HDU 2256Problem of Precision(矩阵快速幂)

题意

求$(\sqrt{2} + \sqrt{3})^{2n} \pmod {1024}$

$n \leqslant 10^9$

Sol

看到题解的第一感受：这玩意儿也能矩阵快速幂？？？

是的，它能qwq。。。。

首先我们把$2$的幂乘进去，变成了

$(5 + 2\sqrt{6})^n$

设$f(n) = A_n + \sqrt{6} B_n$

那么$f(n+1) = (A_n + \sqrt{6} B_n ) * (5 + 2\sqrt{6})$

乘出来得到

$A_{n + 1} = 5 A_n + 12 B_n$

$B_{n + 1} = 2A_n + B B_n$

那么不难得到转移矩阵

$$\begin{pmatrix} 5 & 12 \\ 2 & 5 \end{pmatrix}$$

这样好像就能做了。。

但是实际上后我们最终会得到一个类似于$A_n + \sqrt{6}B_n$的东西，这玩意儿还是不能取模

考虑如何把$\sqrt{6}$的影响消掉。

$(5 + 2 \sqrt{6})^n = A_n + \sqrt{6}B_n$

$(5 - 2 \sqrt{6})^n = A_n - \sqrt{6}B_n$

相加得

$(5 + 2 \sqrt{6})^n + (5 - 2 \sqrt{6})^n = 2A_n$

考虑到$0 < (5 - 2\sqrt{6})^n < 1$

那么

$$\lfloor (5 + 2\sqrt{6})^n \rfloor = 2A_n - 1$$

做完啦qwq

#include<cstdio>
#include<cstring>
#include<iostream>
#define Pair pair<int, int>
#define MP(x, y)
#define fi first
#define se second
// #include<map>
using namespace std;
#define LL long long
const LL MAXN = , mod = ;
inline LL read() {
    char c = getchar(); LL x = , f = ;
    while(c < '' || c > '') {if(c == '-') f = -; c = getchar();}
    while(c >= '' && c <= '') x = x *  + c - '', c = getchar();
    return x * f;
}
int T, N;
struct Matrix {
    LL m[][], N;
    Matrix() {N = ; memset(m, , sizeof(m));}
    Matrix operator * (const Matrix &rhs) const {
        Matrix ans;
        for(int k = ; k <= N; k++)
            for(int i = ; i <= N; i++)
                for(int j = ; j <= N; j++)
                    (ans.m[i][j] += 1ll * m[i][k] * rhs.m[k][j] % mod) % mod;
        return ans;
    }
};

Matrix fp(Matrix a, int p) {
    Matrix base; base.m[][] = ; base.m[][] = ;
    while(p) {
        if(p & ) base = base * a;
        a = a * a; p >>= ;
    }
    return base;
}
int main() {
    T = read();
    while(T--) {
        N = read();
        Matrix a;
        a.m[][] = ; a.m[][] = ;
        a.m[][] = ; a.m[][] = ;
        a = fp(a, N - );
        LL ans = ( * a.m[][] +  * a.m[][]) % mod;
        printf("%I64d\n", ( * ans - ) % mod);
    }
    return ;
}

/**/