uva 10229 - Modular Fibonacci(矩阵快速幂）
题目链接：uva 10229 - Modular Fibonacci
题目大意：给出n和m，求出f(n) %m, f(x)为斐波那契数列。
解题思路：因为n的范围在0~214783647，所以计算量比较大，所以用矩阵快速幂。
{(1, 1), (1, 0)} ^ n *（f[1], f[0]) = (f[n], f[n - 1]).
#include <stdio.h>#include <math.h>long long n, b, k;struct state {long long s[2][2];state(long long a = 0, long long b = 0, long long c = 0, long long d = 0) {s[0][0] = a, s[0][1] = b, s[1][0] = c, s[1][1] = d;}}tmp(1, 0, 0, 1), c(1, 1, 1, 0);state count(const state& p, const state& q) {state f;for (int i = 0; i < 2; i++)for (int j = 0; j < 2; j++)f.s[i][j] = (p.s[i][0] * q.s[0][j] + p.s[i][1] * q.s[1][j]) % b;return f;}state solve(long long k) {if (k == 0) return tmp;else if (k == 1) return c;state a = solve(k / 2);a = count(a, a);if (k % 2) a = count(a, c);return a;}int main () {int m;while (scanf("%lld%d", &n, &m) == 2) {b = pow(2, m);if (n) {state ans = solve(n - 1);k = ans.s[0][0];} elsek = 0;printf("%lld\n", k);}return 0;}