UVA 10655 - Contemplation! Algebra(矩阵快速幂)

UVA 10655 - Contemplation! Algebra

题目链接

题意：给定p, q, n代表p=a+b,q=ab求an+bn

思路：矩阵快速幂，公式变换一下得到(an+bn)(a+b)=an+1+bn+1+ab(an−1+bn−1)，移项一下得到an+1+bn+1=(an+bn)p−q(an−1+bn−1)

这样就可以用矩阵快速幂求解了

代码：

#include <stdio.h>#include <string.h>long long p, q, n;struct mat {long long v[2][2];mat() {memset(v, 0, sizeof(v));}mat operator * (mat c) {mat ans;for (int i = 0; i < 2; i++) {for (int j = 0; j < 2; j++) {for (int k = 0; k < 2; k++) {ans.v[i][j] = ans.v[i][j] + v[i][k] * c.v[k][j];    }   }  }  return ans; }};mat pow_mod(mat x, long long k) {mat ans;ans.v[0][0] = ans.v[1][1] = 1;while (k) {if (k&1) ans = ans * x;x = x * x;k >>= 1; } return ans;}long long solve() {if (n == 0) return 2;if (n == 1) return p;if (n == 2) return p * p - 2 * q;mat a;a.v[0][0] = p; a.v[0][1] = -q; a.v[1][0] = 1;a = pow_mod(a, n - 2);return a.v[0][1] * p + a.v[0][0] * (p * p - 2 * q);}int main() {while (scanf("%lld%lld%lld", &p, &q, &n) == 3) {printf("%lld\n", solve());}return 0;}