矩阵快速幂 UVA 10870 Recurrences

题目传送门

题意：f(n) = a1f(n − 1) + a2f(n − 2) + a3f(n − 3) + . . . + adf(n − d), for n > d，求f (n) % m。训练指南的题目

分析：令：，.则

#include <bits/stdc++.h>

int d, n, m;
int a[16], f[16];

struct Mat {
    int m[17][17];
    int row, col;
    Mat() {
        //row = col = 16;
        memset (m, 0, sizeof (m));
    }
    void init(int sz) {
        row = col = sz;
        for (int i=1; i<row; ++i) {
            m[i][i+1] = 1;
        }
        int c = sz - 1;
        for (int i=2; i<=col; ++i) {
            m[sz][i] = a[c--];
        }
    }
    void change(int sz) {
        row = col = sz;
        for (int i=1; i<=sz; ++i) {
            m[i][i] = 1;
        }
    }
};

Mat operator * (const Mat &a, const Mat &b) {
    Mat ret;
    ret.row = a.row; ret.col = b.col;
    for (int i=1; i<=a.row; ++i) {
        for (int j=1; j<=b.col; ++j) {
            for (int k=1; k<=a.col; ++k) {
                int &r = ret.m[i][j];
                r = (r + 1ll * a.m[i][k] * b.m[k][j]) % m;
            }
        }
    }
    return ret;
}

Mat operator ^ (Mat x, int n) {
    Mat ret; ret.change (d+1);
    while (n) {
        if (n & 1) {
            ret = ret * x;
        }
        x = x * x; n >>= 1;
    }
    return ret;
}

//Running_Time
int main() {
    while (scanf ("%d%d%d", &d, &n, &m) == 3) {
        if (!d && !n && !m) {
            break;
        }
        for (int i=1; i<=d; ++i) {
            scanf ("%d", a+i);
        }
        for (int i=1; i<=d; ++i) {
            scanf ("%d", f+i);
        } 
        if (n <= d) {
            printf ("%d
", f[n] % m);
        } else {
            Mat ans, Fd;
            ans.init (d + 1);
            ans = ans ^ (n - d);
            
            Fd.row = d + 1; Fd.col = 1;
            for (int i=2; i<=d+1; ++i) {
                Fd.m[i][1] = f[i-1];
            }

            ans = ans * Fd;
            printf ("%d
", ans.m[d+1][1]);
        }
    }

    return 0;
}