uva 10518 - How Many Calls?(矩阵快速幂)

xiaoxiao2021-03-01  6

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; }

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