P1349 广义斐波那契数列(矩阵乘法)

题目

p1349 广义斐波那契数列

解析

把普通的矩阵乘法求斐波那契数列改一改，随便一推就出来了
\[\begin{bmatrix}f_2\\f_1 \end{bmatrix}\begin{bmatrix} p&q\\ 1&0\\ \end{bmatrix}^{n-2}=\begin{bmatrix}f_n\\f_{n-1} \end{bmatrix}\]
水题

代码

#include <bits/stdc++.h>
#define int long long

using namespace std;

const int n = 100;

int n, m, a1, a2, p, q;

struct matrix {
    int a[n][n];
    matrix() {
        memset(a, 0, sizeof a);
    }

    void initmatrix() {
        a[1][1] = p, a[1][2] = q;
        a[2][1] = 1;
    }

    matrix operator * (const matrix &oth) const {
        matrix ans;
        for (int k = 1; k <= 2; ++k)
            for (int i = 1; i <= 2; ++i)
                for (int j = 1; j <= 2; ++j)
                    ans.a[i][j] = (ans.a[i][j] + (a[i][k] * oth.a[k][j]) % m) % m;
        return ans;
    }

} init;

matrix qpow(matrix a, int b) {
    matrix ans = init;
    while (b) {
        if (b & 1) ans = ans * a;
        b >>= 1, a = a * a;
    }
    return ans;
}

template<class t>inline void read(t &x) {
    x = 0; int f = 0; char ch = getchar();
    for ( ; !isdigit(ch); ch = getchar()) f |= (ch == '-');
    for ( ; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
    x = f ? -x : x;
    return;
}

signed main() {
    read(p), read(q), read(a1), read(a2), read(n), read(m);
    if (n <= 2) {
        printf("%lld", n == 1 ? a1 : a2);
        return 0;
    }
    init.initmatrix();
    init = qpow(init, n - 3);
    printf("%lld\n", (a2 * init.a[1][1] + a1 * init.a[1][2]) % m);
    return 0;
}