UVA10870 Recurrences —— 矩阵快速幂

题目链接：https://vjudge.net/problem/UVA-10870

题意：

典型的矩阵快速幂的运用。比一般的斐波那契数推导式多了几项而已。

代码如下：

#include <bits/stdc++.h>
#define rep(i,s,t) for(int (i)=(s); (i)<=(t); (i)++)
#define ms(a,b) memset((a),(b),sizeof((a)))
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const double eps = 1e-6;
const int mod = 10000007;
const int maxn = 2e5+10;

struct MA
{
    LL mat[20][20];
    void init()
    {
        rep(i,1,19) rep(j,1,19)
            mat[i][j] = (i==j);
    }
};

LL n,d,m;
LL a[20],f[20];

MA mul(MA x, MA y)
{
    MA tmp;
    ms(tmp.mat,0);
    rep(i,1,d) rep(j,1,d) rep(k,1,d)
        tmp.mat[i][j] += (1LL*x.mat[i][k]*y.mat[k][j])%m, tmp.mat[i][j] %= m;
    return tmp;
}

MA qpow(MA x, LL y)
{
    MA s;
    s.init();
    while(y)
    {
        if(y&1) s = mul(s,x);
        x = mul(x,x);
        y >>= 1;
    }
    return s;
}

int main()
{
    while(scanf("%lld%lld%lld",&d,&n,&m) && (d || n||m))
    {
        rep(i,1,d) scanf("%lld",&a[i]);
        rep(i,1,d) scanf("%lld",&f[i]);

        if(n<=d)
        {
            printf("%lld
", f[n]);
            continue;
        }

        MA x;
        ms(x.mat,0);
        rep(i,1,d) x.mat[1][i] = a[i];
        rep(i,2,d) x.mat[i][i-1] = 1;
        x = qpow(x,n-d);

        LL ans = 0;
        rep(i,1,d)
            ans += (1LL*x.mat[1][i]*f[d-i+1])%m, ans %= m;
        printf("%lld
",ans);
    }
}