Uva10299-modular Fibonacci (Fibonacci series + matrix fast power)

Question Link

Give N and M, and find F (n) % m. f (x) is the Fibonacci sequence.

Idea: Because N is big, it is very likely that it will use the formula to calculate it, so we can use the Matrix to quickly solve the power. | (1, 1), (1, 0) | * | f (n-1), F (n-2) | = | f (N), F (n-1) |, so f (N) equivalent to | F (1), F (0) | multiplied by N-1 | (1, 1), (1, 0) |.

Code:

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>typedef long long ll;using namespace std;ll n, p, k;int m;struct mat{    ll s[2][2];    mat(ll a = 0, ll b = 0, ll c = 0, ll d = 0) {        s[0][0] = a;         s[0][1] = b;         s[1][0] = c;         s[1][1] = d;     }}c(1, 1, 1, 0), tmp(1, 0, 0, 1);mat operator * (const mat& a, const mat& b) {    mat Mat;    for (int i = 0; i < 2; i++)        for (int j = 0; j < 2; j++)            Mat.s[i][j] = (a.s[i][0] * b.s[0][j] + a.s[i][1] * b.s[1][j]) % p;    return Mat; }mat pow_mod(ll k) {    if (k == 0)        return tmp;    else if (k == 1)        return c;    mat a = pow_mod(k / 2);    a = a * a;    if (k % 2)        a = a * c;    return a;}int main() {    while (scanf("%lld%d", &n, &m) != EOF) {        p = pow(2, m);         if (n) {            mat ans = pow_mod(n - 1);             k = ans.s[0][0];        }          else            k = 0;        printf("%lld\n", k);    }     return 0;}

Uva10299-modular Fibonacci (Fibonacci series + matrix fast power)