uva 10870(矩阵快速幂)

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;
}