uva 10870(矩阵快速幂)

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题意：计算f(n)
f(n) = a1 f(n - 1) + a2 f(n - 2) + a3 f(n - 3) + … + ad f(n - d), for n > d.
题解：斐波那契的变形，把2个扩大成d个，然后加了a1…ad的参数，构造矩阵直接矩阵快速幂计算。

#include <stdio.h>#include <string.h>const int N = 20;struct Mat {    long long g[N][N];}res, ori;long long d, n, m;Mat multiply(Mat x, Mat y) {    Mat temp;    for (int i = 0; i < d; i++)        for (int j = 0; j < d; j++) {            temp.g[i][j] = 0;            for (int k = 0; k < d; k++)                temp.g[i][j] = (temp.g[i][j] + x.g[i][k] * y.g[k][j]) % m;        }    return temp;}void calc(long long n) {    while (n) {        if (n & 1)            ori = multiply(ori, res);        n >>= 1;        res = multiply(res, res);    }}int main() {    while (scanf("%lld%lld%lld", &d, &n, &m) && d + n + m) {        memset(res.g, 0, sizeof(res.g));        memset(ori.g, 0, sizeof(ori.g));        for (int i = 0; i < d; i++) {            scanf("%lld", &res.g[i][0]);            res.g[i][0] %= m;            if (i > 0)                res.g[i - 1][i] = 1;        }        for (int i = 0; i < d; i++) {            scanf("%lld", &ori.g[0][d - 1 - i]);            ori.g[0][d - 1 - i] %= m;        }        calc(n - d);        printf("%lld\n", ori.g[0][0]);    }    return 0;}

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