HDU 4549 M斐波那契数列

题目链接：HDU 4549 M斐波那契数列

题目大意：

题解：
先找规律：
\(F_0=a,\)
\(F_1=b,\)
\(F_2=a\times b,\)
\(F_3=a\times b^2,\)
\(F_4=a^2\times b^3,\)
\(F_5=a^3\times b^5,\)
\(...\)
\(F_n=a^{f_{n-1}}\times b^{f_n}\)（\(f_n\)为斐波那契数列）
所以题目就变成了用矩阵快速幂求斐波那契数列。
构造矩阵：

\[\left(\begin{matrix} f_i \\ f_{i-1} \end{matrix}\right) = \left(\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}\right) \times \left(\begin{matrix} f_{i-1} \\ f_{i-2} \end{matrix}\right) \]

所以：

\[\left(\begin{matrix} f_n \\ f_{n-1} \end{matrix}\right) = \left(\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix}\right)^{n-1} \times \left(\begin{matrix} 1 \\ 0 \end{matrix}\right) \]

因为结果需要对\(1e9+7\)取余，由费马小定理可知\(a^{p-1}\equiv 1(\mod p)\)，
由此可得：

\[\begin{aligned} ans&=(a^{f_{n-1}}\%mod\times b^{f_n}\%mod)\%mod \\ &=((a^{f_{n-1}\%(mod-1)}\times a^{(mod-1)^{[f_{n-1}\div (mod-1)]}})\%mod+(b^{f_n\%(mod-1)}\times b^{(mod-1)^{[f_n\div (mod-1)]}})\%mod)\%mod \\ &=(a^{f_{n-1}\%(mod-1)}\%mod+b^{f_n\%(mod-1)}\%mod)\%mod \end{aligned} \]

所以在矩阵快速幂中对\(1e9+6\)取余，在快速幂中对\(1e9+7\)取余。

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
#define ll long long
const int mod = 1e9 + 7;

struct Matrix {  // 矩阵
    int row, col;
    ll num[2][2];
};

Matrix multiply(Matrix a, Matrix b) {  // 矩阵乘法
    Matrix temp;
    temp.row = a.row, temp.col = b.col;
    memset(temp.num, 0, sizeof(temp.num));
    for (int i = 0; i < a.row; ++i)
        for (int j = 0; j < b.col; ++j)
            for (int k = 0; k < a.col; ++k)
                temp.num[i][j] = (temp.num[i][j] + a.num[i][k] * b.num[k][j]) % (mod - 1);
    return temp;
}

Matrix MatrixFastPow(Matrix base, ll k) {  // 矩阵快速幂
    Matrix ans;
    ans.row = ans.col = 2;
    ans.num[0][0] = ans.num[1][1] = 1;
    ans.num[0][1] = ans.num[1][0] = 0;
    while (k) {
        if (k & 1) ans = multiply(ans, base);
        base = multiply(base, base);
        k >>= 1;
    }
    return ans;
}

ll fastPow(ll n, ll k) {  // 快速幂
    ll ans = 1;
    while (k) {
        if (k & 1) ans = ans * n % mod;
        n = n * n % mod;
        k >>= 1;
    }
    return ans;
}

int main() {
    ll a, b, n;
    Matrix base;
    base.row = base.col = 2;
    base.num[0][0] = base.num[0][1] = base.num[1][0] = 1;
    base.num[1][1] = 0;
    while (~scanf("%lld%lld%lld", &a, &b, &n)) {
        if (!n) {
            printf("%lld\n", a);
        } else {
            Matrix ans = MatrixFastPow(base, n - 1);
            printf("%lld\n", fastPow(a, ans.num[1][0]) * fastPow(b, ans.num[0][0]) % mod);
        }
    }
    return 0;
}