BZOJ4002 [JLOI2015]有意义的字符串

据说这两场加起来只要170= =而这是最简单的题目了QAQ

看到$(\frac {b + \sqrt {d} } {2} )^n$，第一反应是共轭根式$(\frac {b - \sqrt {d} } {2} )^n$

首先有$(\frac {b + \sqrt {d} } {2} )^n + (\frac {b - \sqrt {d} } {2} )^n$为整数

由高中课本知识可知，上式其实是一个三项递推数列的通项公式，而数列的递推式非常简单

$$f[x] = b * f[x - 1] - \frac{b^2 - d} {4} * f[x - 2]$$

其中$f[0] = 2, f[1] = b$，这样子直接矩阵乘法就好啦~

再看$A = (\frac {b - \sqrt {d} } {2} )^n \in [-1, 1]$

然后。。。然后发现bzoj上题面错了QAQ，实际上是"对于20%的数据$b=1,d=5$"

故$A > 0$当且仅当$d \not= b^2$且$n$为偶数，此时要将$f[n]$减掉$1$，否则不用减

 /**************************************************************
     Problem: 4002
     User: rausen
     Language: C++
     Result: Accepted
     Time:48 ms
     Memory:1272 kb
 ****************************************************************/

 #include <cstdio>
 #include <iostream>
 #include <cstring>
 #include <algorithm>

 using namespace std;
 typedef unsigned long long ll;
 const ll mod = 7528443412579576937ull;

 template <class T> T sqr(T x) {
     return x * x;
 }

 template <class T> T mul(T x, T y) {
     static T res;
     if (x < y) swap(x, y);
     res = ;
     while (y) {
         if (y & ) res = (res + x) % mod;
         x = (x << ) % mod, y >>= ;
     }
     return res;
 }

 ll b, d, n;

 struct mat {
     ll x[][];

     inline void clear() {
         memset(x, , sizeof(x));
     }
     inline void one() {
         memset(x, , sizeof(x));
         x[][] = x[][] = ;
     }
     inline void pre() {
         this -> clear();
         x[][] = , x[][] = (d - sqr(b)) >> , x[][] = , x[][] = b;
     }

     inline ll* operator [] (int i) {
         return x[i];
     }

     inline mat operator * (const mat &p) const {
         static mat res;
         static int i, j, k;
         for (res.clear(), i = ; i <= ; ++i)
             for (j = ; j <= ; ++j)
                 for (k = ; k <= ; ++k)
                     res[i][j] = (res[i][j] + mul(x[i][k], p.x[k][j])) % mod;
         return res;
     }
     inline mat& operator *= (const mat &p) {
         return *this = *this * p;
     }
 } a;

 inline mat pow(const mat &p, ll y) {
     static mat x, res;
     res.one(), x = p;
     while (y) {
         if (y & ) res *= x;
         x *= x, y >>= ;
     }
     return res;
 }

 int main() {
     cin >> b >> d >> n;
     a.pre();
     a = pow(a, n);
     cout << (((a[][] << ) % mod + mul(b, a[][]) - (d != sqr(b) && !(n & ))) % mod + mod) % mod << endl;
     return ;
 }