标签:style http color os io ar for 代码 amp
题意:给出n和m,求出f(n) % m, f(x)为斐波那契数列。
思路:因为n挺大的,如果直接利用公式计算很有可能会TLE,所以利用矩阵快速幂求解,|(1, 1), (1, 0)| * |f(n - 1), f(n - 2)| = |f(n), f(n - 1)|,所以求f(n)相当于|f(1), f(0)|乘上n - 1次的|(1, 1), (1, 0)|。
代码:
#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(斐波那契数列+矩阵快速幂)
标签:style http color os io ar for 代码 amp
原文地址:http://blog.csdn.net/u011345461/article/details/38900043