_minus1 % m + fn_minus2 % m) %m ; fn_minus2 = fn_minus1; fn_minus1 = fn; } } return fn; } template <typename T> T findPeriod(T n,T m){ // Find the periodicity of Fibanacci series (i.e. Pisano period) vector<T > vMod; for (T i=0; i<n; i++){ T modi = calc_fibModm(i,m); vMod.push_back(modi); if (i>2 && vMod[i] ==1 && vMod[i-1] == 0) {// this might be the start of a new periods as all new period start with 01 sequence // now calculate forward (i-2) new elements and check with the first set // whether all elements are same (to establish periodicity) bool bAllOk= true; for (T j=i-1; j<2*i-3+1; j++){ T modi_forward = calc_fibModm(j,m); if (vMod[j-(i-1)] != modi_forward) { bAllOk=false; break; } } if (bAllOk) { T period = i-1; cout<<"n="<<n<<", m="<<m<<" --> period ="<<period<<endl; return period; } } } // if reached here, that means the periodicty hasn't started yet, increase n string errStr="Not big enough n=("+to_string(n)+") to catch periodicity. Increase n to get periodicity"; throw runtime_error(errStr); return -1; } template <typename T> T get_fibonacci_huge(T n,T m) { // note that Fn % m = F_(n % period) % m if (m==1){ return 0; } else { // general case T period = findPeriod(n,m); return calc_fibModm((n%period), m); } } int main() { long long n, m; // try n=281621358815590 m=30524 output should be 11963 std::cin >> n >> m; std::cout <<"Fn % m ="<< get_fibonacci_huge(n, m) << '\n'; }