Recurrences UVA 10870 （斐波拉契的一般形式推广）

题意：f(n) = a1f(n−1) + a2f(n−2) + a3f(n−3) + ... + adf(n−d), 计算这个f(n)

最重要的是推出矩阵。

#include<cstdio>
#include<cstring>
#define ll long long
ll mod, d, n;
ll a[16];
ll f[16];
struct jz
{
    ll num[16][16];
    jz(){ memset(num, 0, sizeof(num)); }
    jz operator*(const jz&p)const
    {
        jz ans;
        for (int k = 0; k < d; ++k)
        for (int i = 0; i < d;++i)
        for (int j = 0; j < d; ++j)
            ans.num[i][j] = (ans.num[i][j] + num[i][k] * p.num[k][j] % mod) % mod;
        return ans;
    }
}p;
jz POW(jz x, ll n)
{
    jz ans;
    for (int i = 0; i < d; ++i)ans.num[i][i] = 1;
    for (; n;n>>=1, x=x*x)
    if (n & 1)ans = ans*x;
    return ans;
}
void init()
{
    for (int i = 0; i < d; ++i)
        p.num[0][i] = a[i];
    for (int i = 1; i < d; ++i)
        p.num[i][i - 1] = 1;
}
int main()
{
    while (scanf("%lld%lld%lld", &d, &n, &mod) != EOF, d + n + mod)
    {
        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]); }
        else
        {
            init();
            jz ans = POW(p, n - d);
            ll kk = 0;
            for (int i = 0, j = d - 1; i < d; ++i, --j)
            {
                kk = (kk + ans.num[0][i] * f[j] % mod) % mod;
            }
            printf("%lld\n", kk);
        }
    }
    return 0;
}