UVA - 10870 Recurrences 矩阵快速幂

题目大意：有一个规则

f(n) = a1 ＊f(n - 1) + a2 ＊f(n - 2) + a3 ＊f(n - 3) + … + ad＊ f(n - d), for n > d.求f(n)

解题思路：矩阵快速幂水题，类似fibonacci

[code]#include<cstdio>
typedef long long ll;
const int N = 20;
struct Matrix{
    ll mat

;
}A, B, tmp;

ll n, d, m;
ll a
, f
;
void init() {
    for(int i = 0; i < d; i++)
        for(int j = 0; j < d; j++) {
            A.mat[i][j] = B.mat[i][j] = 0;
            if(i == j)
                B.mat[i][j] = 1;
        }
    for(int i = 0; i < d; i++) 
        A.mat[i][0] = a[i];
    for(int i = 1; i < d; i++)
        A.mat[i-1][i] = 1;
}

Matrix matMul(Matrix x, Matrix y) {
    for(int i = 0; i < d; i++)
        for(int j = 0; j < d; j++) {
            tmp.mat[i][j] = 0;
            for(int k = 0; k < d; k++)
                tmp.mat[i][j] = (tmp.mat[i][j] + x.mat[i][k] * y.mat[k][j]) % m;
        }
    return tmp;
}

void solve() {
    while(n) {
        if(n & 1)
            B = matMul(B,A);
        A = matMul(A,A);
        n >>= 1;
    }
}

int main() {
    while(scanf("%lld%lld%lld", &d, &n, &m) != EOF) { 
        if(n + m + d == 0)
            break;
        for(int i = 0; i < d; i++)
            scanf("%lld", &a[i]);
        for(int i = 0; i < d; i++)
            scanf("%lld", &f[i]);
        if(n <= d) {
            printf("%lld\n", f[n - 1] % m);
            continue;
        }
        init();
        n -= d;
        solve();
        ll ans = 0;
        for(int i = 0; i < d; i++) 
            ans = (ans + (B.mat[i][0] * f[d - 1 - i]) % m) % m;
        printf("%lld\n", ans % m);
    }
    return 0;
}