hdu 4565(矩阵快速幂)

题意：给出a,b,n,m按如下公式计算输出Sn

题解：和之前做过的一题很像，推导公式如下
(a+根号b)^n = xn + yn×根号b
–> (xn-1 + yn-1×根号b)×(a+根号b) = (a*xn-1 + b*yn-1) + (xn-1 + a×yn-1)×根号b

然后写成矩阵形式后

(a-根号b)^n = xn - yn×根号b

xn+yn×根号b = xn+yn×根号b + xn-yn×根号b - (xn-yn×根号b) = 2×xn + (a-根号b)

由于题目有给出范围(a-1)^n < b < a^n，所以 0 < a-根号b < 1，那么为了向上取整结果就是2*xn。

#include <stdio.h>#include <string.h>const int N = 3;int a, b, n, m;struct Mat {    int g[N][N];}ori, res;Mat multiply(Mat x, Mat y) {    Mat temp;    memset(temp.g, 0, sizeof(temp.g));     for (int i = 0; i < 2; i++)        for (int j = 0; j < 2; j++)            for (int k = 0; k < 2; k++)                temp.g[i][j] = (temp.g[i][j] + x.g[i][k] * y.g[k][j]) % m;    return temp;    }void calc(int n) {    while (n) {        if (n & 1)            res = multiply(ori, res);        n >>= 1;        ori = multiply(ori, ori);    }}int main() {    while (scanf("%d%d%d%d", &a, &b, &n, &m) == 4) {        memset(res.g, 0, sizeof(res.g));        ori.g[0][0] = ori.g[1][1] = a % m;        ori.g[0][1] = b % m;        ori.g[1][0] = 1;        res.g[0][0] = 1;        calc(n);        printf("%d\n", (2 * res.g[0][0]) % m);    }    return 0;}