uva 10518 - How Many Calls?（矩阵快速幂）
题目链接：uva 10518 - How Many Calls?
公式f(n) = 2 * F(n) - 1, F(n)用矩阵快速幂求。
#include <stdio.h>#include <string.h>long long n;int b;struct state {int s[2][2];state(int a = 0, int b = 0, int c = 0, int 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 cas = 1;while (scanf("%lld%d", &n, &b), n || b) {state ans = solve(n);printf("Case %d: %lld %d %d\n", cas++, n, b,(2 * ans.s[0][0] - 1 + b) % b);}return 0;}